Calculus Visualized - by Dennis F Davis

0:03can you learn calculus in the time it

0:04takes to watch a long movie yes you can

0:08my name is Dennis Davis I'm an engineer

0:11not a mathematician I try to make my

0:14videos visually enlightening and

0:16fast-paced but this video is very long

0:19because it covers almost the entire

0:21first year of calculus using visuals

0:24graphs and

0:27diagrams I'd rather show you calculus

0:29visual then just tell you the rules and

0:31formulas if you see where the rules and

0:34formulas come from that can help you

0:36understand remember and make use of them

0:39to become truly proficient at calculus

0:41you'll need practice which you won't get

0:44just by watching this video in the

0:46description I'll link to some helpful

0:48practice oriented videos by

0:51others whether you're new to calculus

0:53studying it now or just want a refresher

0:56I hope you'll find this video

0:58informative and engaging my only

1:00assumptions are that you're already

1:02familiar with functions and algebra so

1:05here we

1:06go calculus is the study of change and

1:09rates of change of mathematical

1:11functions when we use calculus we

1:14perform operations on functions that

1:17result in different functions this isn't

1:20a new idea at all we perform operations

1:23such as addition on numbers to get new

1:27numbers we operate on sets to get new

1:30sets on matrices to get new

1:34matrices and we can operate on functions

1:37to get new functions in fact you've

1:40probably already done this in algebra

1:42when you had a function and found its

1:44inverse we start with a function f ofx

1:48and write out F in terms of X and Y so

1:51that y equals some function of X then we

1:54rewrite the equation switching X and Y

1:58then solve the second equation for y and

2:01that's the inverse of the first

2:04function so we start with a

2:07function perform an operation on it and

2:10get a new function the point is taking

2:13an existing function and Performing some

2:16operation on it to get a new function

2:18isn't a new or advanced math concept and

2:21that's really what calculus is in fact

2:24calculus is all about performing two

2:27operations on functions that's it that's

2:30calculus the first operation is called

2:34differentiation when we differentiate a

2:36function the new resulting function is

2:38called the derivative of the first

2:40function the derivative is the first

2:43topic we'll cover learning to take the

2:45derivative of a wide variety of function

2:47types is roughly the first semester of

2:51calculus the second calculus operation

2:54is called integration when we integrate

2:56a function the new resulting function is

2:59called the integral of the first

3:01function integration is roughly the

3:04second semester of

3:06calculus and there's a wonderful

3:07relationship between these two

3:09operations that we'll get to at just the

3:11right

3:12time when functions are simple these

3:15operations are simple and calculus is

3:17simple it's when the functions we

3:20operate on get complex that calculus

3:22seems complex so if calculus has a

3:25reputation for being a difficult subject

3:28it's not really calculus

3:30fault we'll start with the derivative

3:33and consider a simple function f ofx = x

3:37-1 this is a linear function so its

3:40graph is a straight line I said calculus

3:43was the study of change and rates of

3:45change and the rate of change of a

3:47straight line is its slope rise over run

3:51deltay over Delta

3:53X I like to color code things in my

3:56videos I'm using blue for x coordinates

3:58and distances and yellow for y and I'll

4:02use pink for slope or rates of change I

4:05won't draw a lot of attention to the

4:06colors and may not mention them again

4:09but their consistent use may let you see

4:11some apparent order or pattern that

4:13might not otherwise be

4:15clear for a straight line it's easy to

4:18find the slope just pick any two points

4:20and subtract their y and x coordinates

4:23to get Delta y over Delta X Delta Y is 3

4:27- 0 this distance the difference between

4:30the yellow y-coordinates and Delta X is

4:344 min-1 this distance the difference

4:37between the blue x coordinates so for

4:40this linear function the slope is 3 over

4:433 or POS 1 for every unit X changes y

4:48changes by one times that amount since

4:50the slope is

4:51one if the slope were positive 1/2 then

4:55when X changes by some amount y changes

4:58by positive 1/2 as much if the slope

5:01were -2 then when X changes y changes by

5:05-2 times as much that is two times as

5:08much but in the opposite direction y

5:11would get smaller as X gets

5:14larger when the graph line is horizontal

5:17the slope is zero because y never

5:19changes so Delta Y is zero when the

5:23graph line is vertical the slope is

5:25undefined because X never changes Delta

5:28X is zero so Delta y over Delta X is

5:31undefined since the denominator is

5:35zero everything's nice and simple when

5:37the function is linear but what about

5:39the function FX = x^2 - 2x + 1 the graph

5:45of this function is a parabola so now

5:47the question what's the slope prompts a

5:50new question in response where do you

5:53mean because the slope is smoothly

5:56changing it's different at different

5:58points

6:04suppose we want to define the slope of

6:06the function at xals 1.6 here now if you

6:10know calculus or more specifically the

6:12rules of differentiation that we'll

6:14cover soon then within a few seconds you

6:17can figure out that the slope at x = 1.6

6:20is

6:211.2 if you don't know calculus yet I

6:24promise that soon you'll know how to do

6:26this but without calculus it's not easy

6:29f after all slope is rise over run

6:31deltay over Delta X but with just one

6:34point where does Delta Y come from or

6:37Delta X slope is the rate of change and

6:40there's no change at a single point

6:42there's only a change between two points

6:45so it's a tricky question we could

6:48eyeball it by drawing our best attempt

6:50at a line tangent to the curve at x =

6:531.6 then measure the slope it won't be

6:56very accurate it's not always easy to

6:58draw an accurate tangent line but

7:01without calculus that might be an option

7:03but let me show another way in the 1600s

7:06the great minds that developed calculus

7:09approach the problem this way I'll zoom

7:12in on our grid at the red point of

7:14Interest where x equals

7:161.6 we want to know the slope at the red

7:19point I warnant you I like to color code

7:21things let's choose a nearby Point green

7:25and find the slope deltay over Delta X

7:28between red red and green so we estimate

7:32the slope at red as the slope of the

7:35pink line between red and green Delta y

7:38over Delta X it'll be pretty close and

7:41the closer green is to Red the better

7:43the

7:45estimate this is the approach that will

7:47lead us to the derivative the x value of

7:49our red point where we want to know the

7:51slope is

7:531.6 we find the Y value by plugging 1.6

7:57into the function and we get 0.

8:0036 let's choose a nearby x value for the

8:03Green Point let's say

8:061.61 so we find the yvalue for the

8:09nearby Green Point by plugging

8:121.61 into our function to get 361

8:161201 remember here's our function y =

8:20x^2 - 2x + 1 we can easily find the

8:24slope deltay over Delta X between two

8:27points given their coordinates

8:34and we get

8:351.21 the slope between red and green

8:39which is our estimate for the slope at

8:41Red calculus will tell us I promise that

8:45the slope at Red is 1.2 so our estimate

8:48is very close the closer we let green

8:51get to Red the closer the estimate will

8:53get to

8:551.2 so let's prove it we'll still choose

8:58a nearby Green Point point but now

9:00instead of choosing an actual small

9:02distance away from Red such as

9:040.1 or

9:07.001 we'll use a variable for the tiny X

9:10distance to green and call it

9:12h the coordinates of the red Point are

9:15still 1.6 comma

9:180.36 but let's consider the coordinates

9:21of the Green Point its x coordinate is

9:231.6 + H and its y-coordinate is the

9:27function's value when we plug in 1.6 + H

9:32for

9:33X now let's find the slope as Delta y

9:36over Delta X Delta Y is the y-coordinate

9:40of the Green Point F of 1.6 + H minus

9:44the y-coordinate of the red point which

9:46is still

9:480.36 I'm shading the components to the

9:50color of the corresponding point and

9:53Delta X is the x coordinate at Green 1.6

9:56+ H minus the x coordinate at red

10:001.6 the denominator Delta X looks pretty

10:04easy to simplify 1.6 + H - 1.6 is simply

10:09H this makes sense we chose H over here

10:13to be Delta X so to find Delta y over

10:16Delta X we'll need to expand this

10:18expression by taking our function f ofx

10:21and rewriting it substituting 1.6 + H in

10:25for

10:26X this is just algebra so I'm showing it

10:30quickly pause if you want to step

10:32through the

10:35details we end up with this expression

10:38for the yalue of the nearby Green Point

10:41so we can now find deltay over Delta x

10:4536 + 1.2 H + h^ 2

10:50-36 all over

10:53H the +36 and minus. 36 cancel leaving

10:58us with 1.2 H + h^2 over H as long as H

11:04is not zero which it's not it's some

11:07tiny tiny change but not zero then we

11:10can cancel an H from each term and the

11:13slope Delta y over Delta X is 1.2 + H

11:18let's remember that we said H was an

11:20arbitrarily small number we're going to

11:22find the limit of Delta y over Delta X

11:25as H approaches zero you may be familiar

11:28with the concept of the limit from

11:30studying discontinuous functions in

11:33algebra we write the limit like this l i

11:36m and underneath a variable right arrow

11:40and a literal value we read this as the

11:43limit as H approaches zero in our case

11:46the limit of 1.2 + H as H approaches

11:50zero as H gets closer and closer to zero

11:54the expression 1.2 + H gets closer and

11:57closer to 1.2

11:59so the limit is 1.2 and that's deltay

12:03over Delta X the slope at the red point

12:06or at least the slope as Delta X we

12:09called it h approaches

12:13zero and now we can continue our

12:16dialogue what's the slope where do you

12:19mean at x =

12:211.6 the slope there is 1.2 that we found

12:25by taking the slope between two very

12:27close points at the limit as the

12:29difference between their x coordinates

12:31approach

12:33zero I want to review this equation

12:36again which is how we estimate the slope

12:38of a function at a point using algebra

12:41you'll see this important expression in

12:43the first chapter of every single

12:45calculus textbook we're going to use it

12:47several more times and I want to make

12:49sure you're comfortable with it it's the

12:52algebra behind estimating the slope at

12:54the red Point as the slope between the

12:56red point and the nearby Green Point the

12:59Delta y over Delta X the difference

13:02between the y-coordinates is Delta y the

13:04numerator of our slope estimate the

13:07difference between the x coordinates is

13:09Delta X the denominator of our slope

13:12estimate we just need to come up with

13:14expressions for these differences the x

13:17coordinate of the red point is simply X

13:20so the red y-coordinate is the

13:22function's value at x f

13:25ofx the x coordinate of the nearby Green

13:28Point is X x + H H is the small Delta x

13:32amount we added to X to get a nearby

13:34point so the green y coordinate is the

13:37function's value at x + h f of X+

13:42H so for the numerator we can plug in

13:44the difference F ofx + hus F ofx I can

13:49include the green and red memory aid y

13:51coordinate of green minus y coordinate

13:54of red for the denominator we get x + H

13:57- x green x coordinate minus red x

14:02coordinate well it doesn't get written

14:04out this way very often since x + H - x

14:07so obviously simplifies to H and H is

14:11the small distance we deliberately chose

14:13for Delta X in the first place so the

14:15denominator is usually just H same thing

14:19so when you see this expression please

14:22think of this triangle and the rise

14:24overrun slope it represents Delta y over

14:27Delta X two quick points first I'm

14:31illustrating a positive slope but if the

14:33slope were negative the expression will

14:35correctly result in a negative estimate

14:37for the slope since f ofx is greater

14:40than F ofx + H subtracting a larger

14:43number from a smaller will result in a

14:45negative number H is always positive

14:48since it's the small Delta X we added to

14:51X so the equation works for positive and

14:54negative

14:55slopes the second point is the key idea

14:58that uses this expression as the

14:59entryway into calculus let me make a

15:02copy of the diagram and expression the

15:05top and bottom start out identical but

15:07I'm going to make changes to the bottom

15:09to transform it step by step into the

15:12key Foundation of

15:14calculus we used the limit a moment ago

15:16to find that the slope of our function

15:18was

15:191.2 when we take the limit as Delta X

15:22approaches 0 something very special

15:24happens first the nearby Green Point

15:27approaches the red Point that's pretty

15:29obvious since H is the distance between

15:31their x coordinates and it's approaching

15:34zero second the expression is no longer

15:37an estimate of the slope it is the slope

15:40or what we call the derivative so let me

15:42change the header the expression at the

15:45limit isn't the algebraic estimate of

15:47the slope it's the algebraic definition

15:50of the derivative and there's one more

15:52important change at the limit Delta Y

15:55and Delta X get new names and symbols

15:57they're called Dy Y and DX the d stands

16:01for differential we'll talk more about

16:03it soon on the top is the estimated

16:06slope of a curve based on nearby points

16:09on the bottom is the definition of the

16:11derivative at the limit is the

16:13horizontal distance between the two

16:15points approaches

16:17zero so here's how we use the limit to

16:20find the derivative at

16:221.6 there was a lot of algebra involved

16:24to get to our answer 1.2 but no calculus

16:28yet

16:29you might not believe me but you will

16:31soon calculus is easier than taking

16:34limits taking limits with algebra is

16:37tedious this is tedious certainly not

16:41difficult but timec consuming and prone

16:43to errors if you're not careful and when

16:46I promised to show you how you could

16:47know almost immediately that the slope

16:49at 1.6 was 1.2 I was not talking about

16:53doing all this algebra in your head

16:56calculus is easier than this and I'll

16:58show you there's just one more key step

17:01the destination will make the journey

17:03worth it so we just found the slope of

17:06the function at x = 1.6 by plugging 1.6

17:10into the function and finding the slope

17:12to a nearby Green Point now let's see

17:15what happens when we generalize we'll

17:18algebraically evaluate Delta y over

17:20Delta X at the variable value X instead

17:24of the specific value 1.6

17:29the Delta y over Delta X formula has the

17:32same Parts the y-coordinate of the

17:34nearby Green Point is f ofx + H the

17:37y-coordinate of the red point is f

17:40ofx the x coordinate of the Green Point

17:43is x + H and the x coordinate of the red

17:46point is X the denominator of course

17:49simplifies to

17:51H so when we expand f of x + H we get x

17:55+ h^ 2 - 2 * x + h H + 1 and you can see

18:01the correspondence to the terms in the

18:04function we just plug in x + H for X

18:08expand and simplify using algebra so

18:11here's the x coordinate of the Green

18:13Point in terms of X it's the first term

18:16in the numerator of our deltay over

18:19Delta X slope

18:21expression next we need to subtract F

18:23ofx which is right here minus f ofx

18:32three terms

18:35cancel and we're left with these terms

18:38as Delta y h^2 + 2 xh - 2

18:43H the denominator is H the arbitrarily

18:47small number we chose for Delta

18:49X we cancel an H from each term and get

18:53Delta y/ Delta x = h + 2x - 2 let's let

18:58take the limit as H approaches Z we get

19:012x -

19:032 now let's pause for a moment and

19:06appreciate what we've just done we have

19:09an expression for the slope of the

19:11function y = x^2 - 2x + 1 in terms of X

19:18have we just found a function that

19:19Returns the slope of another

19:22function let's call the second function

19:25fime of X and try it out more on this

19:28prime notation

19:30shortly a moment ago we found that the

19:32slope at x = 1.6 is 1.2 so let's try out

19:37our new general function frime of X at

19:411.6 we get 2 * 1.6 - 2 3.2 - 2 yes we

19:49get 1.2 just like before let's try this

19:53point at the vertex of the parabola

19:55where xal 1 it looks like the slope we

19:58get should be zero frime of 1 = 2 * 1 -

20:032 yes it's zero and we have indeed found

20:07the function frime of X that Returns the

20:10slope of f ofx for any value of x and

20:14now we can finish our dialogue with the

20:16broader question what's the slope of

20:19this function at any X and the answer

20:22for this function is 2x - 2 because this

20:25will return the slope of x^2 - 2x + 1

20:29for any

20:30X so we've just performed the first

20:33calculus operation

20:35differentiation we started with a

20:37function f ofx = x^2 - 2x + 1 and we

20:42performed an operation on it to yield a

20:44new function the first function's

20:46derivative fime of x = 2x - 2 the

20:50derivative of a function evaluates to

20:53the slope of that function at every x

20:56value we found the derivative the hard

20:59way by taking limits and crunching

21:01through algebra I'll show a simpler way

21:03in a moment but regardless of the method

21:06we've just performed the differentiation

21:08operation to get the derivative of a

21:12function since the derivative of a

21:14function is just another function we can

21:17graph it also the pink line is fime of X

21:21= 2x -

21:222 frime of X is a common notation for

21:26the derivative of f ofx I'll cover some

21:29notation conventions in a moment the

21:32pink line represents the derivative of

21:34the white Parabola so of course the

21:36derivatives value at x = 1.6 is 1.2 the

21:42slope of the parabola at x =

21:461.6 let's go over some vocabulary and

21:49notation concerning derivatives with

21:51slopes we quite naturally speak of Delta

21:54y over Delta X the slope between two

21:57points this ratio gets closer and closer

22:00to the slope at X as Delta X gets closer

22:03and closer to zero in calculus we

22:06introduced new terms for Delta X and

22:08Delta y that have the limit as Delta X

22:11approaches zero concept built in we say

22:15that the limit as Delta X approaches

22:17zero of Delta X is DX we use lowercase D

22:21instead of the Greek letter Delta as

22:23calculus shorthand that means at the

22:26limit so when you see D it means means

22:28at the limit as the change to the

22:30independent variable often X approaches

22:34zero one more time instead of writing

22:36all this you can just write this which

22:39means the same thing it's common to

22:42think of DX as the slightest tiniest

22:44change in X and this thought might serve

22:46you well but DX is really whatever Delta

22:49X becomes as it gets closer and closer

22:51to zero it's an idea that we treat like

22:55a number DX is called the different

22:58differential of X I'll have to draw it

23:00with some sort of thickness so we can

23:02see it but it's actually unimaginably

23:06narrow Dy is the differential of Y which

23:09is Delta Y at the limit but it's not the

23:12limit as Delta y approaches zero like DX

23:15it's also the limit as Delta X

23:17approaches zero Dy is the interesting

23:20thing we observe as Delta X gets closer

23:23and closer to

23:25zero so at the limit as Delta X

23:27approaches zer our ratio Delta y over

23:30Delta X becomes dy/ DX this is the

23:34derivative the ratio Dy / DX is the

23:38differential of Y with respect to X it's

23:42often set out loud as Dy by DX or simply

23:47dydx let me show you some other ways you

23:49might see the derivative

23:52represented this notation denotes a

23:54differential change in something with

23:56respect to X we can pull the something

23:59out of the ratio it means the same thing

24:02this part means the change or derivative

24:04with respect to X and this part is what

24:07we're taking the derivative

24:09of since yal F ofx we could also write d

24:13by DX of f ofx it means the same thing

24:17and since F ofx = x^2 - 2x + 1 we could

24:21also write d by DX of x^2 - 2x + 1 they

24:26all mean the same thing hopefully the

24:29symbology is

24:30clear differential of something with

24:34respect to the corresponding

24:36differential of

24:38X I mentioned this one earlier but

24:40derivatives of functions can also be

24:43represented with the prime or single

24:45quote character fime of X is the

24:48derivative of f ofx so these are all

24:51different representations of the same

24:53calculus concept the derivative of f ofx

24:56with respect to X

24:59we've been using X and Y as our

25:01independent and dependent variables

25:03which is quite natural considering these

25:04are the standard cartisian coordinate

25:06system variables and we often represent

25:09the dependent variable y as a function

25:11of the independent variable X like this

25:14but you should know that other variables

25:16can be used for example in physics and

25:18Engineering the independent variable is

25:21often time if the function we're

25:23operating on represents something that

25:25can change over time for example s equal

25:28F of T could represent the displacement

25:30of a particle from a starting point at

25:32time

25:34T in this case the derivative of the

25:37function is DS by DT not Dy by DX I just

25:42don't want you to get locked into X and

25:43Y and not recognize calculus Concepts

25:46when you see them referencing different

25:49variables as it happens the independent

25:51variable t for time is so common that

25:54there's an additional shorthand

25:55representation for the derivative of a

25:58function function with respect to time F

26:00dot a function label with a DOT over it

26:04represents the derivative of that

26:05function with respect to

26:08time now let's go over the basic rules

26:11of differentiation which are rules and

26:14techniques to find the derivative of

26:16various types of functions and the big

26:18payoff is that the rules will let us

26:20find the derivatives without the timec

26:22consuming process of taking

26:25limits covering the rules of

26:27differentiation will take some time as

26:29this is essentially the entire first

26:31semester of calculus and I'm going to

26:34take limits to show you that the rules

26:35are true and correct and give you

26:37insight into why they

26:40work on the left are five types of

26:43functions each rule is a special

26:45shortcut for taking the derivative of

26:47that function type the shortcuts are a

26:50result of observing the pattern that

26:52reveals itself when we take the limit

26:55you'll see what I mean on the right side

26:57are the the rules or shortcuts for how

26:59to find the derivatives of combinations

27:02of function Types on the left so that

27:05you don't give up hope let me tell you

27:07exactly what we'll be doing I'll start

27:09with these four rules on the top then

27:12I'll show you some super shortcuts

27:13involving these four rules at this point

27:17I'll keep my promise because you'll be

27:19able to differentiate x^2 - 2x + 1

27:22almost instantly in your head and you'll

27:25know that the derivative at x = 1.6 is

27:28is

27:291.2 by this time we'll know enough

27:31calculus to set up and solve some

27:33interesting problems then I'll introduce

27:35the second derivative and higher order

27:37derivatives and then we'll finish the

27:39rules of differentiation with these last

27:42five rules I'll show this agenda again

27:44so we can keep track of our progress

27:46covering these topics we'll finish the

27:48first calculus operation differentiation

27:52after that we'll learn the second

27:53calculus operation integration so let's

27:56Dive In we'll start with the derivative

27:59of a constant if we have a function f

28:02ofx equals a constant such as 3 its

28:05graph would be a horizontal line I'm

28:07going to make constants green no matter

28:10what value we choose for X on the

28:12horizontal axis the function returns

28:14three and the slope is zero this is true

28:17for any constant C since there's no

28:20change to the function's value as X

28:22changes the derivative is

28:25zero we write the generalization like

28:27this

28:28d by DX of C equals 0 where C is any

28:32constant it says that the derivative of

28:35a constant is zero and that's the

28:37constant rule we're not cheating or

28:39saying anything new if we were to again

28:42let two points get closer and closer to

28:44each other and apply the limit to the

28:46slope expression we would get zero we'd

28:49get zero every time because we'd always

28:51get C minus C

28:54here all of these rules of

28:56differentiation are General ations of

28:58what happens to the slope as we take the

29:00limit we simply notice the pattern like

29:03we did here for constants and then use

29:06the pattern instead of taking limits and

29:09that's why calculus at least

29:11differentiation is easier than

29:14algebra the next pattern or rule of

29:16differentiation is called the power rule

29:19heads up the power rule is a big part of

29:22how you can take the derivative of x^2 -

29:242x + 1 in your head I'll the rule with

29:28some examples then I'll show how it can

29:30be visualized I'll tell you the power

29:33rule first then I'll prove it's true in

29:35a few minutes the power rule is applied

29:38the powers of X like X2 X cubed x 4th

29:43and so on it says that the derivative of

29:46x the N is n * X

29:49nus1 it looks horrible but every

29:52calculus student in history just

29:54remembers this about the power rule

29:56bring the exponent downstairs

29:58and subtract one so to find the

30:01derivative of x cubed we take the

30:03exponent 3 and bring it downstairs in

30:06front of X then we subtract one from the

30:09exponent and get 3 x^2 can you see how

30:12that matches the power rule

30:15shortcut what's the derivative of

30:18X2 well we bring the two downstairs to

30:21get 2x to the something for the exponent

30:24we subtract one from the original

30:26exponent two 2 - 1 is 1 and X to ^ of 1

30:31is just X so the derivative of x^2 is

30:362x can you see on your own that the

30:38derivative of x

30:404th is 4X cubed yes the power rule is

30:45pretty easy I'm arranging the exponents

30:48in numerical order so let's go back up

30:50top for the derivative of x to the first

30:53Power bring the one downstairs and the

30:56exponent becomes zero since x to the 0

30:59is 1 the derivative is 1 * 1 which is 1

31:03so the derivative of x is 1 this one's

31:07easy to see on a graph the line

31:09representing yal X is a straight line

31:12whose slope is obviously 1 it all works

31:14out let me scooch over here to get more

31:17room I'll sketch in x to the 0o and x to

31:20the first power so you can see that the

31:22pattern is kept and speaking of pattern

31:25let's go up top again to find the

31:26derivative of x to the 0 the 0er comes

31:29downstairs and the exponent becomes -1

31:33well 0 * anything is 0o so the

31:35derivative is zero and let's notice that

31:38x to the 0 power is 1 so this is a

31:41special case of the constant rule which

31:43says that the derivative of any constant

31:46such as one is zero so the power rule

31:49and constant rule give us the same

31:51derivative for exponent zero pretty neat

31:53it all works together and the rules are

31:56consistent how about this one what's the

31:59derivative of the square < TK of X well

32:02remember that the < TK of X is X raised

32:04to the2 power the power rule isn't

32:07limited to integer exponents we bring

32:10the 1/2 downstairs and subtract one from

32:131/2 to get an exponent of

32:15-2 since x^ -2 is 1 / < TK X this gives

32:21us 12 * 1 the < TK X or 1 / 2 < TK X the

32:28fraction has a radical in the

32:30denominator so we can rationalize by

32:32multiplying the numerator and

32:33denominator by squ < TK of x to get < TK

32:37X over 2X and that's the derivative of

32:40the squ < TK of X using the power rule

32:42which works for all real exponents not

32:45just

32:46integers well I've told you the power

32:48rule but I haven't proven that it's true

32:50now I'll prove it with limits let's take

32:53the derivative of x to the n as the

32:55limit of Delta y over Delta X like this

32:58as H approaches zero just like before we

33:01have our green and red y values the most

33:05tedious part of this proof is expanding

33:07the green binomial in this video's

33:10description I put some links to videos

33:12that go into more detail on the binomial

33:15expansion in short when the green x + H

33:18to the N is expanded the first green

33:21term will always be x to the n and the

33:24last green term will always be H to the

33:26N then moving towards the middle the

33:29second term is X to the n minus1 and H

33:32to the 1st prefixed with the binomial

33:34coefficient which for the second term is

33:36always n none of the other coefficients

33:39are important for this proof but it's

33:41interesting how the coefficients of each

33:43green term follow a pattern revealed by

33:46Pascal's triangle but that's not what

33:48this video is about so see the links in

33:49the description for more information

33:52math is so interesting and easy when you

33:54can see how it all fits together the

33:57second to last screen term will always

33:59have X and H to the N minus1 along with

34:02the binomial coefficient which is also n

34:05for the second to last term all the

34:08terms in the middle denoted by The

34:09Orange Box are the remaining binomial

34:12expansion terms and will have factors of

34:14H to a power of two or greater that will

34:17be important for our proof in a moment

34:20in fact all the green terms except the

34:22first two will have H factors raised to

34:25a power of two or more and there're

34:27still still the Red X to the end term we

34:29subtract at the end let's bring

34:32everything down and encapsulate the

34:34orange we'll always have positive and

34:36negative x to the N so they'll cancel

34:39every time every remaining term will

34:42always have at least one H since the

34:45denominator is H we can cancel an H from

34:48each remaining term this leaves n * x n

34:53-1 plus the orange terms which initially

34:56all had factors of h^2 or higher and now

35:00since we divided three by H they all

35:03have factors of H or higher when we take

35:06the limit as H approaches Z all of these

35:08terms approach zero and we're left with

35:11n * x n -1 and so we've proven the power

35:17rule once more the rule comes from the

35:20slope expression and is the pattern we

35:22notice every time we take the limit so

35:24when we do calculus we don't need to

35:27take the limit we just use the pattern

35:29and the pattern for the derivative of x

35:32the N is n * X

35:36nus1 let me show you a visual

35:38interpretation of the power rule that

35:41should build your intuition the power

35:43rule tells us that the derivative of x^2

35:46is 2x let's consider the function x^2 to

35:49be the area a of an actual square with

35:53sides of length x the derivative of X2

35:56with respect to X X is the amount the

35:59area changes per change in X so if we

36:02let X Change by this differential amount

36:05DX by how much does the area of the

36:08square change well it changes by the

36:11area of these two narrow

36:13strips this one has an area of its

36:16height x * its width DX so X

36:21DX this one on top also has an area of

36:24its height DX times its width X so its

36:29area is also

36:31xdx for the sake of completeness let me

36:34point out this tiny corner piece whose

36:36area is dx^ s it's one of the orange

36:39terms from the binomial expansion that

36:41goes to zero as Delta X approaches zero

36:45so at the limit as Delta X approaches

36:47zero the change in the blue squares area

36:50the differential of a is the

36:52differential of our function x^2 this is

36:55the added area to X DX so the derivative

37:00of x^2 the change in x^2 per change in X

37:04is indeed

37:062x let's look at the derivative of x

37:09cubed visually the power rule tells us

37:12that its derivative is

37:143x^2 let's visualize the function X

37:17cubed as the volume of an actual Cube

37:19whose sides have length x when we

37:22increase X by the tiny differential DX

37:26what happens to the volume of the the

37:28cube when we add DX over here we get

37:31this new volume a thin slab on this face

37:34his volume is the area of the face X2

37:38time its thickness DX so the additional

37:42volume is x s DX you might see where

37:45we're going we get the same additional

37:48volume on these other two faces for a

37:50total of three x^2

37:53DX because I've drawn DX with some

37:56visible thickness you might know notice

37:57these three thin regions having volume X

38:00their length time dx^ 2 their

38:03cross-sectional area so the added volume

38:06is 3 x dx^ 2 this last tiny volume has

38:11sides equal to DX so its volume is DX

38:14cubed these orange terms all go to zero

38:17at the limit but it's interesting to see

38:20that even they have a visual

38:21interpretation on the

38:23diagram so the differential change in

38:25the volume X cubed is 3 x^2 DX this

38:30means the derivative of x cubed is the

38:32change in X cubed per change in X which

38:35is indeed 3x^2 as the power rule tells

38:38us to illustrate the usefulness of the

38:41power rule consider finding the

38:43derivative of x to the 4th with

38:49limits as you can see the power rule is

38:52so much simpler and always gives the

38:54same answer once more when we do

38:56calculus we use these shortcut rules we

38:59don't take limits in a calculus course

39:02or textbook you'll take limits only for

39:04the first week or two to demonstrate

39:06what the derivative means but once you

39:08learn the rules of differentiation

39:10you'll use them and you won't find

39:12limits anymore but I will still find

39:15limits in this video to prove the rules

39:17and illustrate some

39:19points now we'll go over ways to take

39:22the derivative of combinations of

39:24functions the first is the addition rule

39:27it says that the derivative of the sum

39:29of two functions is the sum of their

39:31distinct derivatives or in terms easy to

39:34remember the derivative of the sum is

39:37the sum of the

39:38derivatives here I'm using f and g as

39:41the two functions and the prime symbol

39:44to denote their derivatives an example

39:46should make this simple let's find the

39:48derivative of x^2 + x first the

39:52derivative of x^2 is 2X and the

39:55derivative of x is is 1 so the

39:59derivative of x^2 + x is 2x + 1 the

40:03derivative of the sum equals the sum of

40:06the

40:08derivatives let's look at the graph and

40:10see why this makes sense here are the

40:12two functions we're adding y = x^2 in

40:15white and Y = X in green and the blue

40:19curve is their sum y = x^2 + x let's

40:24take two nearby X values and look look

40:27at the corresponding Delta Y's for the

40:29three

40:30curves here's Delta y for white the

40:33change in X2 over our small Delta X and

40:36here's Delta y for green the change in

40:39green y over our small Delta X and

40:42finally here's Delta y for blue the

40:44change in x^2 + x over our small Delta X

40:49can you see that since blue equals y

40:51plus green that Delta y for blue equals

40:54the sum of the white and green Delta y's

40:57since blue equals white plus green

40:59everywhere any difference in blue must

41:02be equal to the corresponding difference

41:04in white plus green remember as Delta X

41:07approaches zero we use the word

41:09differential to describe the changes so

41:11the differential of the sum equals the

41:13sum of the differentials and that's the

41:16addition rule without elaboration I'll

41:18assert that the same relationship holds

41:20with subtraction that the derivative of

41:23the difference between two functions

41:25equals the difference of their

41:26respective derivatives so I'll write the

41:29addition rule with plus or minus since

41:31it works for addition and

41:33subtraction next is the product or

41:36multiplication rule it says the

41:38derivative of the product of two

41:40functions is the first function times

41:42the derivative of the second plus the

41:44second times the derivative of the

41:47first so the pattern is different than

41:50the addition rule because the derivative

41:52of the product is not the product of the

41:55derivatives like we did for the the

41:57power rule let's visualize the product

41:59rule by considering the product of the

42:01two functions to be a rectangle whose

42:03area is the product f *

42:06G on this graph the axes represent the

42:09values of the functions not the

42:11independent variable X at least not

42:14directly when we increment X by its

42:16differential DX F ofx increases by its

42:20own derivative DF by DX and the area of

42:23the rectangle increases slightly by this

42:26thin strip

42:27at the same time G ofx increases by its

42:30own different derivative DG by DX and

42:33the area of the rectangle increases

42:35slightly by this thin

42:37strip the area of this first strip is

42:40its height G of x times its width DF by

42:45DX and the area of the second strip is

42:48its height DG by DX times its width F

42:52ofx so the derivative of the product of

42:55the two functions is the addition area

42:58of the two rectangular strips and their

43:00dimensions are each function times the

43:03derivative of the other and that's a

43:05nice visual interpretation of the

43:07product

43:09rule now we can look back at our sample

43:12function x^2 - 2x + 1 and consider which

43:16of these rules we'll need to use to find

43:18its

43:18derivative well it's the sum of three

43:21simpler functions so let's start with

43:23the addition rule the derivative of x^2

43:26is 2X X by the power rule for the next

43:29term we need to subtract the derivative

43:31of 2x which is the product of 2 and X so

43:35we'll need the product rule we'll get

43:37back to it in a second and the

43:39derivative of one is zero by the

43:41constant rule interesting to take the

43:43derivative of a simple polinomial we

43:46need all four of the rules we've covered

43:48so far now let's use the product rule to

43:51find the derivative of 2x the product

43:54rule says that the derivative of the

43:55product is the first function times the

43:58derivative of the second plus the second

44:00function times the derivative of the

44:02first the first function is two and the

44:05second function is X the derivative of

44:082x is 2 * the derivative of X Plus x *

44:12the derivative of two the derivative of

44:15x with respect to X is 1 by the power

44:18rule so the first term becomes 2 * 1 the

44:22derivative of two a constant is zero so

44:25the second term becomes x * 0 this all

44:28simplifies to two so the derivative of

44:312x is 2 which makes the derivative of

44:34our original function x^2 - 2x + 1 = 2x

44:39- 2 of course this is the same answer we

44:43got when we took the limit well this is

44:45the function I promised you'd be able to

44:47differentiate in a few seconds but using

44:50these four rules certainly took more

44:51than a few seconds let me show you a

44:54calculus super shortcut based on the

44:56product rule the derivative of any

45:00constant K * x with respect to X is

45:03simply the constant K because when we

45:05use the product rule we get K * the

45:08derivative of x + x * the derivative of

45:11K the derivative of x with respect to X

45:15is always one so the first term will

45:17always be K and the derivative of K with

45:20respect to X is always Zero by the

45:23constant rule so the second term will

45:25always be zero this pattern occurs every

45:28time the product rule is applied to KX

45:31so the derivative of any constant K * X

45:35is always

45:36K this is the kind of pattern the rules

45:39of differentiation let us exploit now

45:42let's go back and find the derivative of

45:44x^2 - 2x + 1 at x = 1.6 by

45:48differentiating left to right 2x - 2

45:53with practice you learn to ignore

45:55constants plug Again The Chosen x value

45:58of 1.6 to get 3.2 minus 2 so 1.2 and

46:04that's how with practice you can know

46:07almost immediately that the derivative

46:09or slope of x^2 - 2x + 1 at x = 1.6 is

46:171.2 okay now for another great shortcut

46:20which is a generalization of the first

46:22one for KX this time we'll take the

46:25derivative of KX to to the N so we have

46:28a power of X with some constant

46:31coefficient K so this is the product of

46:34two functions K and x to the N let's use

46:38the product rule again and see where

46:40this leads us we have the first * the

46:43derivative of the second plus the second

46:45* the derivative of the first the

46:48derivative of x to the N is straight

46:50from the power rule n * X nus1 and the

46:54derivative of constant K is a of course

46:57zero so the second term becomes zero

47:00this leaves the derivative as K * n x

47:04nus1 at first this new shortcut looks a

47:07little cumbersome like the power rule

47:09did but look at What it lets us do when

47:12we bring the exponent downstairs we can

47:14just multiply it by whatever coefficient

47:16is already there so the derivative of 4X

47:20cubed is 12 x^2 we bring the three down

47:24stairs multiply it by the co efficient

47:27of four that's already there to get 12

47:30and then we subtract one from the

47:31exponent this simple super shortcut has

47:34the product rule power rule and constant

47:37rule built in it automatically follows

47:40all the rules so knowing this super

47:43shortcut and using the addition rule we

47:46can easily take the derivative of long

47:48polom we just work left to right

47:51differentiating one term at a time pols

47:54are incredibly simple to

47:58differentiate I need to finish the rules

48:00of differentiation but first we actually

48:03know enough calculus to solve some

48:05interesting problems known as

48:07optimization problems they involve

48:09finding the local minimum and maximum

48:12values for a function for example

48:14suppose this curve represents the net

48:17profit our company would make by

48:19manufacturing and selling X number of a

48:21particular item if we make and sell too

48:24few our profit will be limited by the

48:26low number if we make and sell too many

48:29our supply might exceed the demand and

48:31our extra cost for running more machines

48:34and hiring more people won't be offset

48:36by the higher volume there's some

48:38independent variable here that will

48:40maximize the profit function how do we

48:43find it let's notice that at the maximum

48:46the slope of the function is zero lucky

48:49for us we know calculus we can take the

48:51derivative of the profit function and

48:54find the equation for its slope anywhere

48:57when we set the derivative function

48:59equal to zero we can solve for x to get

49:02the exact point at which the original

49:04function is at its

49:05maximum our profit function f ofx in

49:08thousands of dollars is

49:120.012 x^2 + 9.8 x

49:17-500 where X is the number of units we

49:19manufacture and sell to find the value

49:22for x that maximizes the function we

49:25take the derivative of the function

49:26function and set it equal to Z and solve

49:29for x we know how to take the derivative

49:32of polom we have

49:340.024 x +

49:379.8 set this equal to zero and solve x =

49:4348.3 so we should make 408 units to

49:46maximize our profit if your problem

49:48statement asks you for the profit amount

49:51plug 48 into the original profit

49:53function to get the maximum profit

49:55amount F of 48 equals

49:595.8 and we're told this is in thousands

50:02of dollars so the maximum profit is

50:06$5,800 when we make and sell 48

50:10units let's do another problem suppose

50:13we have a rectangular sheet of metal

50:15that measures 32 CM by 24 cm we want to

50:19make a box by cutting squares out of the

50:22corners and folding the resulting sides

50:24up the box won't have a lid

50:27the shapes we cut out of the corners

50:29need to be squares so that when we fold

50:31the sides up they'll all have the same

50:33height what are the dimensions of the

50:35Box having the greatest volume and what

50:38is the

50:39volume okay we need to express the

50:41volume v as some function of a variable

50:44let's use the length of the square sides

50:46that we cut out of the corners and call

50:48it X of course all of these distances

50:51are X the volume of the Box will be its

50:55width times its depth times its height

50:58the width is this distance which in

51:00centim is 32 minus

51:032x the depth is 24 -

51:082x so when we multiply these Expressions

51:11that gives us 4X cubed - 112 x^2 + 768 x

51:18I'm skimming over the algebra so we can

51:19focus on the calculus to find the x

51:22value that maximizes this function we'll

51:24take the derivative set it equal to 0

51:27and solve for x the derivative of the

51:30polom is 12 x^2 -

51:34224x +

51:37768 this is a quadratic equation and

51:39there are several ways to solve it I

51:42used but won't show the quadratic

51:44formula to get two possible solutions X

51:47= 14.1 cm and X = 4.53 CM we need to

51:53check these numbers for feasibility the

51:55short side of the metal sheet is 24 cm

51:59so we can't cut out squares greater than

52:0112 CM there's not enough metal on that

52:03edge so that leaves 4.53 CM as our

52:06answer for X but that's not the answer

52:10to our problem we're asked for the

52:11dimensions that maximize the Box's

52:13volume and for that maximum volume so we

52:17plug in 4.53 cm for X into our width and

52:21depth formulas we get a width of 22.9 4

52:25cm and the depth depth of

52:3214.94%

52:34so multiplying these Dimensions yields a

52:37volume of

52:381,552 cubic cm any other value for the

52:42square size X will result in a lower

52:46volume let's look at a different

52:48function this curve has two points where

52:51the slope is zero a local maximum here

52:54and a local minimum here when we use the

52:57word local to describe a minimum or

52:59maximum we mean compared to the points

53:01nearby for example the local maximum

53:04identified here isn't the function's

53:07maximum it has higher values out to the

53:09right and similarly there are lower

53:12values closer to the y- axis than the

53:14local minimum identified Here Local

53:17means higher or lower than nearby points

53:19on either side anyway when we set the

53:23derivative equal to zero and solve for x

53:25we get these values but it's important

53:28to be able to tell the difference

53:29between a minimum and a maximum if our

53:32management team wanted to maximize

53:34profits and we recommended action

53:36corresponding to this point that could

53:38be a disaster or if they wanted to

53:40minimize budget or time and we chose

53:43this point setting the derivative equal

53:45to zero and solving for x will give us

53:47the points where the slope is zero which

53:49could be a local maximum or minimum but

53:52how can we know

53:54which let's color code the function

53:56slope green for positive here to the

53:59left of the first zero point the slope

54:01is zero at the maximum of course and

54:04then negative red between the two zero

54:06points then the slope is zero again at

54:09the minimum and positive beyond the

54:11second zero point so the slope changes

54:14signs at the Minima and Maxima this

54:17makes sense if it's zero at a point it

54:19must be passing from positive to

54:21negative or from negative to positive

54:25but notice that at the maximum point the

54:27slope is changing from positive to

54:29negative and at the minimum point the

54:31slope is changing from negative to

54:33positive let's plot the function's

54:37derivative of course it's zero at the

54:39two points where the function slope is

54:41zero at the local Maxima the slope of

54:44the pink derivative will always be

54:46changing from positive to negative which

54:49means its slope is negative as you can

54:51see here and at local Minima the slope

54:54of the pink derivative will always be

54:56changing from negative to positive which

54:59means its slope is positive as you can

55:01see here the slope of the derivative is

55:04positive at locom Minima so what do we

55:07mean by the slope of the derivative

55:09remember in calculus we're just

55:11performing operations on functions that

55:13result in different functions when we

55:16differentiate a function f ofx to get

55:18its derivative fime of x frime of X is

55:22just a new function that we can in turn

55:24differentiate to get it derivative the

55:27derivative of a derivative is called the

55:30second derivative or depending on the

55:32variables the second derivative of y

55:35with respect to

55:36X I'll show more notation in a moment

55:39but F Prime of X is a common way to

55:42denote the second derivative the second

55:45derivative of a function tells us the

55:46rate of change of the slope of that

55:49function and as you might guess the

55:51derivative of the original function is

55:53called the first

55:55derivative here here are six examples of

55:57Curves from various functions f ofx for

56:01this first one the slope frime of X is

56:04positive since the function's value is

56:06increasing and since the rate of change

56:08of the increase is not changing that is

56:11it's a steady increase the second

56:13derivative is

56:15zero in this example the first

56:18derivative fime of x is again positive

56:21since the function value is increasing

56:24and since the rate of increase is itself

56:26increasing the second derivative is also

56:30positive in this example the slope is

56:33increasing but it's changing from a

56:35steep High slope to a shallow low slope

56:38so the second derivative is negative can

56:41you see how these two functions have

56:43different behaviors even though they

56:45both have a positive slope for one the

56:48slope is increasing at an increasing

56:50rate for the other the slope is

56:52increasing but at a decreasing rate

56:57down here the slope frime of X is

57:00negative and since the slope is constant

57:02the second derivative is

57:04zero this function has a negative slope

57:07but the slope is getting less and less

57:09negative so the second derivative is

57:11increasing

57:13positive and this last curve also has a

57:16negative slope and its slope is getting

57:18more and more negative so the second

57:20derivative is decreasing

57:23negative since the first derivative can

57:25be thought of is the rate of change the

57:27second derivative is essentially the

57:29rate of change of the rate of change the

57:32second derivative is useful because it

57:34helps describe the behavior of functions

57:37and it's especially useful in

57:39minimization and maximization problems

57:41to distinguish between local Minima and

57:44Maxima at points where frime of X is

57:47zero X represents a minimum point where

57:50the second derivative is positive and X

57:52represents a maximum point where the

57:54second derivative is negative

58:00now for some second derivative notation

58:02the derivative of the derivative is the

58:04second derivative fpre of x since the

58:08derivative is dy by DX we can write the

58:11second derivative as d by DX of Dy by DX

58:15let's move the function up to the

58:17numerator now what follows is symbolic

58:20shorthand not true algebra we take the d

58:23and Dy at the top and combine them to

58:26get d^2 y because there's two D's and

58:29one y and when we take the two DXs and

58:32combine them we get dx^ 2 so the

58:35symbolic representation of the second

58:37derivative of y with respect to X is d^2

58:41y by DX squared I'm not saying it makes

58:44pure algebraic sense maybe a

58:46mathematician can tell us in the

58:48comments if there's a deeper meaning

58:49behind the symbol I'm just an engineer

58:52and I don't

58:53know there are higher order derivatives

58:55of course of course the derivative of a

58:57function second derivative is the

58:59function's third derivative it's easy to

59:01extend the D by DX pattern to see third

59:04derivative as D cubed y by DX cubed fle

59:09Prime is another representation of the

59:11third

59:12derivative please don't think that

59:14higher order derivatives are any more

59:17difficult or complicated than the first

59:18derivative it's the same differentiation

59:21operation following the same rules of

59:24differentiation well let's get back to

59:26the remaining rules of differentiation

59:29but first let's check our progress we've

59:31covered four important rules of

59:34differentiation the constant rule the

59:36power rule the addition subtraction Rule

59:39and the product rule then we covered

59:41some super shortcuts involving polom

59:44solved two problems and learned about

59:47the second and higher order derivatives

59:50now we'll cover the last five rules of

59:52differentiation in this order

59:57first s and

59:59cosine here's a sine wave the plot of y

1:00:02equal s of theta you don't need to be an

1:00:05expert at trigonometry to differentiate

1:00:07s and cosine but if anything I'm about

1:00:10to say seems unfamiliar I have a YouTube

1:00:12trigonometry course if you want to brush

1:00:14up Linked In the

1:00:16description let's draw the sign

1:00:18function's derivative by plotting a few

1:00:20points and seeing what patterns arise

1:00:23we'll start with a local Minima and

1:00:24Maxima there always easy to see so the

1:00:27derivative will be zero and intersect

1:00:30the Theta axis at these blue

1:00:33points at these points where the sine

1:00:35wave crosses the Theta axis in an

1:00:37upwards Direction the slope is one using

1:00:41the expression for the slope again let

1:00:43me demonstrate quickly and without

1:00:45elaboration that Delta y over Delta

1:00:47Theta we've made Theta our independent

1:00:50variable not X is very close to one when

1:00:53Green Delta Theta is very close to zero

1:00:58the slope at these points is one so

1:01:00we'll plot the blue derivative points

1:01:02here at y equal 1 the y-coordinate of

1:01:06each Blue Point represents the slope of

1:01:08the red sign curve at that value of

1:01:11theta the slope at these points is -1 so

1:01:15the Blue Points go down here we could

1:01:18plot more points but let me jump to the

1:01:20answer and plot the derivative of sin

1:01:22Theta it's this smooth curve

1:01:26if you're familiar with trigonometry

1:01:28you'll recognize this curve as cosine

1:01:30Theta the derivative of sin Theta is

1:01:33cosine Theta pretty neat let me show you

1:01:37a proof it assumes a little trigonometry

1:01:39knowledge but I'll go quickly we'll

1:01:41consider a unit circle and focus on the

1:01:44first quadrant let's put angle Theta in

1:01:46standard position since we're on the

1:01:48unit circle this yellow length the

1:01:51radius of the circle is one this

1:01:53horizontal length is cine Theta and this

1:01:56vertical length is sin Theta it's this

1:01:59vertical red length we're interested in

1:02:01as Theta changes ever so slightly by D

1:02:04Theta what's the change to the red

1:02:06length D sin Theta let's find

1:02:09out the pink Arc has length Theta which

1:02:12seems strange because green Theta

1:02:15represents an angle the number of

1:02:16radians and pink Theta represents a

1:02:19distance the number of radi but since

1:02:22the radius of the circle is one the

1:02:24numbers for Theta green GRE and pink are

1:02:27the same let's see what happens near

1:02:29this point and note that this segment of

1:02:31the circle circumference is very nearly

1:02:33a straight line as our focus of

1:02:35attention get smaller and

1:02:37smaller the derivative of sin Theta is

1:02:40how much this vertical distance changes

1:02:42as Theta changes by the tiny

1:02:44differential of theta D Theta and since

1:02:47pink Theta and green Theta have the same

1:02:50measurement I'm going to make the D

1:02:51Theta label green to match our formula

1:02:54to find the change sin Theta let's draw

1:02:57this right triangle and we can see that

1:02:59red Sin Theta changes by this amount

1:03:02that we can call D sin Theta so in this

1:03:05small triangle we have representations

1:03:07for D sin Theta and D Theta which are

1:03:10the numerator and denominator of the

1:03:12derivative we're trying to find since

1:03:15the trig ratios are the ratios between

1:03:17the various sides of a right triangle

1:03:19our derivative ratio is one of the six

1:03:21trig functions this angle is congruent

1:03:24to Theta I'm telling you this without

1:03:26proof and so D sin Theta and D Theta are

1:03:29the adjacent and hypotenuse of the small

1:03:32triangle respectively and adjacent over

1:03:34hypotenuse corresponds to cosine and so

1:03:37we've shown graphically that the

1:03:39derivative of s is cosine be careful

1:03:43because the opposite is not true the

1:03:45derivative of cosine is not s since the

1:03:48blue cosine curve has the exact same

1:03:51shape as the sign curve it seems

1:03:53reasonable to deduce that the derivative

1:03:55of cosine will also have this shape and

1:03:58let's note that the cosine is out of

1:04:00phase with s to illustrate I'll add

1:04:03Theta axis markers at every pi/ 2

1:04:06radians and we can see that the cosine

1:04:08curve the derivative of s is always pi

1:04:11over two radians to the left of s this

1:04:14is easiest to see by comparing peak-to

1:04:17Peak points where the functions have

1:04:18their maximum

1:04:20values since the derivative of s is out

1:04:23of phase to it by Pi / 2 does it it make

1:04:26sense that the derivative of cosine

1:04:28would be out of phase to it well yes

1:04:30indeed it actually is but as you can see

1:04:33we don't have a function with these Peak

1:04:35values but if we flip the sign curve by

1:04:38taking its negative we get the curve we

1:04:40seek and the derivative of cosine Theta

1:04:44is indeed negative sin

1:04:46Theta so it's the derivative of negative

1:04:49sin Theta but looking at the curve you

1:04:51may see what's coming next the

1:04:53derivative of negative sin Theta is

1:04:56cosine Theta and taking the derivative

1:04:59of cosine Theta gets us back to sin

1:05:02Theta and this four-step cycle comprises

1:05:05the trig related rules of

1:05:08differentiation it might help to

1:05:09remember that the trig functions

1:05:11alternate that is taking the derivative

1:05:13of a sign yields a cosine and vice versa

1:05:17then it's easy to remember that the

1:05:19derivative of s keeps the sign so the

1:05:22derivative of positive sign is positive

1:05:24cosine keep the S and the derivative of

1:05:27negative sin Theta is negative cosine

1:05:30Theta the derivative of s keeps the S on

1:05:34the other hand the derivative of a

1:05:35cosine function flips the sign the

1:05:38derivative of positive cosine Theta is

1:05:41negative sin Theta and the derivative of

1:05:43negative cosine Theta is positive sin

1:05:47Theta as you may know there are four

1:05:49more trig functions but we'll have to

1:05:51skip them for now and cover their

1:05:53derivatives later let's test our

1:05:56knowledge what's the derivative of x Cub

1:05:58* sin x well we have the product of two

1:06:02differentiable functions we can call f

1:06:05and g so we'll use the product

1:06:08rule the derivative of the product

1:06:10equals the first times the derivative of

1:06:12the second plus the second * the

1:06:15derivative of the first so it's simple

1:06:18there's really no intermediary steps

1:06:20just write down the components and

1:06:22that's the derivative of the product X

1:06:25cubed cine x + sin x *

1:06:313x^2 now for the chain

1:06:35Ru the chain rule is how we

1:06:37differentiate composite functions a

1:06:40composite function is a function whose

1:06:42argument includes another function you

1:06:45can think of composite functions as

1:06:47embedded functions where one function is

1:06:49embedded in the other for example sin 2x

1:06:54is a composite function because s is a

1:06:56function and its argument 2x is another

1:07:00function the 2x function is embedded in

1:07:03the sign function as its argument this

1:07:06is very common in math science and

1:07:08engineering so you'll use the chain rule

1:07:10a lot probably more than any other

1:07:13rule let's see why we need the chain

1:07:16rule when we find the derivative of sin

1:07:182x first we know the derivative of sin x

1:07:21with respect to X is cosine X it's true

1:07:25it's one of of our rules of

1:07:26differentiation the one we just covered

1:07:29but we cannot say that the derivative of

1:07:31sin 2x with respect to X is cosine 2X

1:07:35that's false it's close we'll need to

1:07:38adjust a bit with the chain rule to get

1:07:40the derivative of sin 2x but this isn't

1:07:43right here's the pattern the derivative

1:07:46of s something with respect to that

1:07:48something equals cosine of that

1:07:50something all three terms need to match

1:07:53for the differentiation rule to apply

1:07:56and when we try to apply the rule to sin

1:07:582x you can see that they don't match I'm

1:08:01illustrating this with the trig rule but

1:08:04the pattern applies to all the

1:08:06rules if we were to modify the equation

1:08:09to be the derivative of sin 2x with

1:08:11respect to 2x then the derivative would

1:08:14be cosine 2X because all the terms would

1:08:18match but in calculus were not asked

1:08:20very often to find the derivative with

1:08:22respect to a function of X just with

1:08:25respect to X so we need to dig a Little

1:08:28Deeper to differentiate composite

1:08:30functions we can write composite

1:08:32functions like this F of G of x g is

1:08:37called the inner function because it's

1:08:39inside the argument for function f which

1:08:41is the outer function the derivative

1:08:44we're seeking is DF by DX the derivative

1:08:47of the outer function with respect to

1:08:50the argument of the inner function our

1:08:52independent variable X here's the key to

1:08:55to understanding the chain rule a

1:08:57differential change in X will result in

1:09:00a differential change to G DG by

1:09:04DX that differential change to G in turn

1:09:07causes a differential change to F DF by

1:09:12DG and that change to function f DF that

1:09:15occurs as a result of the differential

1:09:17change to X DX is the derivative we want

1:09:21DF by

1:09:23DX this is where the chain rule gets its

1:09:26name the differential change to X

1:09:29ripples out in a chain reaction to cause

1:09:31the differential change in the outermost

1:09:34function here's the chain rule for

1:09:36differentiation the derivative of f of g

1:09:39ofx equals the derivative of f with

1:09:42respect to G times the derivative of G

1:09:45with respect to X it should look

1:09:47familiar it's the Chain Reaction we just

1:09:50traced from the independent variable X

1:09:52Out to the outermost function and it

1:09:55makes sense algebraically because

1:09:57there's a clear cancellation chain that

1:09:59makes the chain rule a lot easier to

1:10:01visualize and

1:10:03understand so let's find the derivative

1:10:06of sin 2x for the first Factor DF by DG

1:10:10we need the derivative of the outer sin

1:10:132x with respect to the inner 2x let's

1:10:16notice that these terms match you'll

1:10:19always get a match like this when you

1:10:20use the chain rule we can use the trig

1:10:23rule that the derivative is cosine of

1:10:25the matching term so DF by DG equal

1:10:28cosine 2X by the way this is the answer

1:10:32we said was not right a moment ago to

1:10:34get the correct answer we need to

1:10:35multiply by the last term DG by DX G is

1:10:402x so we get the derivative of 2x with

1:10:43respect to X it doesn't get much easier

1:10:46than this we have a super shortcut that

1:10:48tells us the derivative of 2x with

1:10:50respect to X is 2 so we use the chain

1:10:53rule to determine that the D derivative

1:10:55of sin 2x is 2 cosine

1:10:592X let's do another problem and find the

1:11:02derivative of the sare < TK of 5x^2 +

1:11:063 the inner function is 5x^2 + 3 the

1:11:10outer function is the square root let's

1:11:13rewrite the expression using an exponent

1:11:15of 1/2 to represent the square root this

1:11:18should make it clear which function is

1:11:20the inner function and which is the

1:11:23outer the first Factor we need to find

1:11:25find is DF by DG the derivative of the

1:11:28outer function with respect to the inner

1:11:30the outer function is 5x^2 + 3 to the 1/

1:11:3412 the inner function is 5x^2 + 3 when

1:11:38you use the chain rule you'll always

1:11:40have matching terms and can use the

1:11:42appropriate rule of

1:11:44differentiation in this case the power

1:11:46rule with exponent

1:11:481/2 with the power rule we bring the

1:11:51exponent downstairs and subtract one

1:11:53from it we get 1/2 times the matching

1:11:56expression which turns out to be the

1:11:58inner function G raised to the -2 and

1:12:02that's the first term in the chain rule

1:12:04DF by DG the second Factor DG by DX is

1:12:09simple too it's the derivative with

1:12:11respect to X of 5x^2 + 3 we use the

1:12:15power rule again for this one 10 x you

1:12:19can simplify the expression using

1:12:21algebra and we found the derivative of

1:12:23this composite function

1:12:26let's do one more chain rule example

1:12:28this time with a composite of three

1:12:30functions so we want to find the

1:12:32derivative of f of G of H of X let's set

1:12:37up the chain the derivative of f with

1:12:40respect to G times the derivative of G

1:12:43with respect to H times the derivative

1:12:45of H with respect to x three functions

1:12:48makes the chain concept even more

1:12:51obvious algebraically the dgs cancel and

1:12:54the DH is cancel leaving us with DF by

1:12:57DX the derivative of the outermost

1:13:00function with respect to the independent

1:13:02variable

1:13:03X so let's find the derivative with

1:13:06respect to X of cine 2 4X let's rewrite

1:13:10the function as cosine of 4x^ squared

1:13:13because cosine squar argument means the

1:13:16cosine of the argument

1:13:18squared the inner function is 4X the

1:13:21middle function is cosine and the outer

1:13:24function is power of two let me expand

1:13:27the derivative chain and show you again

1:13:29how simple this

1:13:32is the chain always starts with the

1:13:35differential of the given outermost

1:13:37function as the numerator of the first

1:13:39factor for this problem the given

1:13:42function is cine of 4x^ 2 so D cosine

1:13:474x^

1:13:48SAR the denominator of the first factor

1:13:51is DG the differential of the middle

1:13:54function which is cosine so D cosine

1:13:57forx as usual when we use the chain rule

1:14:01we have matching terms and the

1:14:03derivative with respect to something of

1:14:04something squared is two of that

1:14:06something so the first term in the chain

1:14:09DF by DG is 2 cosine

1:14:124X the numerator of the second factor is

1:14:15the denominator of the first that's how

1:14:18the chain Works D cosine 4X the

1:14:21denominator of the second factor is DH

1:14:24the differential of the inner function

1:14:26which is 4X so

1:14:29d4x again our terms match and we have

1:14:31the derivative of cosine of something

1:14:34with respect to that something the

1:14:36something is 4X and the derivative of

1:14:38cosine is negative s so the second term

1:14:41DG by DH is NE sin

1:14:454X following the pattern the numerator

1:14:48of the third factor is the denominator

1:14:50of the second differential of

1:14:524X and finally at the end of the chain

1:14:55is the differential of the independent

1:14:57variable X DX the last Factor will be a

1:15:00straightforward derivative the

1:15:02derivative of 4X with respect to X is 4

1:15:05so the last Factor DH by DX is

1:15:094 rearrange the terms if you like and we

1:15:12found the derivative of cosine 2 4X and

1:15:15that's the chain rule the one you'll use

1:15:18most often in real life and very easy

1:15:20with

1:15:23practice next is the quo rule where

1:15:25we'll find the derivative of one

1:15:27function divided by another put

1:15:29differently we're finding how the ratio

1:15:31between two functions of X changes as X

1:15:34changes first the derivative of the

1:15:36ratio is not the ratio of the

1:15:39derivatives that might remind you of the

1:15:41product rule since the derivative of the

1:15:43product is not the product of the

1:15:45derivatives the quotient rule says that

1:15:47the derivative of the ratio is the

1:15:50denominator time the derivative of the

1:15:52numerator minus the numerator times the

1:15:55derivative of the

1:15:57denominator all over the denominator

1:16:00squared we can prove it's true using the

1:16:02product rule and chain rule first we

1:16:05rewrite the quotient as a product with a

1:16:08denominator raised to the -1 power so we

1:16:11have d by DX of f ofx * G ofx

1:16:16the1 so we use the product rule the

1:16:19first * the derivative of the second

1:16:21plus the second * the derivative of the

1:16:24first the derivative has four components

1:16:27each straightforward except this one is

1:16:29a composite of a function raised to a

1:16:31power so we'll just need to apply the

1:16:33chain rule we have the derivative of

1:16:36some function raised to the -1 so the

1:16:38power rule tells us that's -1 * the

1:16:41function raised to the -2 then to finish

1:16:44the chain rule we multiply by the

1:16:46derivative of the function G Prime of X

1:16:49now let's

1:16:52simplify on the left side we have f

1:16:55ofx * G Prime of

1:16:58X all over G of x^2 then we add the

1:17:03right

1:17:04side fime of

1:17:06X over G

1:17:08ofx so we're adding two fractions but

1:17:11their denominators don't match if the

1:17:14denominators matched we could add their

1:17:16numerators if we multiplied the right

1:17:18denominator by G of X then they'd both

1:17:21be g^ 2 of X so let's multiply the right

1:17:24term by G of X over G

1:17:28ofx and that's it if we swap the left

1:17:30and right terms the format will match

1:17:32the quotient rule stated above so we've

1:17:35derived the quotient rule from the

1:17:37product rule and chain

1:17:39rule to remember the chain rule I start

1:17:42with the denominator squared then the

1:17:44numerator expression starts with the

1:17:46denominator not squared then like the

1:17:49product rule we multiply one by the

1:17:52derivative of the other but unlike the

1:17:54product rule we subtract instead of add

1:17:57then the right term is opposite

1:17:59derivative Wise from the left term frime

1:18:02becomes f and g becomes G Prime write it

1:18:05from scratch a few times and you'll know

1:18:07it let's find the derivative of a

1:18:10quotient 3x Cub - x^2 + 2 / cosine

1:18:16X we'll start with the denominator squar

1:18:19cine 2qu of X then for the numerator

1:18:22expression we start with the denominator

1:18:24again not squared cine X then multiply

1:18:28by the derivative of the numerator by

1:18:30the power rule the derivative of 3 x Cub

1:18:33- x^2 + 1 is 9 x^2 -

1:18:382x then we subtract the right expression

1:18:41which is the numerator 3x Cub - x^2 + 1

1:18:46* the derivative of the denominator the

1:18:49derivative of cine X is sinx that's

1:18:53pretty much it these negative signs undo

1:18:55each other and with some trig

1:18:57substitution you can get rid of the

1:18:58cosine ^ 2qu x in the denominator and

1:19:01that's the quotient

1:19:04rule speaking of trig now that we know

1:19:06the quotient rule we can find the

1:19:08derivative of the other four trig

1:19:10functions because they can all be

1:19:11expressed as fractions involving s and

1:19:14cosine I'm using the color coding from

1:19:17my trigonometry series just for this

1:19:19chart for the derivative of tangent

1:19:22Theta the quotient rule is the

1:19:23denominator time the derivative of the

1:19:25numerator minus the numerator * the

1:19:28derivative of the denominator all

1:19:30divided the denominator squared this

1:19:33simplifies to cosine 2 thet plus sin s

1:19:36Theta which is 1 over cosine s thet

1:19:40which is secant squ thet since secant is

1:19:431/

1:19:44cosine I'll show the derivation of the

1:19:46other trig functions using the quotient

1:19:48rule but won't step through the details

1:19:51you might need to know these check with

1:19:53your instructor if you know the quotient

1:19:55Rule and the circle trig identities you

1:19:57can figure these out as you need them

1:20:00practice builds confidence here are the

1:20:03derivatives of the six trig

1:20:08functions let's check our agenda we

1:20:11covered the four-part trig cycle for the

1:20:13derivatives of s and cosine we covered

1:20:17the chain rule to find the derivative of

1:20:19composite functions the role you're

1:20:21likely to use more than any other then

1:20:23we went over the quotient Rule and

1:20:25actually derived it from the product

1:20:27rule and the chain rule we use the

1:20:29quotient rule to show the derivatives of

1:20:31the remaining trig functions since

1:20:33they're all ratios that include s and

1:20:36cosine the last rules of differentiation

1:20:39are for exponentials and

1:20:41logarithms please note that all these

1:20:43Atomic function types can be combined

1:20:46and used in all of these rules that

1:20:48allow us to combine functions in various

1:20:51ways so the next rules are for

1:20:54exponential and

1:20:57logarithms these rules are some of the

1:21:00simplest well they've all been pretty

1:21:01simple right but there's a lot of

1:21:03background to review for it all to make

1:21:05sense exponentials and logarithms are

1:21:08usually covered in Algebra 2 or

1:21:10pre-calculus but I'll do a thorough

1:21:12review of the topics needed to

1:21:14understand the rules of

1:21:16differentiation exponential functions

1:21:18have their independent variable X up in

1:21:21the exponent the number on bottom is

1:21:23called the base I'm color coding the

1:21:25base green as a reminder that it's not a

1:21:28variable like X it's a constant such as

1:21:312 don't confuse the exponential function

1:21:342 ra the power of X with the polom or

1:21:38power function x raed to the power of

1:21:40two they're different functions with

1:21:43different graphs and different

1:21:44derivatives you can remember that

1:21:46exponential functions have their

1:21:48variable in the

1:21:49exponent and in short B to the X means

1:21:53multiply con base B by itself x

1:21:58* let's assume we have B raised to the

1:22:017th power as shown the associative

1:22:03property of multiplication says that we

1:22:06can group The B's together like this and

1:22:08get the same result so B 7th = B 3r * B

1:22:144th in general when their bases are the

1:22:17same we can multiply exponentials by

1:22:21adding their exponents let's graph some

1:22:24exponential

1:22:25when the base B is greater than one the

1:22:27exponential function value gets bigger

1:22:29and bigger as X increases the slope is

1:22:32always positive functions like these are

1:22:35used to model exponential

1:22:37growth when the base B is one the

1:22:40exponential function value Y is always

1:22:43one because one to any exponent is one

1:22:46because 1 times itself any number of

1:22:48times is always going to be

1:22:51one and when the Bas is between zero and

1:22:54one the function value gets smaller and

1:22:56smaller as X increases functions like

1:22:59this are used to model exponential

1:23:02decay an exponential function with base

1:23:05B will always be symmetrical across the

1:23:07y AIS to an exponential function whose

1:23:10base is the reciprocal of B like this

1:23:13example of 2 and/

1:23:1612 all the graphs of y equal sum base B

1:23:19to the X pass through the very busy

1:23:22Point 0a 1 because any base B raised to

1:23:26the zeroth power will always equal

1:23:29one here's an animation showing various

1:23:32exponential curves as the green base B

1:23:34changes when B is greater than one the

1:23:37curve is always increasing the higher

1:23:40the base B the faster the increase and

1:23:43as I mentioned every curve passes

1:23:44through the circled point 0 comma

1:23:471 when base B is 1 the curve flattens

1:23:51out because 1 raised to any power x will

1:23:54always

1:23:55B1 and when B is between 0 and 1 the

1:23:58curve is always decreasing lower base

1:24:01values B decrease

1:24:03faster so we have an exponential

1:24:05function y = b to the X where B is a

1:24:08constant and X is the independent

1:24:10variable so given b x and a calculator

1:24:14we can find

1:24:16y but what if we know Y and B and want

1:24:19to find X for any function when we find

1:24:23x given y instead of Y given X that's

1:24:26called taking the inverse of the

1:24:28function suppose we knew Y =

1:24:315.89 and wanted to find X how would we

1:24:34do it the inverse of the exponential

1:24:37function is the logarithm the logarithm

1:24:40answers the question what's the exponent

1:24:43we write and say the logarithm function

1:24:46like this x = log base 1.47 3 of

1:24:515.89 it means X is the expon onent on

1:24:551.47 3 that results in

1:24:595.89 these equations aren't solved by

1:25:01hand we use a calculator and before

1:25:04calculators slide rules let me show you

1:25:07this again emphasizing the inverse

1:25:09relationship to solve this equation for

1:25:12x we need to isolate X to get x equals

1:25:15something but X is in the exponent how

1:25:18do we get x out of the exponent how do

1:25:21we undo

1:25:23exponentiation by taking the logarithm

1:25:26we'll take the logarithm of both sides

1:25:28making sure that the bases match to undo

1:25:31an exponent of Base 1.47 3 we need to

1:25:35take the logarithm base 1.47 3 let's

1:25:39look at the right side of the equation

1:25:41remember the log function answers the

1:25:43question what's the exponent let's

1:25:46transliterate the right hand side what's

1:25:48the exponent on 1.47 3 that results in

1:25:531.4 473 to the X well the answer is X

1:25:58this is rather like asking what's half

1:26:00of twice X the half and the twice undo

1:26:03each other leaving X and the logarithm

1:26:05base 1.47 3 undo exponentiation base

1:26:101.47 3 so we get X on the right hand

1:26:13side which is exactly why we took the

1:26:15logarithm to isolate the exponent x to

1:26:19keep things even and balanced we need to

1:26:21take the logarithm of the left side too

1:26:24and we get log base 1.47 3 of 5.89 which

1:26:29our calculator will tell us is

1:26:334.58 since exponentials and logarithms

1:26:35are inverse functions of each other

1:26:37their graphs are symmetrical across the

1:26:39line yal

1:26:41X the exponential of Base B is a miror

1:26:44reflection of the logarithm base B

1:26:47across the dotted diagonal line Y = X

1:26:51all inverse function pairs share this

1:26:53characteristic not just exponentials and

1:26:56logarithms so naturally since all

1:26:59exponential graphs pass through the

1:27:00point 0 comma 1 because any base raised

1:27:04to the 0 power is 1 all logarithmic

1:27:07graphs pass through the point 1 comma 0

1:27:10because the exponent to any base that

1:27:12results in one is

1:27:17zero the associative property of

1:27:19multiplication tells us that b 7x can be

1:27:22expressed as B 3x * B 4X let's see what

1:27:27happens when we take the logarithm base

1:27:29B of both sides log base B of 7x is

1:27:34simply 7x like before the log base B and

1:27:38the exponent on B cancel out leaving

1:27:41just the exponent and log base B of

1:27:44these two terms are 3x and 4x

1:27:47respectively so the three terms we get

1:27:49after taking log base B are the three

1:27:52exponents of B 7x 3x and 4x and to write

1:27:57the resulting equation we need to

1:27:59combine these terms by adding not

1:28:01multiplying that shouldn't be surprising

1:28:04logarithms effectively bring exponents

1:28:06down and we already observed this

1:28:08property about multiplying

1:28:11exponentials let's go to an extreme and

1:28:13write B 7x as b x multiplied by itself

1:28:187even times now when we take the

1:28:20logarithm base B of both sides we get

1:28:23seven distinct in log base B of B to the

1:28:26X terms that means that log base B of B

1:28:297x is 7 log Bas B of B to the x or in

1:28:34general log base B of B to the NX is n

1:28:39logs Bas B of B to the X we can take the

1:28:42coefficient of x in the exponent and

1:28:45move it to the coefficient of the

1:28:47logarithm we're almost ready for the

1:28:49rules of differentiation but first

1:28:52another important property of

1:28:53exponential

1:28:54functions any exponential function can

1:28:57be expressed as an equivalent

1:28:59exponential function with any other

1:29:01base so here's the graph of y = 2 to the

1:29:05X again we can get the exact same graph

1:29:07from an exponential equation that has

1:29:09another base such as 3 so Y = 2 x can be

1:29:14expressed as y = 3 raised to the

1:29:17something let's find the something by

1:29:20setting the Expressions equal to each

1:29:22other 3 raised to the Something = 2

1:29:25raised to X let's take the logarithm of

1:29:28both sides to isolate the red something

1:29:31variable we need to be careful which

1:29:33base to use for the logarithm we want to

1:29:36isolate the red something so we'll take

1:29:38the log base 3 of both sides since three

1:29:41is the base whose exponent we want to

1:29:43isolate the left side simplifies to our

1:29:46red variable on the right side we take

1:29:48the exponent out and give us X logs base

1:29:513 of 2 and that's the answer 2 the x is

1:29:55the same function as 3 raised to the log

1:29:58base 3 of 2 * X and the calculator will

1:30:01tell us that log base 3 of 2 is about

1:30:040.63

1:30:09093 here's the pattern for switching

1:30:12bases the old base raised to the X power

1:30:15equals the new base raised to the power

1:30:18of log base new base of old base * X so

1:30:23these are the same functions and the

1:30:25point is it's not the base that

1:30:27determines the shape of the exponential

1:30:29function but a combination of the base

1:30:32and whatever coefficient the independent

1:30:34variable has in the

1:30:36exponent the same curve can be described

1:30:39by lots of exponential functions having

1:30:41whatever base you choose however there's

1:30:45a very special exponential base his

1:30:47value is about

1:30:502.718 it's so special that it has its

1:30:53own symbol lowercase e I'll use green as

1:30:56a reminder that e is a constant like two

1:30:59or three not a

1:31:01variable it's kind of a surprise that

1:31:03the constant e pops up in some simple

1:31:08formulas it's a constant of nature like

1:31:11pi and I'll show you in a moment why e

1:31:14is so useful in calculus and why it's

1:31:16called the natural base here's a graph

1:31:19of the exponential function y = 2 X and

1:31:24here's y = 3 x since e is between 2 and

1:31:283 it shouldn't be too surprising that y

1:31:31= e to the x is between them it's a very

1:31:34special Base number but its curve looks

1:31:37just like any other exponential

1:31:40curve if we have a function y = e to X

1:31:44we can find y given X like any other

1:31:47function and we can invert it to express

1:31:49X in terms of Y using the

1:31:52logarithm if y = e to X then X = log

1:31:56base e of Y well the logarithm base e is

1:32:01also special and it has a special symbol

1:32:04and name the natural logarithm or

1:32:07natural log and for its symbol instead

1:32:09of writing L base e we write Ln I know

1:32:14that seems backwards but it's from the

1:32:16Latin Ln means natural

1:32:19log so log base e of Y is equivalent to

1:32:22this expression which can be pronounced

1:32:25as natural logarithm of Y natural log of

1:32:29Y Ln of Y or even Ln y so once more Ln

1:32:35is a mathematical shorthand for log base

1:32:38e the natural log also pops up

1:32:41surprisingly in some simple

1:32:44formulas we'll see this limit again in a

1:32:47moment when printed Ln can look like one

1:32:51n so when handwritten you'll often C Ln

1:32:54written in script or cursive with a

1:32:56loopy L like this here are some examples

1:33:00I found online it's not a big deal I

1:33:02just don't want you to be confused when

1:33:04you see the style and I suggest you use

1:33:07it just write Ln in cursive like it was

1:33:10a

1:33:12word now we're ready for the

1:33:14differentiation rules for

1:33:17exponentials as usual we'll find the

1:33:19slope at a red Point by finding the

1:33:21slope between the red point and a nearby

1:33:23Green Point whose x coordinate is x + H

1:33:28then we'll take the limit as H

1:33:29approaches zero and see what we

1:33:32get our function is 2 X so we plug that

1:33:36into our limit equation note that we

1:33:38have 2 raised to x + H

1:33:41power we can rewrite this as 2 x * 2 H

1:33:47remember now we have 2 to the X twice in

1:33:50the numerator which we can factor out to

1:33:53get 2 x * 2 H -1 all over H remember

1:33:59we're taking the limit as H approaches 0

1:34:02and 2 to the X won't change as H changes

1:34:05because there's no h in it so we can

1:34:07pull it out of the limit now I told you

1:34:10earlier that this limit is the natural

1:34:12log of this number like

1:34:17this and so the derivative of 2 x is 2 x

1:34:23* the natural logarithm of 2 and in

1:34:26general the derivative of B to the x is

1:34:29the natural log of B * B to the X and

1:34:33that's a differentiation rule for

1:34:35exponents we'll make it stronger in a

1:34:38moment but it's good to know that the

1:34:39derivative of B to the x is the natural

1:34:42log of B times the original exponential

1:34:45function B to the

1:34:47X quick what's the derivative of 7 to

1:34:50the x

1:34:53it's the natural logarithm of the base 7

1:34:57times the original exponential function

1:34:597 to the X

1:35:02easy now what if the base were the

1:35:04natural base e same thing the derivative

1:35:08of e to the x is the natural log of e

1:35:11times the original exponential function

1:35:13e to the X well what's the natural log

1:35:17of e Ln e means the exponent on base e

1:35:22remember the base of the natural

1:35:23logarithm Ln is always e that results in

1:35:27E so Ln e is one because E rais power of

1:35:32one is e this is not a special rule for

1:35:35E any log base B of B that is the

1:35:39logarithm of any number to its own base

1:35:42is one because B raised to the power of

1:35:44one is

1:35:46B so log base e of e is one we just have

1:35:51a special symbol for log base e Ln so Ln

1:35:55E equals 1 we substitute the natural log

1:35:59of e which is one into our derivative

1:36:01and simplify to get the derivative of e

1:36:04to the x is e to the X the only function

1:36:08that's its own derivative pretty neat

1:36:11and that's why e is such a special

1:36:13exponential

1:36:15base now on the screen are two equations

1:36:18or rules for derivatives of exponentials

1:36:21the bottom is just a special case of the

1:36:23top for base e since Ln e is one but

1:36:27there's one more variation to consider

1:36:29and then we'll have a single robust rule

1:36:32that will help us find the derivative of

1:36:34all exponential

1:36:36functions often the exponent will not

1:36:38simply be X but some function of X this

1:36:41is very common in real world

1:36:43applications of exponentials we can't

1:36:46use the highlighted rule above it

1:36:48applies only when the exponent matches

1:36:50the independent variable for B raised to

1:36:53a function of X they don't match so the

1:36:55first rule won't work we need to use the

1:36:58chain rule because we have a function f

1:37:00ofx embedded within an exponential since

1:37:04we just covered the chain Ru I hope

1:37:05you'll excuse me if I jump straight to

1:37:07the conclusion here and say that we

1:37:09account for the daisy chained functions

1:37:11by multiplying by the derivative of the

1:37:14exponent so the derivative with respect

1:37:16to X of B raised to some function of X

1:37:19has three factors the derivative of the

1:37:22exponent

1:37:23times the natural logarithm of the base

1:37:26B times the original exponential

1:37:29function and this is the differentiation

1:37:32rule for exponentials to know because it

1:37:34will work for any base e or otherwise

1:37:38and for any exponent x or some function

1:37:41of X let me build a chart to show this

1:37:43is true we'll put the general

1:37:46exponential function in this cell it

1:37:48corresponds to any base meaning not

1:37:51necessarily the natural base e and a

1:37:54function of X in the exponent as opposed

1:37:56to an exponent of Simply X as you'll see

1:38:00the other three cells are simpler cases

1:38:02of this

1:38:04one in the cell above the base isn't

1:38:07necessarily e and there's no function in

1:38:09the exponent meaning the exponent is

1:38:12simply X this cell represents the

1:38:14special case where the Bas is e and the

1:38:17exponent is a function of

1:38:19X and finally this cell represents the

1:38:22special case where the base is e and the

1:38:25exponent is simply

1:38:27X we're going to find the derivative of

1:38:30each of these exponentials using the

1:38:33differentiation rule for exponents the

1:38:36three Factor derivative from a moment

1:38:38ago I'll step through the rule for each

1:38:40cell and you'll see why we only need one

1:38:43rule we'll start here with the most

1:38:45general form the derivative of the

1:38:48exponential is always the derivative of

1:38:50the

1:38:51exponent times the natural log of the

1:38:54base times the original exponential

1:38:57function and that's the three Factor

1:38:59solution that will always work let's

1:39:02apply the same rule to y = b to X it's

1:39:06essentially the same function except the

1:39:08exponent is X instead of f ofx no

1:39:11problem we just write the three factors

1:39:13one at a time we start with the

1:39:16derivative of the exponent well the

1:39:18derivative of x is one so we can ignore

1:39:21the first Factor the second factor is

1:39:24the natural log of the base B and the

1:39:27last factor is simply the original

1:39:29exponential function B to the

1:39:33X so the derivative of B to the x is lnb

1:39:37* B to X we used the same three-part

1:39:40rule but the first Factor went to one

1:39:42since the derivative of the exponent x

1:39:45is one now let's find the derivative of

1:39:48e raised to the F ofx the derivative of

1:39:51the exponent

1:39:53times the natural log of the base since

1:39:56the base is e Ln e is 1 so the second

1:39:59factor is ignored and the third factor

1:40:02is again the original exponential

1:40:04function so the derivative of e to the f

1:40:07ofx is frime of X time e to the F ofx we

1:40:12Ed the same three-part rule but the

1:40:14second Factor went to one since L and E

1:40:16is

1:40:181 now let's find the derivative of e to

1:40:21the X using the same three-part Factor

1:40:24the derivative of the exponent the

1:40:26exponent is X and the derivative of x is

1:40:291 so the first Factor goes to one the

1:40:32second factor is the natural log of the

1:40:34base the base is e and Ln e is one so

1:40:38the second Factor also goes to one the

1:40:41third factor is the original exponential

1:40:43function e to the X so the derivative of

1:40:46e to the x is e to the X the only

1:40:50function that's its own derivative we

1:40:52use the same same three-part rule but

1:40:54the first Factor went to one since the

1:40:56derivative of the exponent x is one and

1:41:00the second Factor also went to one since

1:41:02Ln e is one and that just left the third

1:41:05factor which is the original exponential

1:41:08function the point of this chart is that

1:41:10exponential functions come in several

1:41:12varieties but you don't need to know

1:41:14several rules just this one it will give

1:41:18you the correct derivative for all

1:41:20exponentials as long as you know that

1:41:22the derivative with respect the x is one

1:41:25and that L and E is one one rule to ring

1:41:29them

1:41:33all next is logarithms the rules for

1:41:37logarithms can also be derived with the

1:41:39chain rule but again for the sake of

1:41:41expediency please excuse me if I jump to

1:41:44the rules like we did for exponentials

1:41:46we'll use the most general form of the

1:41:48logarithm function having any base B not

1:41:52necessarily e

1:41:53and the argument can be some function of

1:41:55X not necessarily plain

1:41:58X the general differentiation rule for

1:42:01logarithms also has three factors that

1:42:03kind of correlate to the factors for

1:42:05exponentials the first factor is the

1:42:08derivative of the function that's the

1:42:10same the second factor is one over the

1:42:13natural log of the base this is the

1:42:16reciprocal of the factor for

1:42:17exponentials and the third factor is one

1:42:20over the argument function f ofx the

1:42:23third Factor isn't really that similar

1:42:25to the third factor in the exponential

1:42:27rule but with practice you'll get it the

1:42:30fractions are often combined so you

1:42:32might see the rule like this I'll keep

1:42:34it as three separate factors in the

1:42:36logarithm chart so you can see each

1:42:38factor

1:42:39clearly we'll start with the general

1:42:41cell again in the lower leftand corner

1:42:44and apply the new three Factor rule for

1:42:46logarithms the derivative of the

1:42:48logarithms argument frime of

1:42:51x * 1 over the natural log of the

1:42:55base * 1 over the argument like before

1:42:59we'll use this rule as our pattern it'll

1:43:02work for all the logarithm

1:43:04Expressions let's apply it to Y = log

1:43:07base B of X here the logarithms argument

1:43:10is X not a function of X the first

1:43:13factor is the derivative of the argument

1:43:15the derivative of x is one so we can

1:43:18ignore this

1:43:19Factor the second factor is one over the

1:43:22natural log of the base 1 / L and

1:43:26B and the third factor is the reciprocal

1:43:29of the argument 1

1:43:31/x so the derivative with respect to X

1:43:34of log base B of X is 1 / natural log of

1:43:38B *

1:43:39X let's apply the rule to this cell and

1:43:42find the derivative of the natural log

1:43:44of some function of X the first factor

1:43:47is the derivative of the argument frime

1:43:50of

1:43:50X the second factor is 1 over the

1:43:53natural log of the base the base of Ln

1:43:56the natural log is e and 1 / Ln e is 1

1:44:01so the second Factor goes to

1:44:03one and the third factor is one over the

1:44:07argument so the derivative of the

1:44:09natural log of some function of X is the

1:44:12derivative of the function divided by

1:44:14the

1:44:16function now let's apply the rule to the

1:44:18derivative of Ln X the first factor is

1:44:22the derivative of the argument X so it's

1:44:24one and can be

1:44:26ignored the second factor is one over

1:44:28the natural log of the base the natural

1:44:31log of Base e is one so the second

1:44:33Factor can also be ignored this leaves

1:44:36the third Factor one over the argument

1:44:39so the derivative of Ln X is 1

1:44:43/x so once again a single rule for the

1:44:46most General logarithm will work for any

1:44:49of these special cases and these are the

1:44:52differentiation rules for exponentials

1:44:54and logarithms that you should

1:44:58know and so finally we've covered all

1:45:01these rules of differentiation it's a

1:45:04lot of material it's pretty much the

1:45:06entire first semester of calculus

1:45:08remember calculus is all about

1:45:10performing two operations on functions

1:45:13and we've just covered the first

1:45:14operation

1:45:16differentiation when we differentiate a

1:45:18function the result is the function's

1:45:21derivative the other operation is called

1:45:23integration when we integrate a function

1:45:26the result is the function's

1:45:28integral as a preface to start learning

1:45:30about integration I need to introduce

1:45:32the

1:45:37anti-derivative suppose we're given a

1:45:39function and told that it's a derivative

1:45:41fime of x what then is f ofx in other

1:45:45words what's the function whose

1:45:47derivative is fime of x if frime of X is

1:45:52the derivative of f ofx then f ofx is

1:45:55the anti-derivative of frime of X the

1:45:58second calculus operation integration

1:46:01depends on being able to find

1:46:03anti-derivatives in fact the integral is

1:46:06the

1:46:07anti-derivative this is the important

1:46:09topic for the second half of calculus

1:46:11because differentiation and integration

1:46:14are almost exact opposites of each other

1:46:18but let's start simply suppose fime of X

1:46:21= x^2 how can we find out what f ofx is

1:46:25just to be clear we're not trying to

1:46:27find the derivative of X2 the power rule

1:46:30tells us the derivative of x^2 is 2x

1:46:33easy instead we need to use the power

1:46:36rule backwards we're looking for the

1:46:38function whose derivative is x^2 or the

1:46:41anti-derivative of

1:46:43x^2 as a reminder here's the power rule

1:46:46shortcut for when X the N has a

1:46:48coefficient K the derivative with

1:46:50respect to X of K * X the N is KNN * X

1:46:56nus1 remember we bring the exponent

1:46:58downstairs multiply by any coefficient

1:47:01that's already there and then reduce the

1:47:03exponent by one so imagine that we did

1:47:07this to some function f ofx and the

1:47:09result was frime of X =

1:47:12x^2 what function f ofx did we start

1:47:16with let's muscle through the power rule

1:47:19backwards to take the derivative of a

1:47:21power weed reduce the exponent by one so

1:47:24to take the anti-derivative of a power

1:47:27we'll need to increase it by one so we

1:47:29have something X cubed let's use our

1:47:32imaginations for a moment and see what

1:47:34happens when the green question mark is

1:47:36one that would make F ofx = to X cub and

1:47:40fime of X the derivative of x cubed = to

1:47:453x^2 hm 3x^2 is 3 times larger than

1:47:49frime of X we're targeting so we need a

1:47:52factor factor that will reduce it by 1/3

1:47:55so the green coefficient must be 1/3

1:47:58it's always easy to check

1:47:59anti-derivatives just take its

1:48:01derivative and see if you get the

1:48:02function you started with in this case

1:48:05the derivative of 1/3 x cubed does

1:48:08indeed equal x^2 so the anti-derivative

1:48:11of x^2 is 1/3 x

1:48:14cub in general the anti-derivative of k

1:48:17x n is x n + 1 * the coefficient k / n

1:48:24+1 you may see this written as k x n + 1

1:48:29all over n +1 same thing now there's

1:48:33just one glaring challenge staring right

1:48:35at us this formula won't work when the

1:48:38denominator of the fraction is zero let

1:48:41me build a chart that shows the

1:48:43anti-derivatives of simple powers of x

1:48:45to highlight the pattern and challenge

1:48:48let's start with the derivative of x

1:48:50cubed to find the anti-derivative we add

1:48:53one to the exponent and then divide by

1:48:55that new number 1/4 x 4th in a few

1:48:59moments we're going to make a small

1:49:00adjustment to this expression and the

1:49:02others on this chart so the

1:49:04anti-derivative of X cubed isn't exactly

1:49:071/4 x to 4th but you still need to know

1:49:10this anti-derivative formula for Powers

1:49:13if the derivative is x^2 then the

1:49:15anti-derivative must be 1/3 x cubed plus

1:49:19the adjustment we'll cover it soon I

1:49:21won't mention it again until then this

1:49:24anti-derivative 1/3 x cubed was the

1:49:27example we mused through to get our

1:49:28anti-derivative formula next is the

1:49:31derivative of x to the first Power which

1:49:33is just X following the formula the

1:49:37anti-derivative of X must be 12 x^2 so

1:49:41far so good here the derivative is X to

1:49:44the 0 which is 1 the anti-derivative of

1:49:471 is X since the derivative of x is 1 of

1:49:52course using the formula yields X since

1:49:541 1 * x 1 is simply

1:49:57X now when the derivative is X the ne 1

1:50:01then this is the same as 1 /x and here's

1:50:04where the formula breaks down because it

1:50:06results in a denominator of zero so this

1:50:09is an undefined expression we'll come

1:50:12back to it but let me add a few more to

1:50:13the chart to show that the

1:50:14anti-derivative formula for Powers works

1:50:17for every other power positive or

1:50:19negative it also works for fractional

1:50:21Powers but I don't show any on the chart

1:50:24it works for everything except when the

1:50:26exponent is -1 which corresponds to 1 /x

1:50:29we cannot use this formula to find the

1:50:31anti-derivative of 1

1:50:33/x but 1 /x does have an anti-derivative

1:50:37what is it you might remember this chart

1:50:40where we covered the differentiation

1:50:42rules for logarithms here we showed that

1:50:45the derivative of the natural log of x

1:50:47was 1 /x that must mean that the

1:50:50anti-derivative of 1 /x is the natural

1:50:53log of x almost this one needs another

1:50:56slight adjustment in addition to this

1:50:58slight adjustment but we'll cover the

1:51:01special natural logarithm adjustment

1:51:03right now here's the graph of frime of X

1:51:06as x^ -1 or 1/x I graphed it in pink

1:51:11because it's the slope the slope of what

1:51:14it's anti-derivative which is the

1:51:15natural log of x shown in the white

1:51:18curve here's where the challenge arises

1:51:21the natural log function is the defined

1:51:23only for positive X values the right

1:51:25half of the graph but 1 /x is defined

1:51:28for positive and negative values but not

1:51:31zero so we need a white function curve

1:51:34on the left side of the graph that

1:51:36answers the anti-derivative question

1:51:38what function has this

1:51:40derivative let's note that the pink

1:51:42curve 1 /x is symmetrical across the

1:51:45origin and so can be spun around 180°

1:51:48without changing this means that every

1:51:51pink Point has a twin across the origin

1:51:54with opposite coordinates and so the

1:51:56white function will have reflected

1:51:57points across the y AIS with the same

1:52:00slope thus our white function the

1:52:02anti-derivative of 1 /x is the natural

1:52:05log of the absolute value of x this

1:52:09function is the anti-derivative that

1:52:10works perfectly for both positive and

1:52:13negative X values so here's the general

1:52:16anti-derivative rule for Powers the top

1:52:19rule the bump up the exponent rule works

1:52:22all the time except when n the exponent

1:52:25on X is -1 when n is -1 then that's the

1:52:29anti-derivative of K overx which is K *

1:52:33the natural log of the absolute value of

1:52:37x we still have this adjustment I

1:52:39promised so let me explain what it is

1:52:41and why it's needed I'll illustrate by

1:52:44taking the anti-derivative of a polom

1:52:47we're starting with pols because they're

1:52:49easy and straightforward we'll cover the

1:52:51anti-derivative of other function types

1:52:54later the additional rule for

1:52:56differentiation says that the derivative

1:52:58of the sum is equal to the sum of the

1:53:00derivatives and the same rule applies to

1:53:02anti-derivatives the anti-derivative of

1:53:05the sum is equal to the sum of the

1:53:07anti-derivatives so we can find the

1:53:09anti-derivative of this function term by

1:53:12term we'll go through these quickly the

1:53:14anti-derivative of 8X cued is something

1:53:17X 4th coefficient 8 / 4 is 2 so 2 x x 4

1:53:23it's always easy to check by going

1:53:25backwards the derivative of 2x 4 is

1:53:28equal to 8 x Cub so we can be confident

1:53:31that the first term is right next term

1:53:34the anti-derivative of 3x^2 is something

1:53:37X cubed coefficient 3 over the new

1:53:40exponent 3 is 1 so the second term is

1:53:43plus X cubed the term minus 6X has

1:53:47anti-derivative something

1:53:49x^2 -6 / the new exponent 2 is -3 so the

1:53:54third term of the anti-derivative is

1:53:57minus

1:53:583x^2 and the last term is 1 which we can

1:54:01think of as 1 x to the 0 so the exponent

1:54:04of the anti-derivative is 1 and the

1:54:06coefficient is 1 over 1 or 1 so just

1:54:09plain X and this checks out the

1:54:11derivative with respect to X of X is 1

1:54:15well it looks like we're done we know

1:54:17that frime of X is the derivative of F

1:54:19ofx and that f ofx is the

1:54:21anti-derivative of frime of X here's the

1:54:25anti-derivative we found and we can

1:54:27validate by taking the derivative term

1:54:29by

1:54:32term perfect we nailed it but now let's

1:54:35consider the same function f ofx except

1:54:38it has + one all the derivative terms

1:54:41are the

1:54:43same and we add the derivative of one

1:54:46but the derivative of one is zero by the

1:54:48constant rule so the derivative of this

1:54:51new f ofx is the same as the previous

1:54:54one and it's the same if we add two or

1:54:58subtract one or add or subtract any

1:55:01constant all of these functions have the

1:55:04same derivative the derivative we

1:55:06started with 8X Cub + 3x^2 - 6X +

1:55:111 here's what's going on here's a graph

1:55:15of the anti-derivative we found 2 x 4 +

1:55:19x Cub - 3x^2 + x X and here's the graph

1:55:23of the derivative we started with 8 x

1:55:26Cub + 3x^2 - 6X + 1 remember we started

1:55:32with this pink derivative and we backed

1:55:33into the white equation by taking the

1:55:36anti-derivative the pink derivative

1:55:38curve looks reasonable it says the

1:55:40derivative is zero at these three points

1:55:43where the white slope is

1:55:44zero now let's look at the next equation

1:55:47up the one that ends with + one note

1:55:51that this simply translates the white

1:55:53curve straight up by one unit the slopes

1:55:56don't change everything is just shifted

1:55:58up this makes sense both equations yield

1:56:02the same pink derivative same with the

1:56:04plus two

1:56:05version and the

1:56:07minus1 we can add any positive or

1:56:10negative constant to a function without

1:56:12changing its derivative because adding a

1:56:15constant just moves the entire curve up

1:56:17or down without changing its slope

1:56:20anywhere thus each member of this family

1:56:23of functions has the same

1:56:26derivative so when we find the

1:56:28anti-derivative of the right function

1:56:30which one do we choose well we designate

1:56:33the whole family by adding plus C where

1:56:37Capital C represents any constant and is

1:56:40called the constant of integration this

1:56:42is true for all anti-derivatives not

1:56:45just the anti-derivative of powers and

1:56:47polinomial every function has exactly

1:56:50one derivative but every function has an

1:56:52infinite number of anti-derivatives

1:56:54because they can be shifted up or down

1:56:57by constant C and still have the same

1:57:00derivative so we specify the whole

1:57:02family of functions by including the

1:57:04plus C constant of integration when we

1:57:06find anti-derivatives so the adjustment

1:57:10I promised is the constant of

1:57:11integration plus C it should always be

1:57:14included in anti-derivative functions

1:57:17and so our general anti-derivative rule

1:57:19for Powers needs to be updated to

1:57:21include the constant of

1:57:26integration regarding anti-derivative

1:57:28notation when we're given a function

1:57:30called fime of X it seems pretty clear

1:57:33that its anti-derivative would simply be

1:57:35the function name f without the prime

1:57:38indicator since taking the derivative

1:57:41adds a prime symbol taking the

1:57:43anti-derivative reasonably it seems

1:57:45removes a prime

1:57:47symbol but when a function doesn't have

1:57:49a prime symbol how do we denote its

1:57:51anti- derivative well the convention is

1:57:54that we capitalize the function letter

1:57:57so capital F ofx is the anti-derivative

1:58:00of lowercase f ofx and the derivative of

1:58:03capital F ofx is lowercase f

1:58:07ofx when we intend to denote the

1:58:09derivative of a function f ofx we're

1:58:12already familiar with expressing this as

1:58:14frime of X and as d by DX of f

1:58:18ofx for the anti-derivative of f ofx we

1:58:21can Express this as uppercase F ofx as

1:58:24already mentioned or we can use this new

1:58:27calculus

1:58:31notation this is the integral symbol

1:58:34it's not a Greek letter it's an

1:58:35elongated s for some we'll get into the

1:58:38details in a few minutes the function

1:58:41whose anti-derivative we're finding is

1:58:43called the integrant when I introduce

1:58:46the anti-derivative concept I called

1:58:48this function frime of x to emphasize

1:58:50that it's already the Der ative and that

1:58:52we are going backwards from

1:58:54it when you see an integral expression

1:58:57the integrant won't have a prime symbol

1:58:59as a reminder the integral symbol means

1:59:02find the anti-derivative of the

1:59:04integrand and the differential DX

1:59:07denotes the variable of integration this

1:59:10expression has an intuitive

1:59:11interpretation that we'll be ready for

1:59:13in just a

1:59:14moment here's our sample function and we

1:59:16were asked what function has this

1:59:18derivative we applied the power rule for

1:59:21anti-derivative term by term and got

1:59:23this function the anti-derivative we

1:59:26express the anti-derivative operation

1:59:28with the integral symbol like

1:59:34this the

1:59:35integral of a function yields its

1:59:40anti-derivative remember to include the

1:59:42constant of

1:59:45integration so the second calculus

1:59:47operation integration is finding the

1:59:50anti-derivative but it's so much more

1:59:52more than that and has interesting

1:59:53applications that make it suitable for

1:59:55solving lots of real world

1:59:58problems I want to show you another more

2:00:01insightful interpretation of integration

2:00:04but I need to start in kind of a strange

2:00:06way I'll draw a coordinate system where

2:00:09the horizontal axis denotes time in

2:00:11seconds and the vertical axis denotes

2:00:14velocity in meters per

2:00:16second suppose we have a particle