Calculus Visualized - by Dennis F Davis

https://youtu.be/MO-AExWdl4Q?si=LWeo1v8S2hc_K-UQ

1,270,390 views Jun 20, 2024

This 3-hour video covers most concepts in the first two semesters of calculus, primarily Differentiation and Integration. The visual and animation style is intended to be informative and provide insights into the logical reasoning behind the rules and formulas in calculus. Timestamps 0:00 Can you learn calculus in 3 hours? 2:23 Calculus is all about performing two operations on functions 3:32 Rate of change as slope of a straight line 5:37 The dilemma of the slope of a curvy line 7:04 The slope between very close points 11:30 The limit 15:36 The derivative (and differentials of x and y) 21:48 Differential notation 27:58 The constant rule of differentiation 29:15 The power rule of differentiation 35:36 Visual interpretation of the power rule 39:19 The addition (and subtraction) rule of differentiation 41:34 The product rule of differentiation 43:08 Combining rules of differentiation to find the derivative of a polynomial 44:53 Differentiation super-shortcuts for polynomials 48:03 Solving optimization problems with derivatives 52:46 The second derivative 59:59 Trig rules of differentiation (for sine and cosine) 1:05:54 Knowledge test: product rule example 1:06:32 The chain rule for differentiation (composite functions) 1:15:23 The quotient rule for differentiation 1:19:04 The derivative of the other trig functions (tan, cot, sec, cos) 1:20:55 Algebra overview: exponentials and logarithms 1:33:12 Differentiation rules for exponents 1:41:33 Differentiation rules for logarithms 1:45:32 The anti-derivative (aka integral) 1:46:21 The power rule for integration 1:48:32 The power rule for integration won't work for 1/x 1:52:37 The constant of integration +C 1:57:26 Anti-derivative notation 1:59:59 The integral as the area under a curve (using the limit) 2:12:05 Evaluating definite integrals 2:13:11 Definite and indefinite integrals (comparison) 2:15:30 The definite integral and signed area 2:18:10 The Fundamental Theorem of Calculus visualized 2:21:37 The integral as a running total of its derivative 2:24:25 The trig rule for integration (sine and cosine) 2:27:07 Definite integral example problem 2:32:53 u-Substitution 2:42:05 Integration by parts 2:56:03 The DI method for using integration by parts

0:03 · can you learn calculus in the time it takes to watch a long movie yes you can my name is Dennis Davis I'm an engineer not a mathematician I try to make my videos visually enlightening and fast-paced but this video is very long

0:19 · because it covers almost the entire first year of calculus using visuals graphs and diagrams I'd rather show you calculus visual then just tell you the rules and formulas if you see where the rules and formulas come from that can help you understand remember and make use of them

0:39 · to become truly proficient at calculus you'll need practice which you won't get just by watching this video in the description I'll link to some helpful practice oriented videos by others whether you're new to calculus studying it now or just want a refresher I hope you'll find this video informative and engaging my only assumptions are that you're already familiar with functions and algebra so here we go calculus is the study of change and

1:09 · rates of change of mathematical functions when we use calculus we perform operations on functions that result in different functions this isn't a new idea at all we perform operations such as addition on numbers to get new numbers we operate on sets to get new

1:30 · sets on matrices to get new matrices and we can operate on functions to get new functions in fact you've probably already done this in algebra when you had a function and found its inverse we start with a function f ofx and write out F in terms of X and Y so that y equals some function of X then we rewrite the equation switching X and Y

1:58 · then solve the second equation for y and that's the inverse of the first function so we start with a function perform an operation on it and get a new function the point is taking an existing function and Performing some operation on it to get a new function isn't a new or advanced math concept and that's really what calculus is in fact calculus is all about performing two operations on functions that's it that's

2:30 · calculus the first operation is called differentiation when we differentiate a function the new resulting function is called the derivative of the first function the derivative is the first topic we'll cover learning to take the derivative of a wide variety of function types is roughly the first semester of

2:51 · calculus the second calculus operation is called integration when we integrate a function the new resulting function is called the integral of the first function integration is roughly the second semester of calculus and there's a wonderful relationship between these two operations that we'll get to at just the right time when functions are simple these operations are simple and calculus is simple it's when the functions we operate on get complex that calculus

3:22 · seems complex so if calculus has a reputation for being a difficult subject it's not really calculus fault we'll start with the derivative and consider a simple function f ofx = x -1 this is a linear function so its graph is a straight line I said calculus was the study of change and rates of change and the rate of change of a straight line is its slope rise over run

3:51 · deltay over Delta X I like to color code things in my videos I'm using blue for x coordinates and distances and yellow for y and I'll use pink for slope or rates of change I won't draw a lot of attention to the colors and may not mention them again but their consistent use may let you see some apparent order or pattern that might not otherwise be clear for a straight line it's easy to

4:18 · find the slope just pick any two points and subtract their y and x coordinates to get Delta y over Delta X Delta Y is 3 0 this distance the difference between the yellow y-coordinates and Delta X is 4 min-1 this distance the difference between the blue x coordinates so for this linear function the slope is 3 over 3 or POS 1 for every unit X changes y

4:48 · changes by one times that amount since the slope is one if the slope were positive 1/2 then when X changes by some amount y changes by positive 1/2 as much if the slope were -2 then when X changes y changes by

5:05 · -2 times as much that is two times as much but in the opposite direction y would get smaller as X gets larger when the graph line is horizontal the slope is zero because y never changes so Delta Y is zero when the

5:23 · graph line is vertical the slope is undefined because X never changes Delta X is zero so Delta y over Delta X is undefined since the denominator is zero everything's nice and simple when the function is linear but what about the function FX = x^2 - 2x + 1 the graph

5:45 · of this function is a parabola so now the question what's the slope prompts a new question in response where do you mean because the slope is smoothly changing it's different at different points

6:04 · suppose we want to define the slope of the function at xals 1.6 here now if you know calculus or more specifically the rules of differentiation that we'll cover soon then within a few seconds you can figure out that the slope at x = 1.6

6:20 · is 1.2 if you don't know calculus yet I promise that soon you'll know how to do this but without calculus it's not easy f after all slope is rise over run deltay over Delta X but with just one point where does Delta Y come from or Delta X slope is the rate of change and

6:40 · there's no change at a single point there's only a change between two points so it's a tricky question we could eyeball it by drawing our best attempt at a line tangent to the curve at x = 1.6 then measure the slope it won't be

6:56 · very accurate it's not always easy to draw an accurate tangent line but without calculus that might be an option but let me show another way in the 1600s the great minds that developed calculus approach the problem this way I'll zoom in on our grid at the red point of Interest where x equals 1.6 we want to know the slope at the red point I warnant you I like to color code things let's choose a nearby Point green

7:25 · and find the slope deltay over Delta X between red red and green so we estimate the slope at red as the slope of the pink line between red and green Delta y over Delta X it'll be pretty close and the closer green is to Red the better the estimate this is the approach that will lead us to the derivative the x value of our red point where we want to know the slope is 1.6 we find the Y value by plugging 1.6 into the function and we get 0.

8:00 · 36 let's choose a nearby x value for the Green Point let's say 1.61 so we find the yvalue for the nearby Green Point by plugging 1.61 into our function to get 361 1201 remember here's our function y = x^2 - 2x + 1 we can easily find the slope deltay over Delta X between two points given their coordinates

8:34 · and we get 1.21 the slope between red and green which is our estimate for the slope at Red calculus will tell us I promise that the slope at Red is 1.2 so our estimate is very close the closer we let green get to Red the closer the estimate will get to 1.2 so let's prove it we'll still choose

8:58 · a nearby Green Point point but now instead of choosing an actual small distance away from Red such as 0.1 or .001 we'll use a variable for the tiny X distance to green and call it h the coordinates of the red Point are still 1.6 comma 0.36 but let's consider the coordinates of the Green Point its x coordinate is 1.6 + H and its y-coordinate is the function's value when we plug in 1.6 + H

9:32 · for X now let's find the slope as Delta y over Delta X Delta Y is the y-coordinate of the Green Point F of 1.6 + H minus the y-coordinate of the red point which is still 0.36 I'm shading the components to the color of the corresponding point and Delta X is the x coordinate at Green 1.6 + H minus the x coordinate at red

10:00 · 1.6 the denominator Delta X looks pretty easy to simplify 1.6 + H - 1.6 is simply H this makes sense we chose H over here to be Delta X so to find Delta y over Delta X we'll need to expand this expression by taking our function f ofx and rewriting it substituting 1.6 + H in

10:25 · for X this is just algebra so I'm showing it quickly pause if you want to step through the details we end up with this expression for the yalue of the nearby Green Point so we can now find deltay over Delta x 36 + 1.2 H + h^ 2

10:50 · -36 all over H the +36 and minus. 36 cancel leaving us with 1.2 H + h^2 over H as long as H is not zero which it's not it's some tiny tiny change but not zero then we can cancel an H from each term and the slope Delta y over Delta X is 1.2 + H

11:18 · let's remember that we said H was an arbitrarily small number we're going to find the limit of Delta y over Delta X as H approaches zero you may be familiar with the concept of the limit from studying discontinuous functions in algebra we write the limit like this l i m and underneath a variable right arrow and a literal value we read this as the limit as H approaches zero in our case the limit of 1.2 + H as H approaches

11:50 · zero as H gets closer and closer to zero the expression 1.2 + H gets closer and closer to 1.2 so the limit is 1.2 and that's deltay over Delta X the slope at the red point or at least the slope as Delta X we called it h approaches

12:13 · zero and now we can continue our dialogue what's the slope where do you mean at x = 1.6 the slope there is 1.2 that we found by taking the slope between two very close points at the limit as the difference between their x coordinates approach zero I want to review this equation again which is how we estimate the slope of a function at a point using algebra

12:41 · you'll see this important expression in the first chapter of every single calculus textbook we're going to use it several more times and I want to make sure you're comfortable with it it's the algebra behind estimating the slope at the red Point as the slope between the red point and the nearby Green Point the Delta y over Delta X the difference between the y-coordinates is Delta y the numerator of our slope estimate the difference between the x coordinates is Delta X the denominator of our slope

13:12 · estimate we just need to come up with expressions for these differences the x coordinate of the red point is simply X so the red y-coordinate is the function's value at x f ofx the x coordinate of the nearby Green Point is X x + H H is the small Delta x amount we added to X to get a nearby point so the green y coordinate is the function's value at x + h f of X+

13:42 · H so for the numerator we can plug in the difference F ofx + hus F ofx I can include the green and red memory aid y coordinate of green minus y coordinate of red for the denominator we get x + H x green x coordinate minus red x

14:02 · coordinate well it doesn't get written out this way very often since x + H - x so obviously simplifies to H and H is the small distance we deliberately chose for Delta X in the first place so the denominator is usually just H same thing

14:19 · so when you see this expression please think of this triangle and the rise overrun slope it represents Delta y over Delta X two quick points first I'm illustrating a positive slope but if the slope were negative the expression will correctly result in a negative estimate for the slope since f ofx is greater than F ofx + H subtracting a larger number from a smaller will result in a negative number H is always positive

14:48 · since it's the small Delta X we added to X so the equation works for positive and negative slopes the second point is the key idea that uses this expression as the entryway into calculus let me make a copy of the diagram and expression the top and bottom start out identical but I'm going to make changes to the bottom to transform it step by step into the

15:12 · key Foundation of calculus we used the limit a moment ago to find that the slope of our function was 1.2 when we take the limit as Delta X approaches 0 something very special happens first the nearby Green Point approaches the red Point that's pretty obvious since H is the distance between their x coordinates and it's approaching zero second the expression is no longer

15:37 · an estimate of the slope it is the slope or what we call the derivative so let me change the header the expression at the limit isn't the algebraic estimate of the slope it's the algebraic definition of the derivative and there's one more important change at the limit Delta Y and Delta X get new names and symbols they're called Dy Y and DX the d stands

16:01 · for differential we'll talk more about it soon on the top is the estimated slope of a curve based on nearby points on the bottom is the definition of the derivative at the limit is the horizontal distance between the two points approaches zero so here's how we use the limit to find the derivative at 1.6 there was a lot of algebra involved to get to our answer 1.2 but no calculus

16:28 · yet you might not believe me but you will soon calculus is easier than taking limits taking limits with algebra is tedious this is tedious certainly not difficult but timec consuming and prone to errors if you're not careful and when I promised to show you how you could know almost immediately that the slope at 1.6 was 1.2 I was not talking about

16:53 · doing all this algebra in your head calculus is easier than this and I'll show you there's just one more key step the destination will make the journey worth it so we just found the slope of the function at x = 1.6 by plugging 1.6

17:10 · into the function and finding the slope to a nearby Green Point now let's see what happens when we generalize we'll algebraically evaluate Delta y over Delta X at the variable value X instead of the specific value 1.6

17:29 · the Delta y over Delta X formula has the same Parts the y-coordinate of the nearby Green Point is f ofx + H the y-coordinate of the red point is f ofx the x coordinate of the Green Point is x + H and the x coordinate of the red point is X the denominator of course simplifies to H so when we expand f of x + H we get x + h^ 2 - 2 * x + h H + 1 and you can see

18:01 · the correspondence to the terms in the function we just plug in x + H for X expand and simplify using algebra so here's the x coordinate of the Green Point in terms of X it's the first term in the numerator of our deltay over Delta X slope expression next we need to subtract F ofx which is right here minus f ofx

18:32 · three terms cancel and we're left with these terms as Delta y h^2 + 2 xh - 2 H the denominator is H the arbitrarily small number we chose for Delta X we cancel an H from each term and get Delta y/ Delta x = h + 2x - 2 let's let

18:58 · take the limit as H approaches Z we get 2x - 2 now let's pause for a moment and appreciate what we've just done we have an expression for the slope of the function y = x^2 - 2x + 1 in terms of X

19:18 · have we just found a function that Returns the slope of another function let's call the second function fime of X and try it out more on this prime notation shortly a moment ago we found that the slope at x = 1.6 is 1.2 so let's try out our new general function frime of X at 1.6 we get 2 * 1.6 - 2 3.2 - 2 yes we

19:49 · get 1.2 just like before let's try this point at the vertex of the parabola where xal 1 it looks like the slope we get should be zero frime of 1 = 2 * 1 - 2 yes it's zero and we have indeed found the function frime of X that Returns the slope of f ofx for any value of x and

20:14 · now we can finish our dialogue with the broader question what's the slope of this function at any X and the answer for this function is 2x - 2 because this will return the slope of x^2 - 2x + 1 for any X so we've just performed the first calculus operation differentiation we started with a function f ofx = x^2 - 2x + 1 and we

20:42 · performed an operation on it to yield a new function the first function's derivative fime of x = 2x - 2 the derivative of a function evaluates to the slope of that function at every x value we found the derivative the hard way by taking limits and crunching through algebra I'll show a simpler way in a moment but regardless of the method we've just performed the differentiation operation to get the derivative of a

21:12 · function since the derivative of a function is just another function we can graph it also the pink line is fime of X = 2x - 2 frime of X is a common notation for the derivative of f ofx I'll cover some notation conventions in a moment the pink line represents the derivative of the white Parabola so of course the derivatives value at x = 1.6 is 1.2 the

21:42 · slope of the parabola at x = 1.6 let's go over some vocabulary and notation concerning derivatives with slopes we quite naturally speak of Delta y over Delta X the slope between two points this ratio gets closer and closer to the slope at X as Delta X gets closer and closer to zero in calculus we introduced new terms for Delta X and Delta y that have the limit as Delta X approaches zero concept built in we say

22:15 · that the limit as Delta X approaches zero of Delta X is DX we use lowercase D instead of the Greek letter Delta as calculus shorthand that means at the limit so when you see D it means means at the limit as the change to the independent variable often X approaches

22:34 · zero one more time instead of writing all this you can just write this which means the same thing it's common to think of DX as the slightest tiniest change in X and this thought might serve you well but DX is really whatever Delta X becomes as it gets closer and closer to zero it's an idea that we treat like a number DX is called the different differential of X I'll have to draw it with some sort of thickness so we can see it but it's actually unimaginably

23:06 · narrow Dy is the differential of Y which is Delta Y at the limit but it's not the limit as Delta y approaches zero like DX it's also the limit as Delta X approaches zero Dy is the interesting thing we observe as Delta X gets closer and closer to zero so at the limit as Delta X approaches zer our ratio Delta y over Delta X becomes dy/ DX this is the

23:34 · derivative the ratio Dy / DX is the differential of Y with respect to X it's often set out loud as Dy by DX or simply dydx let me show you some other ways you might see the derivative represented this notation denotes a differential change in something with respect to X we can pull the something out of the ratio it means the same thing

24:02 · this part means the change or derivative with respect to X and this part is what we're taking the derivative of since yal F ofx we could also write d by DX of f ofx it means the same thing and since F ofx = x^2 - 2x + 1 we could also write d by DX of x^2 - 2x + 1 they

24:26 · all mean the same thing hopefully the symbology is clear differential of something with respect to the corresponding differential of X I mentioned this one earlier but derivatives of functions can also be represented with the prime or single quote character fime of X is the derivative of f ofx so these are all different representations of the same calculus concept the derivative of f ofx

24:56 · with respect to X we've been using X and Y as our independent and dependent variables which is quite natural considering these are the standard cartisian coordinate system variables and we often represent the dependent variable y as a function of the independent variable X like this

25:14 · but you should know that other variables can be used for example in physics and Engineering the independent variable is often time if the function we're operating on represents something that can change over time for example s equal F of T could represent the displacement of a particle from a starting point at time T in this case the derivative of the function is DS by DT not Dy by DX I just

25:42 · don't want you to get locked into X and Y and not recognize calculus Concepts when you see them referencing different variables as it happens the independent variable t for time is so common that there's an additional shorthand representation for the derivative of a function function with respect to time F dot a function label with a DOT over it represents the derivative of that function with respect to

26:08 · time now let's go over the basic rules of differentiation which are rules and techniques to find the derivative of various types of functions and the big payoff is that the rules will let us find the derivatives without the timec consuming process of taking

26:25 · limits covering the rules of differentiation will take some time as this is essentially the entire first semester of calculus and I'm going to take limits to show you that the rules are true and correct and give you insight into why they work on the left are five types of functions each rule is a special shortcut for taking the derivative of that function type the shortcuts are a

26:50 · result of observing the pattern that reveals itself when we take the limit you'll see what I mean on the right side are the the rules or shortcuts for how to find the derivatives of combinations of function Types on the left so that you don't give up hope let me tell you exactly what we'll be doing I'll start with these four rules on the top then I'll show you some super shortcuts involving these four rules at this point I'll keep my promise because you'll be able to differentiate x^2 - 2x + 1

27:22 · almost instantly in your head and you'll know that the derivative at x = 1.6 is is 1.2 by this time we'll know enough calculus to set up and solve some interesting problems then I'll introduce the second derivative and higher order derivatives and then we'll finish the rules of differentiation with these last five rules I'll show this agenda again so we can keep track of our progress covering these topics we'll finish the first calculus operation differentiation

27:52 · after that we'll learn the second calculus operation integration so let's Dive In we'll start with the derivative of a constant if we have a function f ofx equals a constant such as 3 its graph would be a horizontal line I'm going to make constants green no matter what value we choose for X on the horizontal axis the function returns three and the slope is zero this is true

28:17 · for any constant C since there's no change to the function's value as X changes the derivative is zero we write the generalization like this d by DX of C equals 0 where C is any constant it says that the derivative of a constant is zero and that's the constant rule we're not cheating or saying anything new if we were to again let two points get closer and closer to each other and apply the limit to the slope expression we would get zero we'd

28:49 · get zero every time because we'd always get C minus C here all of these rules of differentiation are General ations of what happens to the slope as we take the limit we simply notice the pattern like we did here for constants and then use the pattern instead of taking limits and that's why calculus at least differentiation is easier than

29:14 · algebra the next pattern or rule of differentiation is called the power rule heads up the power rule is a big part of how you can take the derivative of x^2 - 2x + 1 in your head I'll the rule with some examples then I'll show how it can be visualized I'll tell you the power rule first then I'll prove it's true in a few minutes the power rule is applied the powers of X like X2 X cubed x 4th

29:43 · and so on it says that the derivative of x the N is n * X nus1 it looks horrible but every calculus student in history just remembers this about the power rule bring the exponent downstairs and subtract one so to find the derivative of x cubed we take the exponent 3 and bring it downstairs in front of X then we subtract one from the exponent and get 3 x^2 can you see how

30:12 · that matches the power rule shortcut what's the derivative of X2 well we bring the two downstairs to get 2x to the something for the exponent we subtract one from the original exponent two 2 - 1 is 1 and X to ^ of 1 is just X so the derivative of x^2 is 2x can you see on your own that the derivative of x 4th is 4X cubed yes the power rule is

30:45 · pretty easy I'm arranging the exponents in numerical order so let's go back up top for the derivative of x to the first Power bring the one downstairs and the exponent becomes zero since x to the 0 is 1 the derivative is 1 * 1 which is 1

31:03 · so the derivative of x is 1 this one's easy to see on a graph the line representing yal X is a straight line whose slope is obviously 1 it all works out let me scooch over here to get more room I'll sketch in x to the 0o and x to the first power so you can see that the pattern is kept and speaking of pattern let's go up top again to find the derivative of x to the 0 the 0er comes downstairs and the exponent becomes -1

31:33 · well 0 * anything is 0o so the derivative is zero and let's notice that x to the 0 power is 1 so this is a special case of the constant rule which says that the derivative of any constant such as one is zero so the power rule

31:49 · and constant rule give us the same derivative for exponent zero pretty neat it all works together and the rules are consistent how about this one what's the derivative of the square < TK of X well remember that the < TK of X is X raised to the2 power the power rule isn't limited to integer exponents we bring the 1/2 downstairs and subtract one from 1/2 to get an exponent of -2 since x^ -2 is 1 / < TK X this gives

32:21 · us 12 * 1 the < TK X or 1 / 2 < TK X the fraction has a radical in the denominator so we can rationalize by multiplying the numerator and denominator by squ < TK of x to get < TK

32:37 · X over 2X and that's the derivative of the squ < TK of X using the power rule which works for all real exponents not just integers well I've told you the power rule but I haven't proven that it's true now I'll prove it with limits let's take the derivative of x to the n as the limit of Delta y over Delta X like this as H approaches zero just like before we have our green and red y values the most

33:05 · tedious part of this proof is expanding the green binomial in this video's description I put some links to videos that go into more detail on the binomial expansion in short when the green x + H to the N is expanded the first green term will always be x to the n and the last green term will always be H to the N then moving towards the middle the second term is X to the n minus1 and H

33:32 · to the 1st prefixed with the binomial coefficient which for the second term is always n none of the other coefficients are important for this proof but it's interesting how the coefficients of each green term follow a pattern revealed by Pascal's triangle but that's not what this video is about so see the links in the description for more information math is so interesting and easy when you can see how it all fits together the second to last screen term will always have X and H to the N minus1 along with

34:02 · the binomial coefficient which is also n for the second to last term all the terms in the middle denoted by The Orange Box are the remaining binomial expansion terms and will have factors of H to a power of two or greater that will

34:17 · be important for our proof in a moment in fact all the green terms except the first two will have H factors raised to a power of two or more and there're still still the Red X to the end term we subtract at the end let's bring everything down and encapsulate the orange we'll always have positive and negative x to the N so they'll cancel every time every remaining term will always have at least one H since the denominator is H we can cancel an H from

34:48 · each remaining term this leaves n * x n -1 plus the orange terms which initially all had factors of h^2 or higher and now since we divided three by H they all have factors of H or higher when we take the limit as H approaches Z all of these terms approach zero and we're left with n * x n -1 and so we've proven the power

35:17 · rule once more the rule comes from the slope expression and is the pattern we notice every time we take the limit so when we do calculus we don't need to take the limit we just use the pattern and the pattern for the derivative of x the N is n * X

35:36 · nus1 let me show you a visual interpretation of the power rule that should build your intuition the power rule tells us that the derivative of x^2 is 2x let's consider the function x^2 to be the area a of an actual square with sides of length x the derivative of X2 with respect to X X is the amount the area changes per change in X so if we

36:02 · let X Change by this differential amount DX by how much does the area of the square change well it changes by the area of these two narrow strips this one has an area of its height x * its width DX so X

36:21 · DX this one on top also has an area of its height DX times its width X so its area is also xdx for the sake of completeness let me point out this tiny corner piece whose area is dx^ s it's one of the orange terms from the binomial expansion that goes to zero as Delta X approaches zero

36:45 · so at the limit as Delta X approaches zero the change in the blue squares area the differential of a is the differential of our function x^2 this is the added area to X DX so the derivative of x^2 the change in x^2 per change in X

37:04 · is indeed 2x let's look at the derivative of x cubed visually the power rule tells us that its derivative is 3x^2 let's visualize the function X cubed as the volume of an actual Cube whose sides have length x when we increase X by the tiny differential DX what happens to the volume of the the cube when we add DX over here we get

37:31 · this new volume a thin slab on this face his volume is the area of the face X2 time its thickness DX so the additional volume is x s DX you might see where we're going we get the same additional volume on these other two faces for a total of three x^2 DX because I've drawn DX with some visible thickness you might know notice these three thin regions having volume X

38:00 · their length time dx^ 2 their cross-sectional area so the added volume is 3 x dx^ 2 this last tiny volume has sides equal to DX so its volume is DX cubed these orange terms all go to zero at the limit but it's interesting to see that even they have a visual interpretation on the diagram so the differential change in the volume X cubed is 3 x^2 DX this

38:30 · means the derivative of x cubed is the change in X cubed per change in X which is indeed 3x^2 as the power rule tells us to illustrate the usefulness of the power rule consider finding the derivative of x to the 4th with

38:49 · limits as you can see the power rule is so much simpler and always gives the same answer once more when we do calculus we use these shortcut rules we don't take limits in a calculus course or textbook you'll take limits only for the first week or two to demonstrate what the derivative means but once you learn the rules of differentiation you'll use them and you won't find limits anymore but I will still find

39:15 · limits in this video to prove the rules and illustrate some points now we'll go over ways to take the derivative of combinations of functions the first is the addition rule it says that the derivative of the sum of two functions is the sum of their distinct derivatives or in terms easy to

39:34 · remember the derivative of the sum is the sum of the derivatives here I'm using f and g as the two functions and the prime symbol to denote their derivatives an example should make this simple let's find the derivative of x^2 + x first the derivative of x^2 is 2X and the derivative of x is is 1 so the derivative of x^2 + x is 2x + 1 the

40:03 · derivative of the sum equals the sum of the derivatives let's look at the graph and see why this makes sense here are the two functions we're adding y = x^2 in white and Y = X in green and the blue curve is their sum y = x^2 + x let's

40:24 · take two nearby X values and look look at the corresponding Delta Y's for the three curves here's Delta y for white the change in X2 over our small Delta X and here's Delta y for green the change in green y over our small Delta X and finally here's Delta y for blue the change in x^2 + x over our small Delta X

40:49 · can you see that since blue equals y plus green that Delta y for blue equals the sum of the white and green Delta y's since blue equals white plus green everywhere any difference in blue must be equal to the corresponding difference in white plus green remember as Delta X

41:07 · approaches zero we use the word differential to describe the changes so the differential of the sum equals the sum of the differentials and that's the addition rule without elaboration I'll assert that the same relationship holds with subtraction that the derivative of the difference between two functions equals the difference of their respective derivatives so I'll write the addition rule with plus or minus since it works for addition and

41:33 · subtraction next is the product or multiplication rule it says the derivative of the product of two functions is the first function times the derivative of the second plus the second times the derivative of the first so the pattern is different than the addition rule because the derivative of the product is not the product of the

41:55 · derivatives like we did for the the power rule let's visualize the product rule by considering the product of the two functions to be a rectangle whose area is the product f * G on this graph the axes represent the values of the functions not the independent variable X at least not directly when we increment X by its differential DX F ofx increases by its own derivative DF by DX and the area of

42:23 · the rectangle increases slightly by this thin strip at the same time G ofx increases by its own different derivative DG by DX and the area of the rectangle increases slightly by this thin strip the area of this first strip is its height G of x times its width DF by

42:45 · DX and the area of the second strip is its height DG by DX times its width F ofx so the derivative of the product of the two functions is the addition area of the two rectangular strips and their dimensions are each function times the derivative of the other and that's a nice visual interpretation of the product rule now we can look back at our sample function x^2 - 2x + 1 and consider which

43:16 · of these rules we'll need to use to find its derivative well it's the sum of three simpler functions so let's start with the addition rule the derivative of x^2 is 2X X by the power rule for the next term we need to subtract the derivative of 2x which is the product of 2 and X so

43:35 · we'll need the product rule we'll get back to it in a second and the derivative of one is zero by the constant rule interesting to take the derivative of a simple polinomial we need all four of the rules we've covered so far now let's use the product rule to find the derivative of 2x the product rule says that the derivative of the product is the first function times the derivative of the second plus the second function times the derivative of the first the first function is two and the

44:05 · second function is X the derivative of 2x is 2 * the derivative of X Plus x * the derivative of two the derivative of x with respect to X is 1 by the power rule so the first term becomes 2 * 1 the derivative of two a constant is zero so the second term becomes x * 0 this all simplifies to two so the derivative of 2x is 2 which makes the derivative of our original function x^2 - 2x + 1 = 2x

44:39 · 2 of course this is the same answer we got when we took the limit well this is the function I promised you'd be able to differentiate in a few seconds but using these four rules certainly took more than a few seconds let me show you a calculus super shortcut based on the product rule the derivative of any constant K * x with respect to X is simply the constant K because when we use the product rule we get K * the derivative of x + x * the derivative of

45:11 · K the derivative of x with respect to X is always one so the first term will always be K and the derivative of K with respect to X is always Zero by the constant rule so the second term will always be zero this pattern occurs every time the product rule is applied to KX so the derivative of any constant K * X

45:35 · is always K this is the kind of pattern the rules of differentiation let us exploit now let's go back and find the derivative of x^2 - 2x + 1 at x = 1.6 by differentiating left to right 2x - 2 with practice you learn to ignore constants plug Again The Chosen x value of 1.6 to get 3.2 minus 2 so 1.2 and

46:04 · that's how with practice you can know almost immediately that the derivative or slope of x^2 - 2x + 1 at x = 1.6 is 1.2 okay now for another great shortcut which is a generalization of the first one for KX this time we'll take the derivative of KX to to the N so we have a power of X with some constant coefficient K so this is the product of two functions K and x to the N let's use

46:38 · the product rule again and see where this leads us we have the first * the derivative of the second plus the second * the derivative of the first the derivative of x to the N is straight from the power rule n * X nus1 and the

46:54 · derivative of constant K is a of course zero so the second term becomes zero this leaves the derivative as K * n x nus1 at first this new shortcut looks a little cumbersome like the power rule did but look at What it lets us do when we bring the exponent downstairs we can just multiply it by whatever coefficient is already there so the derivative of 4X cubed is 12 x^2 we bring the three down

47:24 · stairs multiply it by the co efficient of four that's already there to get 12 and then we subtract one from the exponent this simple super shortcut has the product rule power rule and constant rule built in it automatically follows all the rules so knowing this super shortcut and using the addition rule we can easily take the derivative of long polom we just work left to right differentiating one term at a time pols

47:54 · are incredibly simple to differentiate I need to finish the rules of differentiation but first we actually know enough calculus to solve some interesting problems known as optimization problems they involve finding the local minimum and maximum values for a function for example

48:14 · suppose this curve represents the net profit our company would make by manufacturing and selling X number of a particular item if we make and sell too few our profit will be limited by the low number if we make and sell too many our supply might exceed the demand and our extra cost for running more machines and hiring more people won't be offset by the higher volume there's some independent variable here that will maximize the profit function how do we find it let's notice that at the maximum

48:46 · the slope of the function is zero lucky for us we know calculus we can take the derivative of the profit function and find the equation for its slope anywhere when we set the derivative function equal to zero we can solve for x to get the exact point at which the original function is at its maximum our profit function f ofx in thousands of dollars is 0.012 x^2 + 9.8 x

49:17 · -500 where X is the number of units we manufacture and sell to find the value for x that maximizes the function we take the derivative of the function function and set it equal to Z and solve for x we know how to take the derivative of polom we have 0.024 x + 9.8 set this equal to zero and solve x =

49:43 · 48.3 so we should make 408 units to maximize our profit if your problem statement asks you for the profit amount plug 48 into the original profit function to get the maximum profit amount F of 48 equals 5.8 and we're told this is in thousands of dollars so the maximum profit is $5,800 when we make and sell 48

50:10 · units let's do another problem suppose we have a rectangular sheet of metal that measures 32 CM by 24 cm we want to make a box by cutting squares out of the corners and folding the resulting sides up the box won't have a lid

50:27 · the shapes we cut out of the corners need to be squares so that when we fold the sides up they'll all have the same height what are the dimensions of the Box having the greatest volume and what is the volume okay we need to express the volume v as some function of a variable let's use the length of the square sides that we cut out of the corners and call it X of course all of these distances

50:51 · are X the volume of the Box will be its width times its depth times its height the width is this distance which in centim is 32 minus 2x the depth is 24 - 2x so when we multiply these Expressions that gives us 4X cubed - 112 x^2 + 768 x

51:18 · I'm skimming over the algebra so we can focus on the calculus to find the x value that maximizes this function we'll take the derivative set it equal to 0 and solve for x the derivative of the polom is 12 x^2 - 224x + 768 this is a quadratic equation and there are several ways to solve it I used but won't show the quadratic formula to get two possible solutions X = 14.1 cm and X = 4.53 CM we need to

51:53 · check these numbers for feasibility the short side of the metal sheet is 24 cm so we can't cut out squares greater than 12 CM there's not enough metal on that edge so that leaves 4.53 CM as our answer for X but that's not the answer to our problem we're asked for the dimensions that maximize the Box's volume and for that maximum volume so we plug in 4.53 cm for X into our width and

52:21 · depth formulas we get a width of 22.9 4 cm and the depth depth of 14.94% so multiplying these Dimensions yields a volume of 1,552 cubic cm any other value for the square size X will result in a lower

52:46 · volume let's look at a different function this curve has two points where the slope is zero a local maximum here and a local minimum here when we use the word local to describe a minimum or maximum we mean compared to the points nearby for example the local maximum

53:04 · identified here isn't the function's maximum it has higher values out to the right and similarly there are lower values closer to the y- axis than the local minimum identified Here Local means higher or lower than nearby points on either side anyway when we set the

53:23 · derivative equal to zero and solve for x we get these values but it's important to be able to tell the difference between a minimum and a maximum if our management team wanted to maximize profits and we recommended action corresponding to this point that could be a disaster or if they wanted to minimize budget or time and we chose this point setting the derivative equal to zero and solving for x will give us the points where the slope is zero which could be a local maximum or minimum but

53:52 · how can we know which let's color code the function slope green for positive here to the left of the first zero point the slope is zero at the maximum of course and then negative red between the two zero points then the slope is zero again at the minimum and positive beyond the second zero point so the slope changes signs at the Minima and Maxima this makes sense if it's zero at a point it must be passing from positive to negative or from negative to positive

54:25 · but notice that at the maximum point the slope is changing from positive to negative and at the minimum point the slope is changing from negative to positive let's plot the function's derivative of course it's zero at the two points where the function slope is zero at the local Maxima the slope of the pink derivative will always be changing from positive to negative which means its slope is negative as you can see here and at local Minima the slope

54:54 · of the pink derivative will always be changing from negative to positive which means its slope is positive as you can see here the slope of the derivative is positive at locom Minima so what do we mean by the slope of the derivative remember in calculus we're just performing operations on functions that result in different functions when we differentiate a function f ofx to get its derivative fime of x frime of X is

55:22 · just a new function that we can in turn differentiate to get it derivative the derivative of a derivative is called the second derivative or depending on the variables the second derivative of y with respect to X I'll show more notation in a moment but F Prime of X is a common way to

55:42 · denote the second derivative the second derivative of a function tells us the rate of change of the slope of that function and as you might guess the derivative of the original function is called the first derivative here here are six examples of Curves from various functions f ofx for

56:01 · this first one the slope frime of X is positive since the function's value is increasing and since the rate of change of the increase is not changing that is it's a steady increase the second derivative is zero in this example the first derivative fime of x is again positive since the function value is increasing and since the rate of increase is itself increasing the second derivative is also

56:30 · positive in this example the slope is increasing but it's changing from a steep High slope to a shallow low slope so the second derivative is negative can you see how these two functions have different behaviors even though they both have a positive slope for one the slope is increasing at an increasing rate for the other the slope is increasing but at a decreasing rate

56:57 · down here the slope frime of X is negative and since the slope is constant the second derivative is zero this function has a negative slope but the slope is getting less and less negative so the second derivative is increasing positive and this last curve also has a negative slope and its slope is getting more and more negative so the second derivative is decreasing

57:23 · negative since the first derivative can be thought of is the rate of change the second derivative is essentially the rate of change of the rate of change the second derivative is useful because it helps describe the behavior of functions and it's especially useful in minimization and maximization problems to distinguish between local Minima and

57:44 · Maxima at points where frime of X is zero X represents a minimum point where the second derivative is positive and X represents a maximum point where the second derivative is negative

58:00 · now for some second derivative notation the derivative of the derivative is the second derivative fpre of x since the derivative is dy by DX we can write the second derivative as d by DX of Dy by DX

58:15 · let's move the function up to the numerator now what follows is symbolic shorthand not true algebra we take the d and Dy at the top and combine them to get d^2 y because there's two D's and one y and when we take the two DXs and combine them we get dx^ 2 so the symbolic representation of the second derivative of y with respect to X is d^2 y by DX squared I'm not saying it makes

58:44 · pure algebraic sense maybe a mathematician can tell us in the comments if there's a deeper meaning behind the symbol I'm just an engineer and I don't know there are higher order derivatives of course of course the derivative of a function second derivative is the function's third derivative it's easy to extend the D by DX pattern to see third derivative as D cubed y by DX cubed fle

59:09 · Prime is another representation of the third derivative please don't think that higher order derivatives are any more difficult or complicated than the first derivative it's the same differentiation operation following the same rules of

59:24 · differentiation well let's get back to the remaining rules of differentiation but first let's check our progress we've covered four important rules of differentiation the constant rule the power rule the addition subtraction Rule and the product rule then we covered some super shortcuts involving polom solved two problems and learned about the second and higher order derivatives now we'll cover the last five rules of differentiation in this order

59:57 · first s and cosine here's a sine wave the plot of y equal s of theta you don't need to be an expert at trigonometry to differentiate s and cosine but if anything I'm about to say seems unfamiliar I have a YouTube trigonometry course if you want to brush up Linked In the description let's draw the sign function's derivative by plotting a few points and seeing what patterns arise we'll start with a local Minima and Maxima there always easy to see so the

1:00:27 · derivative will be zero and intersect the Theta axis at these blue points at these points where the sine wave crosses the Theta axis in an upwards Direction the slope is one using the expression for the slope again let me demonstrate quickly and without elaboration that Delta y over Delta Theta we've made Theta our independent variable not X is very close to one when Green Delta Theta is very close to zero

1:00:58 · the slope at these points is one so we'll plot the blue derivative points here at y equal 1 the y-coordinate of each Blue Point represents the slope of the red sign curve at that value of theta the slope at these points is -1 so

1:01:15 · the Blue Points go down here we could plot more points but let me jump to the answer and plot the derivative of sin Theta it's this smooth curve if you're familiar with trigonometry you'll recognize this curve as cosine Theta the derivative of sin Theta is cosine Theta pretty neat let me show you

1:01:37 · a proof it assumes a little trigonometry knowledge but I'll go quickly we'll consider a unit circle and focus on the first quadrant let's put angle Theta in standard position since we're on the unit circle this yellow length the radius of the circle is one this horizontal length is cine Theta and this vertical length is sin Theta it's this vertical red length we're interested in as Theta changes ever so slightly by D Theta what's the change to the red length D sin Theta let's find

1:02:09 · out the pink Arc has length Theta which seems strange because green Theta represents an angle the number of radians and pink Theta represents a distance the number of radi but since the radius of the circle is one the numbers for Theta green GRE and pink are the same let's see what happens near this point and note that this segment of the circle circumference is very nearly a straight line as our focus of attention get smaller and smaller the derivative of sin Theta is

1:02:40 · how much this vertical distance changes as Theta changes by the tiny differential of theta D Theta and since pink Theta and green Theta have the same measurement I'm going to make the D Theta label green to match our formula to find the change sin Theta let's draw this right triangle and we can see that red Sin Theta changes by this amount that we can call D sin Theta so in this small triangle we have representations for D sin Theta and D Theta which are

1:03:10 · the numerator and denominator of the derivative we're trying to find since the trig ratios are the ratios between the various sides of a right triangle our derivative ratio is one of the six trig functions this angle is congruent to Theta I'm telling you this without proof and so D sin Theta and D Theta are the adjacent and hypotenuse of the small triangle respectively and adjacent over hypotenuse corresponds to cosine and so we've shown graphically that the derivative of s is cosine be careful

1:03:43 · because the opposite is not true the derivative of cosine is not s since the blue cosine curve has the exact same shape as the sign curve it seems reasonable to deduce that the derivative of cosine will also have this shape and

1:03:58 · let's note that the cosine is out of phase with s to illustrate I'll add Theta axis markers at every pi/ 2 radians and we can see that the cosine curve the derivative of s is always pi over two radians to the left of s this

1:04:14 · is easiest to see by comparing peak-to Peak points where the functions have their maximum values since the derivative of s is out of phase to it by Pi / 2 does it it make sense that the derivative of cosine would be out of phase to it well yes indeed it actually is but as you can see we don't have a function with these Peak values but if we flip the sign curve by taking its negative we get the curve we seek and the derivative of cosine Theta

1:04:44 · is indeed negative sin Theta so it's the derivative of negative sin Theta but looking at the curve you may see what's coming next the derivative of negative sin Theta is cosine Theta and taking the derivative of cosine Theta gets us back to sin Theta and this four-step cycle comprises the trig related rules of differentiation it might help to remember that the trig functions alternate that is taking the derivative of a sign yields a cosine and vice versa

1:05:17 · then it's easy to remember that the derivative of s keeps the sign so the derivative of positive sign is positive cosine keep the S and the derivative of negative sin Theta is negative cosine Theta the derivative of s keeps the S on

1:05:34 · the other hand the derivative of a cosine function flips the sign the derivative of positive cosine Theta is negative sin Theta and the derivative of negative cosine Theta is positive sin Theta as you may know there are four more trig functions but we'll have to skip them for now and cover their derivatives later let's test our knowledge what's the derivative of x Cub * sin x well we have the product of two

1:06:02 · differentiable functions we can call f and g so we'll use the product rule the derivative of the product equals the first times the derivative of the second plus the second * the derivative of the first so it's simple there's really no intermediary steps just write down the components and that's the derivative of the product X cubed cine x + sin x *

1:06:31 · 3x^2 now for the chain Ru the chain rule is how we differentiate composite functions a composite function is a function whose argument includes another function you can think of composite functions as embedded functions where one function is embedded in the other for example sin 2x

1:06:54 · is a composite function because s is a function and its argument 2x is another function the 2x function is embedded in the sign function as its argument this is very common in math science and engineering so you'll use the chain rule a lot probably more than any other rule let's see why we need the chain rule when we find the derivative of sin 2x first we know the derivative of sin x with respect to X is cosine X it's true

1:07:25 · it's one of of our rules of differentiation the one we just covered but we cannot say that the derivative of sin 2x with respect to X is cosine 2X that's false it's close we'll need to adjust a bit with the chain rule to get the derivative of sin 2x but this isn't

1:07:43 · right here's the pattern the derivative of s something with respect to that something equals cosine of that something all three terms need to match for the differentiation rule to apply and when we try to apply the rule to sin 2x you can see that they don't match I'm

1:08:01 · illustrating this with the trig rule but the pattern applies to all the rules if we were to modify the equation to be the derivative of sin 2x with respect to 2x then the derivative would be cosine 2X because all the terms would

1:08:18 · match but in calculus were not asked very often to find the derivative with respect to a function of X just with respect to X so we need to dig a Little Deeper to differentiate composite functions we can write composite functions like this F of G of x g is

1:08:37 · called the inner function because it's inside the argument for function f which is the outer function the derivative we're seeking is DF by DX the derivative of the outer function with respect to the argument of the inner function our independent variable X here's the key to to understanding the chain rule a differential change in X will result in a differential change to G DG by DX that differential change to G in turn

1:09:07 · causes a differential change to F DF by DG and that change to function f DF that occurs as a result of the differential change to X DX is the derivative we want DF by DX this is where the chain rule gets its

1:09:26 · name the differential change to X ripples out in a chain reaction to cause the differential change in the outermost function here's the chain rule for differentiation the derivative of f of g ofx equals the derivative of f with respect to G times the derivative of G

1:09:45 · with respect to X it should look familiar it's the Chain Reaction we just traced from the independent variable X Out to the outermost function and it makes sense algebraically because there's a clear cancellation chain that makes the chain rule a lot easier to visualize and understand so let's find the derivative of sin 2x for the first Factor DF by DG

1:10:10 · we need the derivative of the outer sin 2x with respect to the inner 2x let's notice that these terms match you'll always get a match like this when you use the chain rule we can use the trig rule that the derivative is cosine of the matching term so DF by DG equal cosine 2X by the way this is the answer we said was not right a moment ago to get the correct answer we need to multiply by the last term DG by DX G is

1:10:40 · 2x so we get the derivative of 2x with respect to X it doesn't get much easier than this we have a super shortcut that tells us the derivative of 2x with respect to X is 2 so we use the chain rule to determine that the D derivative of sin 2x is 2 cosine 2X let's do another problem and find the derivative of the sare < TK of 5x^2 + 3 the inner function is 5x^2 + 3 the

1:11:10 · outer function is the square root let's rewrite the expression using an exponent of 1/2 to represent the square root this should make it clear which function is the inner function and which is the outer the first Factor we need to find find is DF by DG the derivative of the outer function with respect to the inner the outer function is 5x^2 + 3 to the 1/ 12 the inner function is 5x^2 + 3 when

1:11:38 · you use the chain rule you'll always have matching terms and can use the appropriate rule of differentiation in this case the power rule with exponent 1/2 with the power rule we bring the exponent downstairs and subtract one from it we get 1/2 times the matching expression which turns out to be the inner function G raised to the -2 and that's the first term in the chain rule DF by DG the second Factor DG by DX is

1:12:09 · simple too it's the derivative with respect to X of 5x^2 + 3 we use the power rule again for this one 10 x you can simplify the expression using algebra and we found the derivative of this composite function let's do one more chain rule example this time with a composite of three functions so we want to find the derivative of f of G of H of X let's set

1:12:37 · up the chain the derivative of f with respect to G times the derivative of G with respect to H times the derivative of H with respect to x three functions makes the chain concept even more obvious algebraically the dgs cancel and the DH is cancel leaving us with DF by DX the derivative of the outermost function with respect to the independent variable X so let's find the derivative with respect to X of cine 2 4X let's rewrite

1:13:10 · the function as cosine of 4x^ squared because cosine squar argument means the cosine of the argument squared the inner function is 4X the middle function is cosine and the outer function is power of two let me expand

1:13:27 · the derivative chain and show you again how simple this is the chain always starts with the differential of the given outermost function as the numerator of the first factor for this problem the given function is cine of 4x^ 2 so D cosine

1:13:47 · 4x^ SAR the denominator of the first factor is DG the differential of the middle function which is cosine so D cosine forx as usual when we use the chain rule we have matching terms and the derivative with respect to something of something squared is two of that something so the first term in the chain DF by DG is 2 cosine

1:14:12 · 4X the numerator of the second factor is the denominator of the first that's how the chain Works D cosine 4X the denominator of the second factor is DH the differential of the inner function which is 4X so d4x again our terms match and we have the derivative of cosine of something with respect to that something the something is 4X and the derivative of cosine is negative s so the second term DG by DH is NE sin

1:14:45 · 4X following the pattern the numerator of the third factor is the denominator of the second differential of 4X and finally at the end of the chain is the differential of the independent variable X DX the last Factor will be a straightforward derivative the derivative of 4X with respect to X is 4

1:15:05 · so the last Factor DH by DX is 4 rearrange the terms if you like and we found the derivative of cosine 2 4X and that's the chain rule the one you'll use most often in real life and very easy with practice next is the quo rule where we'll find the derivative of one function divided by another put differently we're finding how the ratio between two functions of X changes as X

1:15:34 · changes first the derivative of the ratio is not the ratio of the derivatives that might remind you of the product rule since the derivative of the product is not the product of the derivatives the quotient rule says that the derivative of the ratio is the denominator time the derivative of the numerator minus the numerator times the derivative of the denominator all over the denominator

1:16:00 · squared we can prove it's true using the product rule and chain rule first we rewrite the quotient as a product with a denominator raised to the -1 power so we have d by DX of f ofx * G ofx

1:16:16 · the1 so we use the product rule the first * the derivative of the second plus the second * the derivative of the first the derivative has four components each straightforward except this one is a composite of a function raised to a power so we'll just need to apply the chain rule we have the derivative of some function raised to the -1 so the power rule tells us that's -1 * the function raised to the -2 then to finish the chain rule we multiply by the derivative of the function G Prime of X

1:16:49 · now let's simplify on the left side we have f ofx * G Prime of X all over G of x^2 then we add the right side fime of X over G ofx so we're adding two fractions but

1:17:11 · their denominators don't match if the denominators matched we could add their numerators if we multiplied the right denominator by G of X then they'd both be g^ 2 of X so let's multiply the right term by G of X over G

1:17:28 · ofx and that's it if we swap the left and right terms the format will match the quotient rule stated above so we've derived the quotient rule from the product rule and chain rule to remember the chain rule I start with the denominator squared then the numerator expression starts with the denominator not squared then like the product rule we multiply one by the derivative of the other but unlike the product rule we subtract instead of add

1:17:57 · then the right term is opposite derivative Wise from the left term frime becomes f and g becomes G Prime write it from scratch a few times and you'll know it let's find the derivative of a quotient 3x Cub - x^2 + 2 / cosine

1:18:16 · X we'll start with the denominator squar cine 2qu of X then for the numerator expression we start with the denominator again not squared cine X then multiply by the derivative of the numerator by the power rule the derivative of 3 x Cub x^2 + 1 is 9 x^2 -

1:18:38 · 2x then we subtract the right expression which is the numerator 3x Cub - x^2 + 1 * the derivative of the denominator the derivative of cine X is sinx that's pretty much it these negative signs undo each other and with some trig substitution you can get rid of the cosine ^ 2qu x in the denominator and that's the quotient

1:19:04 · rule speaking of trig now that we know the quotient rule we can find the derivative of the other four trig functions because they can all be expressed as fractions involving s and cosine I'm using the color coding from my trigonometry series just for this chart for the derivative of tangent Theta the quotient rule is the denominator time the derivative of the numerator minus the numerator * the derivative of the denominator all divided the denominator squared this simplifies to cosine 2 thet plus sin s

1:19:36 · Theta which is 1 over cosine s thet which is secant squ thet since secant is 1/ cosine I'll show the derivation of the other trig functions using the quotient rule but won't step through the details you might need to know these check with your instructor if you know the quotient Rule and the circle trig identities you can figure these out as you need them practice builds confidence here are the derivatives of the six trig

1:20:08 · functions let's check our agenda we covered the four-part trig cycle for the derivatives of s and cosine we covered the chain rule to find the derivative of composite functions the role you're likely to use more than any other then we went over the quotient Rule and actually derived it from the product rule and the chain rule we use the quotient rule to show the derivatives of the remaining trig functions since they're all ratios that include s and

1:20:36 · cosine the last rules of differentiation are for exponentials and logarithms please note that all these Atomic function types can be combined and used in all of these rules that allow us to combine functions in various ways so the next rules are for exponential and

1:20:57 · logarithms these rules are some of the simplest well they've all been pretty simple right but there's a lot of background to review for it all to make sense exponentials and logarithms are usually covered in Algebra 2 or pre-calculus but I'll do a thorough review of the topics needed to understand the rules of differentiation exponential functions have their independent variable X up in the exponent the number on bottom is called the base I'm color coding the base green as a reminder that it's not a

1:21:28 · variable like X it's a constant such as 2 don't confuse the exponential function 2 ra the power of X with the polom or power function x raed to the power of two they're different functions with different graphs and different derivatives you can remember that exponential functions have their variable in the exponent and in short B to the X means multiply con base B by itself x

1:21:58 · * let's assume we have B raised to the 7th power as shown the associative property of multiplication says that we can group The B's together like this and get the same result so B 7th = B 3r * B

1:22:14 · 4th in general when their bases are the same we can multiply exponentials by adding their exponents let's graph some exponential when the base B is greater than one the exponential function value gets bigger and bigger as X increases the slope is always positive functions like these are used to model exponential growth when the base B is one the exponential function value Y is always one because one to any exponent is one

1:22:46 · because 1 times itself any number of times is always going to be one and when the Bas is between zero and one the function value gets smaller and smaller as X increases functions like this are used to model exponential decay an exponential function with base B will always be symmetrical across the y AIS to an exponential function whose base is the reciprocal of B like this example of 2 and/ 12 all the graphs of y equal sum base B

1:23:19 · to the X pass through the very busy Point 0a 1 because any base B raised to the zeroth power will always equal one here's an animation showing various exponential curves as the green base B changes when B is greater than one the curve is always increasing the higher the base B the faster the increase and as I mentioned every curve passes through the circled point 0 comma 1 when base B is 1 the curve flattens

1:23:51 · out because 1 raised to any power x will always B1 and when B is between 0 and 1 the curve is always decreasing lower base values B decrease faster so we have an exponential function y = b to the X where B is a constant and X is the independent variable so given b x and a calculator

1:24:14 · we can find y but what if we know Y and B and want to find X for any function when we find x given y instead of Y given X that's called taking the inverse of the function suppose we knew Y = 5.89 and wanted to find X how would we

1:24:34 · do it the inverse of the exponential function is the logarithm the logarithm answers the question what's the exponent we write and say the logarithm function like this x = log base 1.47 3 of

1:24:51 · 5.89 it means X is the expon onent on 1.47 3 that results in 5.89 these equations aren't solved by hand we use a calculator and before calculators slide rules let me show you this again emphasizing the inverse relationship to solve this equation for x we need to isolate X to get x equals

1:25:15 · something but X is in the exponent how do we get x out of the exponent how do we undo exponentiation by taking the logarithm we'll take the logarithm of both sides making sure that the bases match to undo an exponent of Base 1.47 3 we need to take the logarithm base 1.47 3 let's

1:25:39 · look at the right side of the equation remember the log function answers the question what's the exponent let's transliterate the right hand side what's the exponent on 1.47 3 that results in 1.4 473 to the X well the answer is X

1:25:58 · this is rather like asking what's half of twice X the half and the twice undo each other leaving X and the logarithm base 1.47 3 undo exponentiation base 1.47 3 so we get X on the right hand side which is exactly why we took the logarithm to isolate the exponent x to keep things even and balanced we need to take the logarithm of the left side too and we get log base 1.47 3 of 5.89 which

1:26:29 · our calculator will tell us is 4.58 since exponentials and logarithms are inverse functions of each other their graphs are symmetrical across the line yal X the exponential of Base B is a miror reflection of the logarithm base B across the dotted diagonal line Y = X

1:26:51 · all inverse function pairs share this characteristic not just exponentials and logarithms so naturally since all exponential graphs pass through the point 0 comma 1 because any base raised to the 0 power is 1 all logarithmic graphs pass through the point 1 comma 0 because the exponent to any base that results in one is

1:27:17 · zero the associative property of multiplication tells us that b 7x can be expressed as B 3x * B 4X let's see what happens when we take the logarithm base B of both sides log base B of 7x is

1:27:34 · simply 7x like before the log base B and the exponent on B cancel out leaving just the exponent and log base B of these two terms are 3x and 4x respectively so the three terms we get after taking log base B are the three exponents of B 7x 3x and 4x and to write

1:27:57 · the resulting equation we need to combine these terms by adding not multiplying that shouldn't be surprising logarithms effectively bring exponents down and we already observed this property about multiplying exponentials let's go to an extreme and write B 7x as b x multiplied by itself

1:28:18 · 7even times now when we take the logarithm base B of both sides we get seven distinct in log base B of B to the X terms that means that log base B of B 7x is 7 log Bas B of B to the x or in

1:28:34 · general log base B of B to the NX is n logs Bas B of B to the X we can take the coefficient of x in the exponent and move it to the coefficient of the logarithm we're almost ready for the rules of differentiation but first another important property of exponential functions any exponential function can be expressed as an equivalent exponential function with any other base so here's the graph of y = 2 to the

1:29:05 · X again we can get the exact same graph from an exponential equation that has another base such as 3 so Y = 2 x can be expressed as y = 3 raised to the something let's find the something by setting the Expressions equal to each other 3 raised to the Something = 2

1:29:25 · raised to X let's take the logarithm of both sides to isolate the red something variable we need to be careful which base to use for the logarithm we want to isolate the red something so we'll take the log base 3 of both sides since three is the base whose exponent we want to isolate the left side simplifies to our red variable on the right side we take the exponent out and give us X logs base 3 of 2 and that's the answer 2 the x is

1:29:55 · the same function as 3 raised to the log base 3 of 2 * X and the calculator will tell us that log base 3 of 2 is about 0.63 093 here's the pattern for switching bases the old base raised to the X power equals the new base raised to the power of log base new base of old base * X so

1:30:23 · these are the same functions and the point is it's not the base that determines the shape of the exponential function but a combination of the base and whatever coefficient the independent variable has in the exponent the same curve can be described by lots of exponential functions having whatever base you choose however there's a very special exponential base his value is about 2.718 it's so special that it has its own symbol lowercase e I'll use green as

1:30:56 · a reminder that e is a constant like two or three not a variable it's kind of a surprise that the constant e pops up in some simple formulas it's a constant of nature like pi and I'll show you in a moment why e is so useful in calculus and why it's called the natural base here's a graph of the exponential function y = 2 X and here's y = 3 x since e is between 2 and

1:31:28 · 3 it shouldn't be too surprising that y = e to the x is between them it's a very special Base number but its curve looks just like any other exponential curve if we have a function y = e to X we can find y given X like any other function and we can invert it to express X in terms of Y using the logarithm if y = e to X then X = log

1:31:56 · base e of Y well the logarithm base e is also special and it has a special symbol and name the natural logarithm or natural log and for its symbol instead of writing L base e we write Ln I know

1:32:14 · that seems backwards but it's from the Latin Ln means natural log so log base e of Y is equivalent to this expression which can be pronounced as natural logarithm of Y natural log of Y Ln of Y or even Ln y so once more Ln

1:32:35 · is a mathematical shorthand for log base e the natural log also pops up surprisingly in some simple formulas we'll see this limit again in a moment when printed Ln can look like one n so when handwritten you'll often C Ln

1:32:54 · written in script or cursive with a loopy L like this here are some examples I found online it's not a big deal I just don't want you to be confused when you see the style and I suggest you use it just write Ln in cursive like it was a word now we're ready for the differentiation rules for exponentials as usual we'll find the slope at a red Point by finding the slope between the red point and a nearby Green Point whose x coordinate is x + H

1:33:28 · then we'll take the limit as H approaches zero and see what we get our function is 2 X so we plug that into our limit equation note that we have 2 raised to x + H power we can rewrite this as 2 x * 2 H remember now we have 2 to the X twice in the numerator which we can factor out to get 2 x * 2 H -1 all over H remember

1:33:59 · we're taking the limit as H approaches 0 and 2 to the X won't change as H changes because there's no h in it so we can pull it out of the limit now I told you earlier that this limit is the natural log of this number like this and so the derivative of 2 x is 2 x

1:34:23 · * the natural logarithm of 2 and in general the derivative of B to the x is the natural log of B * B to the X and that's a differentiation rule for exponents we'll make it stronger in a moment but it's good to know that the derivative of B to the x is the natural log of B times the original exponential function B to the X quick what's the derivative of 7 to

1:34:50 · the x it's the natural logarithm of the base 7 times the original exponential function 7 to the X easy now what if the base were the natural base e same thing the derivative of e to the x is the natural log of e times the original exponential function e to the X well what's the natural log of e Ln e means the exponent on base e

1:35:22 · remember the base of the natural logarithm Ln is always e that results in E so Ln e is one because E rais power of one is e this is not a special rule for E any log base B of B that is the logarithm of any number to its own base is one because B raised to the power of one is B so log base e of e is one we just have

1:35:51 · a special symbol for log base e Ln so Ln E equals 1 we substitute the natural log of e which is one into our derivative and simplify to get the derivative of e to the x is e to the X the only function

1:36:08 · that's its own derivative pretty neat and that's why e is such a special exponential base now on the screen are two equations or rules for derivatives of exponentials the bottom is just a special case of the top for base e since Ln e is one but

1:36:27 · there's one more variation to consider and then we'll have a single robust rule that will help us find the derivative of all exponential functions often the exponent will not simply be X but some function of X this is very common in real world applications of exponentials we can't use the highlighted rule above it applies only when the exponent matches the independent variable for B raised to