﻿WEBVTT

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<v ->When we wanna take derivatives,</v>

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we've looked at a couple different ways of doing that.

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Most commonly, people have seen

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the Power Rule at this point.

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What we're gonna see though, in the next few slides

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is that there are other techniques for taking derivatives,

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and these get into some pretty complicated rules

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that many people memorize as they go

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into higher calculus studies.

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So, the first thing is it says

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we're gonna look at the Chain Rule.

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There are many rules for differentiation,

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depending on the function.

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One method, the Chain Rule,

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applies to a function within a function,

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or a composition of functions.

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So, if I have f of x is equal to f of g of x,

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then if I wanna use the Chain Rule, here's how it works:

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d y d x is equal to d y over d u times d u over d x.

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And here's an example: let's say my function is

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one plus x squared quantity to the fifth,

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so I have two functions going on.

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My first function is x to the fifth,

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and my second function that's going on

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is one plus x squared.

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That's a composition of functions,

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it's where one function is the input

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into the other function.

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So to calculate a derivative using the Chain Rule,

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what I'm gonna do is a substitution using the letter u.

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I'm gonna let u represent my inside function,

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or let u be one plus x squared,

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and then y would be equal to u to the fifth.

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Pause real quick and make sure that makes sense to you

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because it's called a u or dummy substitution,

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that inside now instead of being called one plus x squared,

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is just called the letter u.

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So when I go to do the derivative,

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d y d x would be the derivative of u squared

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over d u times the derivative of one plus x squared d x.

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And the derivative of u to the fifth

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would be five u to the fourth

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multiplied by the derivative of u.

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Derivative of u there's back to

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x is instead of x squared, it's gonna be two x.

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And then, if I simplify that, I'll have

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ten x u to the fourth,

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now don't leave your answer there,

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it's a common mistake, is that students

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will leave u in their answer.

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They did the derivative, they think they're done,

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but keep in mind u is something that I introduced,

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it wasn't in the original problem.

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So, to do this correctly, I need to get rid of that u.

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Or I'll go back and instead of having ten x u

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to the fourth, more appropriately,

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it would be ten x times the quantity

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one plus x squared to the fourth.

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There are similar rules that

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happen with products and quotients.

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On your test, you're gonna wanna look out for trick answers.

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If you're asked to take a derivative

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and you notice it's a compositional functions

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or a product of functions or a quotient of functions,

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and if you don't have these rules memorized,

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you're just gonna wanna know that there is a rule,

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and there's probably gonna be a trick answer

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presented for you.

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If I'm asked to do the derivative of a product,

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the product of f times g,

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I would do f times the derivative of g

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plus g times the derivative of f.

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It's like the first function as it is

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times the derivative of the second

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plus the second function as it is

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times derivative of the first.

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Or if I wanna do a quotient,

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let's say I wanna do the quotient of f divided by g,

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I would do g times the derivative of f

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minus f times the derivative of g

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divided by g squared.

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So again, it's not just derivative of the top,

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derivative of the bottom, and dividing them,

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that would be a trick answer.

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The correct way to do a derivative

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of a quotient is on your screen.

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Another type of standard technique for taking derivatives

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is if you have a logarithmic function.

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Here we're looking at the natural log,

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and then also we'll see what happens

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if you have an exponential function.

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The derivative of the natural log

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of absolute value of x is one over x,

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and so looking ahead, if you're working

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with an integration problem,

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which of course is where you're doing this backwards,

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if you see one over x, you know

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the anti-derivative is the natural log

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of absolute value of x.

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Back to derivatives, the derivative of e to the x

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is equal to e to the x.

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That's kinda cool.

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And the last one, the derivative of a to the x

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is equal to a to the x times the natural log of a.

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Again, these are rules that people might memorize

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as they go into higher level calculus studies.

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For you, you're gonna wanna be aware

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that there are some kind of trick answers.

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Like, if you're given that quotient

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a trick answer would be just the quotient

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of a derivatives which is not correct.

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Now moving on, there's trigonometric derivatives as well.

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So, on your screen you'll see that

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derivative of sine is cosine,

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derivative of tangent is secant squared,

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derivative of secant is secant x tangent x,

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the derivative of sine negative one of x

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is one over the square root of one minus x squared.

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Down the right hand column, the derivative of cosine x

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is negative sine x,

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derivative of cotangent x is negative cosecant x,

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derivative of cosecant x

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is negative cosecant x cotangent x

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and the derivative of tangent negative one

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would be one over one plus x squared.

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So again, these are rules that you could memorize,

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or just be aware that there are some

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unique properties for higher level derivatives.

