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Welcome back.

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I'm now going to introduce you to a new tool for

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solving derivatives.

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Really between this rule, which is the product rule, and the

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chain rule and just knowing a lot of function derivatives,

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you'll be ready to tackle almost any derivative problem.

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Let's start with the chain rule.

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Let's say that f of x is equal to h of x times g of x.

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This is the product rule.

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In the chain rule it was f of x is equal to h of g of x.

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Right?

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I don't know if you remember that.

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In this case, f of x is equal to h of x times g of x.

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If that's the case, then f prime of x is equal to the

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derivative of the first function times a second

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function plus the first function times the derivative

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of the second function.

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Pretty straightforward.

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Let's apply it.

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Let's say that-- I don't like this brown color, let me pick

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something more pleasant.

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Maybe mauve.

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Let's say that f of x is equal to 5x to the fifth minus x to

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the seventh times 20x squared plus 3x to them mine 7.

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So one way we could have done it, we could just

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multiply this out.

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This wouldn't be too bad, and then take the derivative

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like any polynomial.

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But let's use this product rule that I've just shown you.

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So the product rules that says, let me take the derivative of

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the first expression, or h of x if we wanted to map

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it into this rule.

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The derivative of that is pretty straightforward.

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5 times 5 is 25.

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25x to the fourth, right?

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Then minus 7, x to the sixth.

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We're just going to multiply it times this second expression,

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doing nothing different to it.

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Maybe I'll just do it in a different color.

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Times 20x plus 3x minus 7.

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And then to that we will add the derivative of

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this second function.

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The derivative of that second function is 40x minus

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21x to the minus eighth.

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And that times this first function.

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I guess I'll switch back to mauve, I think

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you get the point.

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5x to the fifth minus x to the seventh.

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All we did here was we said OK, f of x is made of these two

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expressions and they are multiplied by each other.

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If I want to take the derivative of it, I take the

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derivative of the first one and multiply it by the second one.

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And then I add that to the derivative of the second

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one and multiply it by the first one.

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Let's do some more examples and I think that will

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hit the point home.

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Clear image.

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Change the colors and I'm back in business.

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Let me think of a good problem.

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Let me do another one like this, and then I'll actually

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introduce ones and the product rule and the chain rule.

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So let's say that f of x is equal to 10x to the third plus

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5x squared minus 7 times 20x to the eighth minus 7.

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Then we say f prime of x, what's the derivative of

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this first expression.

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It's 30x squared plus 10x.

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And I just multiply it times this expression, right?

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20x to the eighth minus 7.

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And I add that to the derivative of this second

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expression, this is all on one line but I ran out of space,

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160x to the seventh, right?

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8 times 20 is 160.

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And then the derivative of 7 is zero.

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So it's just 160x to the seventh times this

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first expression.

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10x to the third plus 5x squared minus seven.

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There we go.

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And you could simplify it.

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You could multiply this out if you wanted or you could

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distribute this out if you wanted, maybe try to

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condense the terms.

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But that's really just algebra.

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So this is using the product rule.

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I'm going to do one more example where I'll show you,

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I'm going to use the product and the chain rule and

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I think this will optimally confuse you.

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I want to make sure I have some space.

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Here I'm going to use a slightly different notation.

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Instead of saying f of x and then what's f prime of x, I'm

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going to say y is equal to x squared plus 2x to the fifth

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times 3x to the minus three plus x squared to the minus 7.

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And I want to find the rate at which y changes relative to x.

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So I want to find dy over dx.

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This is just like, if this was f of x, it's just

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like saying f prime of x.

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This is just a [UNINTELLIGIBLE]

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notation.

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So what do I do in the chain rule?

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First I want the derivative of this term.

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Let me use colors to make it not too confusing.

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So what's the derivative of this term?

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We are going to use the chain rule first.

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So we take the derivative of the inside which is 2x plus 2

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and multiply times the derivative of the

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larger expression.

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But we keep x squared plus 3x there so it's times 5 times

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something to the fourth.

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And that something is x squared plus 2x.

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So there we took the derivative of this first term right here

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and then the product rules says we take the derivative of the

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first term, we just multiply it by the second term.

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So the second term is just 3x to the minus 3 plus x squared

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and all that to the minus 7.

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We did that and then to that we add plus the derivative of this

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second term times this first term.

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We're going to use the chain rule again.

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What's the derivative of the second term?

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I'll switch back to the light blue.

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Light blue means the derivative of one of the terms.

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So we take the derivative of the inside, the derivative of

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inside is minus 3 times 3 is minus 9, x go down one to

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the minus 4, plus 2x.

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And now we take the derivative of the whole thing.

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Times minus 7 times something to the minus 8, and that

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something is this inside.

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3x to the minus 3 plus x squared.

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And then we multiply this thing, this whole thing which

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is the derivative of the second term times the first term.

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Times, and I'm just going to keep going, times x squared

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plus 2x to the fifth.

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So this is a really, I mean you might want to

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simplify at this point.

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You can take this minus 7 and multiply it

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out and all of that.

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But I think this gives you the idea.

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And if you have to multiply this out and then do the

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derivative if it's just a polynomial, this would

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take you forever.

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But using the chain rule, you're actually able to, even

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though we ended up with a pretty complicated answer,

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we got the right answer.

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And now we could actually evaluate the slope of this very

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complicated function at any point just by substituting the

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point into this fairly complicated expression.

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But at least we could do it.

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I think you're going to find that the chain and the product

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rules become even more useful once we start doing derivatives

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of expressions other than polynomials.

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I'm going to teach you about trigonometric functions and

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natural log and logarithm and exponential functions.

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Actually, I'll probably do that in the next presentation.

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So I will see you soon.

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