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Course: Calculus, all content (2017 edition) > Unit 2
Lesson 20: Chain rule- Chain rule
- Worked example: Derivative of cos³(x) using the chain rule
- Worked example: Derivative of ln(√x) using the chain rule
- Worked example: Derivative of √(3x²-x) using the chain rule
- Chain rule intro
- Chain rule overview
- Worked example: Chain rule with table
- Chain rule with tables
- Quotient rule from product & chain rules
- Chain rule with the power rule
- Applying the chain rule graphically 1 (old)
- Applying the chain rule graphically 2 (old)
- Applying the chain rule graphically 3 (old)
- Chain rule
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Chain rule with the power rule
An example of Sal differentiating (2x³+5x²-7)⁸ using the chain rule and the power rule.
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Video transcript
- [Voiceover] So we've got the function f of x is equal to two x to the third plus five x squared minus seven. All of that to the eighth power. And we want to find the derivative of our function f with respect to x. Now the key here is to
realize that this function can be viewed as a
composition of two functions. How do we do that? Well let me diagram it out. So let's say we want to start with, and I'll do it down here
so we have some space. So we're gonna start with an x and what's the first thing that we would do? If you were just trying to
evaluate it given some x? Well, the first thing
you would take two times that x to the third power
plus five times that x squared and then minus seven. So, what if we imagined a function here that just did that first part. That just evaluated two x to the third plus five x squared
minus seven for your x. So let's call that the function u. So whatever you input
into that function u, you're gonna get two times that input to the third power, plus
five times that input to the second power, minus seven. And so when you do that,
when you input with an x, what do you output? What do you output here? Well, you're going to output u of x, which is equal to two
x to the third power, plus five x squared minus seven. And I'll do it all in one color just so I don't have to keep changing colors. So two x to the third power, plus five x squared, minus seven. That is u of x. Now what's the next thing you're gonna do? You're not done evaluating f of x yet. You would then take that value and then input into another function. You would then take the
eight power of that value. So then, we will take that and input it into another function. Let's call that function v. And that function,
whatever input you give it, and I'm using these squares just to say whatever input goes into that function. You're going to take
it to the eighth power. And so in this case, what do you get? What do you end up with? Well, you end up with v of u of x. V of u of x. Or you could view this as
v of two x to the third, plus five x squared, minus seven. Or, you could view this
as two x to the third, plus five x squared, minus seven. All of that to the eighth power. All of that to the eighth power
and that's what f of x is. So as we just saw, f of x can be viewed as the composition of v and u. This is f of x. So, if we write f of x. If we write f of x being equal to v of u of x, then we see very clearly the
chain rule is very useful here. The chain rule tells us that f prime of x is going to be the derivative
of v, with respect to u. So it's going to be v prime of, not x, but v prime of u of x. The derivative of v, with respect to u, times the derivative of
u, with respect to x. So u prime of x. So we know a few things already. So let's, let me just write
things down very clearly. So we know that u of x is equal
to two x to the third power, plus five x squared, minus seven. What is u prime of x? Here we're just going
to use some derivative properties and the power rule. Three times two is six x. Three minus one is two, six x squared. Two times five is 10. Take one off that exponent, it's gonna be 10 x to the first power, or just 10 x. And the derivative of a
constant is just zero, so we can just ignore that. So that's u prime of x. Now, we know that v, if
we inputted an x into v, so v of x would be equal
to x to the eighth power. V prime of x, well we just
use the power rule again. That's eight x to the seventh power. And so v prime of u of x, so if you were to input
u of x into v prime, well it's going to be equal to eight times u of x to the seventh power. Whatever you input into v prime, you're gonna take it to the seventh power and multiply it by eight. So u of x and that's the same thing. That is the same thing. This is equal to eight times
this entire expression. U of x is two x to the third power, plus five x squared, minus seven. So there you have it. F prime of x. It is equal to this, which we just figured
out v prime of u of x is all of this business. So it's equal to eight,
times two x to the third, plus five x squared, minus seven. All of that to the seventh power, times u prime of x. U prime of x, we figured out is that. So times six x squared, plus ten x. Now, as you get more
practice with the chain rule, you'll recognize this faster and actually you won't have
to right all of this down, you'll be able to do a
lot of it in your head. You'll say, okay, I'm
gonna take the derivative of the outside function, the
blue function you could say, with respect to what I have in the inside. So, if I was taking the
derivative of x to the eighth, it would be eight x to the seventh. That's with respect to x. But if I'm taking the derivative of this, with respect to the inside, well, where I had the x's before, I would
just have this u of x. So it's gonna be eight times
this to the seventh power. And I multiply that times
the derivative of the inside. Which is six x squared plus 10 x.