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  • What I want to do in this video is start with the abstract--

  • actually, let me call it formula for the chain rule,

  • and then learn to apply it in the concrete setting.

  • So let's start off with some function,

  • some expression that could be expressed

  • as the composition of two functions.

  • So it can be expressed as f of g of x.

  • So it's a function that can be expressed as a composition

  • or expression that can be expressed

  • as a composition of two functions.

  • Let me get that same color.

  • I want the colors to be accurate.

  • And my goal is to take the derivative of this business,

  • the derivative with respect to x.

  • And what the chain rule tells us is

  • that this is going to be equal to the derivative

  • of the outer function with respect to the inner function.

  • And we can write that as f prime of not x, but f prime

  • of g of x, of the inner function.

  • f prime of g of x times the derivative

  • of the inner function with respect to x.

  • Now this might seem all very abstract and math-y.

  • How do you actually apply it?

  • Well, let's try it with a real example.

  • Let's say we were trying to take the derivative

  • of the square root of 3x squared minus x.

  • So how could we define an f and a g

  • so this really is the composition

  • of f of x and g of x?

  • Well, we could define f of x.

  • If we defined f of x as being equal to the square root of x,

  • and if we defined g of x as being equal to 3x squared

  • minus x, then what is f of g of x?

  • Well, f of g of x is going to be equal to-- I'm

  • going to try to keep all the colors accurate,

  • hopefully it'll help for the understanding.

  • f of g of x is equal to-- where everywhere you see the x,

  • you replace with the g of x-- the principal root of g of x,

  • which is equal to the principal root of-- we

  • defined g of x right over here-- 3x squared minus x.

  • So this thing right over here is exactly

  • f of g of x if we define f of x in this way

  • and g of x in this way.

  • Fair enough.

  • So let's apply the chain rule.

  • What is f prime of g of x going to be equal to,

  • the derivative of f with respect to g?

  • Well, what's f prime of x?

  • f prime of x is equal to-- this is the same thing

  • as x to the 1/2 power, so we can just apply the power rule.

  • So it's going to be 1/2 times x to the--

  • and then we just take 1 away from the exponent, 1/2 minus 1

  • is negative 1/2.

  • And so what is f prime of g of x?

  • Well, wherever in the derivative we saw an x,

  • we can replace it with a g of x.

  • So it's going to be 1/2 times-- instead

  • of an x to the negative 1/2, we can write a g of x to the 1/2.

  • And this is just going to be equal to-- let

  • me write it right over here.

  • It's going to be equal to 1/2 times

  • all of this business to the negative 1/2 power.

  • So 3x squared minus x, which is exactly what we

  • need to solve right over here. f prime of g of x

  • is equal to this.

  • So this part right over here I will--

  • let me square it off in green.

  • What we're trying to solve right over here,

  • f prime of g of x, we've just figured out

  • is exactly this thing right over here.

  • So the derivative of f of the outer function with respect

  • to the inner function.

  • So let me write it.

  • It is equal to 1/2 times g of x to the negative 1/2,

  • times 3x squared minus x.

  • This is exactly this based on how we've defined

  • f of x and how we've defined g of x.

  • Conceptually, if you're just looking

  • at this, the derivative of the outer thing,

  • you're taking something to the 1/2 power.

  • So the derivative of that whole thing

  • with respect to your something is going to be 1/2 times

  • that something to the negative 1/2 power.

  • That's essentially what we're saying.

  • But now we have to take the derivative of our something

  • with respect to x.

  • And that's more straightforward.

  • g prime of x-- we just use the power rule for each

  • of these terms-- is equal to 6x to the first,

  • or just 6x minus 1.

  • So this part right over here is just going to be 6x minus 1.

  • Just to be clear, this right over here

  • is this right over here and we're multiplying.

  • And we're done.

  • We have just applied the power rule.

  • So just to review, it's the derivative

  • of the outer function with respect to the inner.

  • So instead of having 1/2x to the negative 1/2,

  • it's 1/2 g of x to the negative 1/2,

  • times the derivative of the inner function with respect

  • to x, times the derivative of g with respect

  • to x, which is right over there.

What I want to do in this video is start with the abstract--

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A2 初級

Worked example: Derivative of Ã(3x_-x) using the chain rule | AP Calculus AB | Khan Academy

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    yukang920108 に公開 2022 年 09 月 09 日
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