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# Worked example: Derivative of √(3x²-x) using the chain rule

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.C (LO)
,
FUN‑3.C.1 (EK)

## Video transcript

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.
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