If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:5:31

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 I guess you could 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 can be expressed as the product of two functions of two functions so it can be expressed as f of G of X 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 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 derivative of the inner function with respect to X now this might seem all very abstract in Matthew how do you actually apply it let's try it with a real example let's say we were trying to take the derivative let's say we're trying to take the derivative of the square root the square root of 3x squared 3x squared minus X so how could we define an F and a G so that 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 G of X as being equally as being equal to 3x squared minus X then what is f of G of X well F F of G of X is going to be equal to I'm going to try to keep all the colors accurate and hopefully help for the understanding F of G of X is equal to where everywhere you see the X you replaced with the G of X the principal root of G of X G of X which is equal to which is equal to the principal root of we defined G of X right over here 3 x squared minus X of 3 x 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 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 F prime of G of X well wherever we're in the derivative we saw an X we can replace it with the 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 we can write a G of X to the 1/2 and this is just going to be equal to this is just going to be equal to let me write it right over here it's going to be equal to 1/2 1/2 times all of this business to the negative 1/2 power so 3x squared 3x squared minus X which is exactly so which is exactly what we need to solve right over here F prime of G of X 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 I'm what we're trying to solve right over you're 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 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 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 one-half power so the derivative of that whole thing with respect to Y or something is going to be one-half times that's something to the negative one-half power that's essentially what we're saying but now we have to take the derivative of our something with respect to X 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 6 X to the first or just 6x minus one so this part right over here is just going to be 6x minus one 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 1/2 X 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