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

Differentiating related functions intro

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.D (LO)
,
CHA‑3.D.1 (EK)
,
CHA‑3.D.2 (EK)
Sometimes we have an equation that relates to functions of the same variable. Using the chain rule, we can find the derivatives of those functions with respect to that variable.

Want to join the conversation?

Video transcript

- [Instructor] We are told the differentiable functions X and Y are related by the following equation, Y is equal to square root of X. And it's interesting, they're telling us that they are both differentiable functions, even X is a function, must be a function of something else. Well, they tell us that the derivative of X with respect to T is 12 and they want us to find the derivative of Y with respect to T when X is equal to nine. So, let's just make sure we can understand this. So, they're telling us that both X and Y are functions. Arguably they are both functions of T. Y is a function of X but then X is a function of T so Y could also be a function of T. One way to think about it is if X is equal to F of T, then Y is equal to the square root of X which would just be F of T. Another way to think about it if you took T as your input into your function F, you're going to produce X and then if you took that as your input into the square root function you are going to produce Y. So, you could just view this as just one big box here, that Y is a function of Y but now let's actually answer their question. To tackle it we just have to apply the chain rule. The chain rule tells us that the derivative of Y with respect to T is going to be equal to the derivative of Y with respect to X times the derivative of X with respect to T. So, let's apply it to this particular situation. We're gonna have the derivative of Y with respect to T is equal to the derivative of Y with respect to X. Well, what's that? Well, Y is equal to the principle root of X. You could also write this as Y is equal to X to the one half power. We could just use the power rule. The derivative of Y with respect to X is one half X to the negative one half. So, let me write that down. One half X to the negative one half and then times the derivative of X with respect to T times the derivative of X with respect to T. Well, let's see, we wanna find what we have here in orange, that's what the questions asks us. They tell us when X is equal to nine and the derivative of X with respect to T is equal to 12. So, we have all of the information necessary to solve for this. So, this is going to be equal to one half times nine to the negative one half, nine to the negative one half, times DX/DT, the derivative of X with respect to T is equal to 12, times 12. So, let's see, nine to the one half would be three, nine to the negative one half would be 1/3, so this is 1/3, so this will all simplify to 1/2 times 1/3 is 1/6, so we could have a six in the denominator and then we are going to have a 12 in the numerator. So, 12/6, so the derivative of Y with respect to T when X is equal to nine and derivative of X with respect to 12 is two.