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AP Calc: CHA‑3 (EU), CHA‑3.D (LO), CHA‑3.D.1 (EK), CHA‑3.D.2 (EK)

- [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.

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