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CHA‑3 (EU)

, CHA‑3.D (LO)

, CHA‑3.D.1 (EK)

, CHA‑3.D.2 (EK)

we are told the differentiable functions x and y are related by the following equation y is equal to square root of x it's interesting there's telling us that they're 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 what X is equal to 9 so let's just make sure we can understand this so they're telling us that both x and y are functions arguably they're 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 is just one big box here that Y is a function of T 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 going to 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 principal 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 1/2 X to the negative 1/2 so let me write that down 1/2 X to the negative 1/2 and then time's the derivative of X with respect to T times the derivative of X with respect to T let's see we want to find what we have here in orange that's what the question asks us they tell us when X is equal to 9 and the derivative of X with respect to T is equal to 12 so we have all the information necessary to solve for this so this is going to be equal to 1/2 times 9 to the negative 1/2 9 to the negative 1/2 times DX DT the derivative of X with respect to T is equal to 12 times 12 so let's see 9 to the 1/2 would be 3 9 to the negative 1/2 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 can have a 6 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 9 and derivative of X with respect to T is 12 is 2

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