Differentiating related functions intro | Advanced derivatives | AP Calculus AB | Khan Academy

We are told the differentiable functions x and y are related by the following equation: y is equal to the square root of x. It's interesting, they're 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 when x is equal to 9.

So let's just make sure we can understand this. 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 as 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 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.

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.

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, so we multiply by 12. Let’s see, nine to the one-half would be three, and nine to the negative one-half would be one-third. So this is one-third.

This will all simplify to one-half times one-third, which is one-sixth. So we could have a six in the denominator, and then we are going to have a twelve in the numerator, so twelve sixths. So the derivative of y with respect to t when x is equal to 9 and the derivative of x with respect to t is 12 is 2.