Worked example: Differentiating related functions | AP Calculus AB | Khan Academy

The differentiable functions X and Y are related by the following equation: the sine of X plus cosine of Y is equal to square root of 2. They also tell us that the derivative of X with respect to T is equal to 5. They also ask us to find the derivative of Y with respect to T when Y is equal to π/4 and 0 is less than X is less than π/2.

So given that they are telling us the derivative of X with respect to T, and we want to find the derivative of Y with respect to T, it's a safe assumption that both X and Y are functions of T. You could even rewrite this equation right over here: you could rewrite it as sin(X), which is a function of T, plus cosine(Y), which is a function of T, is equal to square root of 2.

Now, it might confuse you a little bit. You're not used to seeing X as a function of a third variable or Y as a function of something other than X. But remember, X and Y are just variables. This could be F(T) and this could be G(T) instead of X(T) and Y(T), and that might feel a little bit more natural to you.

But needless to say, if we want to find dY/dT, what we want to do is take the derivative with respect to T of both sides of this equation. So let's do that. We're going to do it on the left-hand side, so it's going to be... we're going to take that with respect to T, the derivative of that with respect to T. We're going to take the derivative of that with respect to T, and then we're going to take the derivative of the right-hand side, this constant, with respect to T.

So let's think about each of these things. So what is this? Let me do it in a new color, the stuff that I'm doing in this aqua color right over here. How could I write that? So I'm taking the derivative with respect to T. I have sin of something which is itself a function of T, so I would just apply the chain rule here. I'm first going to take the derivative with respect to X of sin(X).

I could write sin(X(T)), but I'll just revert back to sin(X) here for simplicity. Then I would then multiply that times the derivative of the inside, you could say with respect to T, times the derivative of X with respect to T. This might be a little bit counterintuitive to how you've applied the chain rule before when we only dealt with X's and Y's, but all that's happening is I'm taking the derivative of the outside of sin of something with respect to the something—in this case, it is X—and then I'm taking the derivative of the something—in this case, X—with respect to T.

Well, we can do the same thing here for this second term here. So I'd want to take the derivative with respect to Y of cosine(Y), and then I would multiply that times the derivative of Y with respect to T. Then all of that is going to be equal to what? Well, the derivative with respect to T of a constant, square root of 2, is a constant; it's not going to change as T changes, so its derivative, its rate of change, is zero.

All right, so now we just have to figure out all of these things. So first of all, the derivative with respect to X of sin(X) is cos(X) times the derivative of X with respect to T. I'll just write that out here, the derivative of X with respect to T. Then we're going to have... it's going to be a plus here, the derivative of Y with respect to T.

So plus the derivative of Y with respect to T. I'm just swapping the order here so that this goes out front. Now, what's the derivative of cosine(Y) with respect to Y? Well, that is negative sine(Y), and so actually, let me just put a sin(Y) here, and then I want to have a negative. So let me erase this and put a negative there, and that is all going to be equal to zero.

What can we figure out now? They've told us that the derivative of X with respect to T is equal to 5; they tell us that right over here, so this is equal to 5. We want to find the derivative of Y with respect to T. They tell us what Y is; Y is π/4. This Y is π/4, so we know this is π/4.

Let's see, we have to figure out what... we still have two unknowns here. We don't know what X is, and we don't know what the derivative of Y with respect to T is, and this is what we need to figure out. So what would X be when Y is π/4? Well, to figure that out, we can go back to this original equation right over here. So when Y is π/4, you get: sin(X) plus cosine(π/4) is equal to square root of 2 times sin(π/4).

If we revert to our unit circle, we're in the first quadrant. If you think in degrees, that's a 45-degree angle; that's going to be square root of 2 over 2. So we can subtract square root of 2 over 2 from both sides which is going to give us sin(X) is equal to... well, if you take square root of 2 over 2 from square root of 2, you're taking half of it away, so you're going to have half of it less.

So sin(X) is equal to square root of 2 over 2. And so what X value, when I take the sine of it—and remember, the angle, if we're thinking in the unit circle, is going to be in that first quadrant—X is an angle in this case, right over here. Well, that's going to be once again π/4.

So this tells us that X is equal to π/4 when Y is equal to π/4. And so we know that this is π/4 as well. So let me just rewrite this because it's getting a little bit messy.

So we know that 5 times cos(π/4) minus dY/dT (the derivative of Y with respect to T, which is what we want to figure out) times sin(π/4) is equal to zero. Let me put some parentheses here just to clarify things a little bit.

All right, so let's see now. It's just a little bit of algebra. Cos(π/4) we already know is square root of 2 over 2, and sin(π/4) is also square root of 2 over 2.

And let's see, if we divide both sides of this equation by square root of 2 over 2, well, what's that going to give us? Well, then this square root of 2 over 2 divided by square root of 2 over 2 is going to be 1. And then 0 divided by square root of 2 over 2 is just still going to be zero.

So this whole thing simplifies to 5 times 1, which is just 5, minus the derivative of Y with respect to T is equal to zero.

And so there you have it. You add the derivative of Y with respect to T to both sides, and we get the derivative of Y with respect to T is equal to 5 when all of these other things are true: when the derivative of X with respect to T is 5, and the derivative, and I should say Y, is equal to π/4.