Second derivatives (implicit equations): find expression | AP Calculus AB | Khan Academy

Let's say that we're given the equation that (y^2 - x^2 = 4), and our goal is to find the second derivative of (y) with respect to (x). We want to find an expression for it in terms of (x) and (y). So pause this video and see if you can work through this.

All right, now let's do it together.

Now, some of you might have wanted to solve for (y) and then use some traditional techniques, but here we have a (y^2), and so it might involve a plus or minus square root. Some of y'all might have realized, "Hey, we can do a little bit of implicit differentiation," which is really just an application of the chain rule. So let's do that.

Let's first find the first derivative of (y) with respect to (x). To do that, I'll just take the derivative with respect to (x) of both sides of this equation. And then what do we get? Well, the derivative with respect to (x) of (y^2) — we're going to use the chain rule here. First, we can take the derivative of (y^2) with respect to (y), which is going to be equal to (2y), and then that times the derivative of (y) with respect to (x).

Once again, this comes straight out of the chain rule. Then from that, we will subtract — what's the derivative of (x^2) with respect to (x)? Well, that's just going to be (2x). And then, last but not least, what is the derivative of a constant with respect to (x)? Well, it doesn't change, so it's just going to be equal to (0).

All right, now we can solve for our first derivative of (y) with respect to (x). Let's do that. We can add (2x) to both sides, and we would get (2y) times the derivative of (y) with respect to (x) is equal to (2x). Now, I can divide both sides by (2y), and I am going to get that the derivative of (y) with respect to (x) is equal to (\frac{x}{y}).

Now, the next step is let's take the derivative of both sides of this with respect to (x), and then we can hopefully find our second derivative of (y) with respect to (x). To help us there, actually let me rewrite this, and I always forget the quotient rule — although it might be a useful thing for you to remember — but I could rewrite this as a product, which will help me at least. So I'm going to rewrite this as the derivative of (y) with respect to (x) is equal to (x \cdot y^{-1}).

Now, if we want to find the second derivative, we apply the derivative operator on both sides of this equation — the derivative with respect to (x). Our left-hand side is exactly what we eventually wanted to get, so the second derivative of (y) with respect to (x).

And what do we get here on the right-hand side? Well, we can apply the product rule. So first we can say the derivative of (x) with respect to (x) — well, that is just going to be (1) times the other thing, so times (y^{-1}). Then we have plus (x) times the derivative of (y^{-1}).

So, plus (x) times what’s the derivative of (y^{-1})? Well, first we can find the derivative of (y^{-1}) with respect to (y) — we'll just leverage the power rule there — so that's going to be (-1 \cdot y^{-2}).

Then we would multiply that times the derivative of (y) with respect to (x) — just an application of the chain rule times (\frac{dy}{dx}).

And remember, we know what the derivative of (y) with respect to (x) is. We already solved for that; it is (\frac{x}{y}). So this over here is going to be (\frac{x}{y}), and now we just have to simplify this expression.

This is going to be equal to — and I'll try to do it part by part — that part right over there is just going to be (\frac{1}{y}), and then all of this business — let's see if I can simplify that — this negative is going to go out front, so minus, and then I’m going to have (x \cdot x) in the numerator, and then it's going to be divided by (y^2) and then divided by another (y). So it's going to be minus (\frac{x^2}{y^3}), or another way to think about it: (x^2 \cdot y^{-3}).

And we are done! We have just figured out the second derivative of (y) with respect to (x) in terms of (x) and (y).