Second derivatives (implicit equations): find expression

- [Instructor] Let's say that we're given the equation that y squared minus x squared is equal to four. And our goal is to find the second derivative of y with respect to x, and we want to find an expression for it in terms of x's and y's. 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 squared, and so it might involve a plus or a minus square root. And so 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. And 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 squared, we're gonna use the chain rule here. First, we can take the derivative of y squared with respect to y, which is going to be equal to two y, and then that times the derivative of y with respect to x. Once again, this comes straight out of the chain rule. And then, from that, we will subtract, what's the derivative of x squared with respect to x? Well, that's just going to be two x. 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 zero. All right, now we can solve for our first derivative of y with respect to x. Let's do that. We can add two x to both sides, and we would get two y times the derivative of y with respect to x is equal to two x. And now I can divide both sides by two y, and I am going to get that the derivative of y with respect to x is equal to x, x over 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. And 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 times y to the negative one power, y to the negative one power. And now, if we want to find the second derivative, we apply the derivative operator on both sides of this equation, derivative with respect to x. And 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 one times the other thing, so times y to the negative one power, y to the negative one power. And then we have plus x times the derivative of y to the negative one. So plus x, what's the, times, what's the derivative of y to the negative one power? Well, first, we can find the derivative of y to the negative one power with respect to y. We'll just leverage the power rule there. So that's going to be negative one times y to the negative two power. And then we would multiply that times the derivative of y with respect to x, just an application of the chain rule, times dy/dx. And remember, we know what the derivative of y with respect to x is. We already solved for that. It is x over y. So this over here is going to be x over y. And so 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 a one over 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 times x in the numerator. And then it's going to be divided by y squared and then divided by another y. So it's going to be minus x squared over y to the third, over y to the third, or another way to think about it, x squared times y to the negative three. And we are done. We have just figured out the second derivative of y with respect to x in terms of x's and y's.