 Once we have the chain rule, we can do something called implicit differentiation, and this arises in the following context. Sometimes I have a graph where the x and y coordinates are not given in the form y equals some computation I can do on x, but instead they're related by some equation. As a simple example, x squared plus y squared equals 25, that's the equation that gives me a circle. Or I have a somewhat more complicated expression, x cubed plus y cubed equals a times x times y, and that describes a curve that's called the folium of Descartes. And it's an interesting curve that shows up in almost every calculus book, in part because it is a challenge curve. Descartes posed this curve as a challenge to Fermat and said, can you find the slope of the tangent line to it? I bet you can't. And of course Fermat said, ah, no problem, and applied his method and was able to solve the problem that Descartes posed. There's another type of curve that looks like x squared plus y squared squared equals some constant squared times x squared minus y squared, and this describes a curve that's called a Lemniscate, and it looks like an infinity symbol. And there's various other curves that we can describe this way. Now, while we could try and solve these equations and express y as a function of x, it would require a lot of work and might not even be possible. So for the circle, it isn't too bad. For the folium, we have to solve a cubic equation, and that's a lot more difficult. For the Lemniscate, we have to solve an equation of degree 4, and that is again considerably more difficult. So the question is, well, do we have to do this if we're just trying to find the slope of the line tangent to the graph? And the answer is no. We can still find the derivative if we apply the chain rule. And one of the nice things about this is if you know a little bit of calculus, you can get around a lot of messy algebra. Well, let's see how that works for something where we might have an idea of what the actual solution is. Let's consider the graph of the circle, x squared plus y squared equals 4, and we want to find the slope of the line tangent to the graph at some specific point on the circle. So here's where the chain rule and our differential notation is extremely helpful in guiding our work on this problem. So we'll use the chain rule. We'll differentiate both sides of the equation. I want to find the derivative with respect to x, so I'll differentiate both sides, and I can write down what I am doing, and a little bit of analysis goes a long way. Here I have, on the right-hand side, I have derivative of 4, not going to be a problem. On the left-hand side, I have what type of function is this? I take x squared, take y squared, I add. The last thing I have going on over here on the left-hand side is a sum. So this is the derivative of a sum, and I can rewrite that as the sum of the derivatives. Again, if I want to apply the chain rule, and you always want to apply the chain rule, the important thing is we'll drop out everything except the last thing we do. So in both of these, the last thing I do is I square it. So I have the derivative of a square, derivative of a square, derivative of 4, and now I can differentiate. So applying the chain rule, derivative of a square is 2 times this times derivative in both cases. And now put things back where we found them. Originally we had an x in the square, so now this is 2x dx, this is 2y dy, still derivative of 4, and now I can take the final steps. Derivative of x, derivative of y is what I'm looking for. Derivative of constant is going to be 0, so I let those values be found. And now I have an equation that relates the derivative of y with respect to x, and I can apply a little bit of algebra and solve for that value. And now this tells me if I want to find dy dx, I need to know what x is, I need to know what y is. Well, because I'm looking at a specific point, I do actually have those x and y values. So I'll just substitute those in. dy dx equals minus x over y, 4 over negative 3, otherwise known as 4 thirds. And there's the slope of the line tangent to the graph of the circle at this particular point. How about the folium? Remember that this was the curve that Descartes challenged Fermat to find the tangent to, and Fermat found the tangent using his own methods. We can find the slope of the tangent using the tools of calculus. So we have our equation giving the folium, and we're interested in a particular point. So we'll use the chain rule and differentiate both sides of the equation. Again, a little bit of analysis at the beginning goes a long way towards helping us solve the problem. Over on the left hand side, I have x cubed plus y cubed. That's a sum. Derivative of the sum is the sum of the derivatives. Over on the right hand side, I have 9x times y, and that's a product. Derivative of a product, I have to use the product rule. So I'll go ahead and expand those out. So there's my derivative of a sum is the sum of the derivatives. Here's my product rule where I'm going to treat 9x as the first function and y as the second. So that's first derivative second plus second derivative of the first. So again, now I have the derivative of something cubed. So let's go ahead and drop everything except for those last functions. Derivative of a cubed, 3 something squared, and chain rule always applies derivative of the thing inside. So there's my derivatives, 3 something squared, derivative of whatever. 3 something squared, derivative of whatever, and here I just differentiated 9x, derivative of y, dy dx. And skin to garden rule, put everything back where you found it, and I need to find the derivative of x and the derivative of y. Derivative of x is just 1, derivative of y, dy dx. Now I do want to find the derivative slope of the line tangent, so that means I want to solve for dy dx. So I'll just solve for dy dx, and that's just a matter of getting all the dy dxs on one side. I have a useful rule in mathematics in general. If you don't know what to do next, try factoring. Everything over on the right hand side has a dy dx in it, so I can factor that out. And now I can solve for dy dx by dividing through by 9x minus 3y squared. And that gives me my expression for dy dx in terms of x and y. And if I only knew what x and y were, then I could figure out what dy dx is. So I'll substitute in those values, x is 2, y is 4, so I'll substitute those into my expression for the value of dy dx. And after all the dust clears, I get dy dx equal to 12.