 The chain rule is important as a method of finding derivatives, but what makes it a power tool of differentiation is the use of what's called implicit differentiation. The reason this is useful is that it's often easier to write down some relationship between x and y than it is to write down a formula that gives you the value of y in terms of x. Now, in some cases, if we have this equation that relates x and y, we might be able to solve for y. But this is not one of those cases where we can easily solve for y, and, fortunately, we don't have to. Rather than solving for y, we can differentiate in place using the chain rule, assuming that y is a function of x. Now, in order to do this most efficiently, it helps to introduce something called operator notation, and this goes back to one of the ways we have of indicating the derivative. And, again, dy dx should not be read as a fraction, but it's convenient to read it as d over dx applied to y, where d over dx means take the derivative of what follows with respect to x. So, for example, d over dx of x squared is take the derivative of x squared with respect to x, and you get 2x. The advantage of operating notation is that we can write down an expression without knowing what it actually is equal to, so we can write d over dx of y, and we know this is take the derivative of y with respect to x, which we can write as dy dx. We can combine this operator notation with the use of the chain rule. So, d over dx of y cubed. So, if I want to take the derivative of y cubed, I look at the last thing that we do, which is we cube something, and I'll ignore everything except for this last operation. So, this derivative of y cubed becomes the derivative of cubed. And our chain rule says find the derivative normally, and then multiply by the derivative of the inside function. So, the derivative of cubed is 3 squared times the derivative of... We'll invoke the kindergarten rule and put things back where you found them, and in this case, what was in the was a y, so we put the y back into every set of parentheses. And we still have this unresolved derivative. So, now we find the derivative of y with respect to x, and while we don't know what that is, we can at least write it down, dy over dx. An interesting thing happens if we try to do this with a function that we can differentiate already, such as x cubed. Let's look at the derivative of x cubed this way. So, again, we'll ignore everything except for the last function. This is a cubed function, so I'll drop that out and say that I want to find the derivative of... cubed. And the chain rule says if you want to find the derivative of... cubed, it's 3 squared times the derivative of... Putting everything back where we found it, we have an unresolved derivative, which we can write as dx over dx. Again, this differential notation does not represent a fraction. It represents an operation. In this case, find the derivative of x with respect to x. However, differential notation is a useful form of notation because it suggests a much simpler way to get this result. If we take dx over dx to be 1, which it is, we get the following useful result. You can never go wrong by applying the chain rule. In particular, the worst thing that happens when you apply the chain rule when you don't need to apply the chain rule is you'll end up with a couple of factors of dx over dx, but if dx over dx is replaced with 1, we still get the correct derivative. How about x cubed plus y cubed equals xy? So we'll differentiate with respect to x. Over on the left-hand side, we have a sum, so we can differentiate that as a sum. So that gives us the derivative of x cubed plus the derivative of y cubed. And again, this is the derivative of something cubed. So we'll differentiate. We'll get 3 something squared times the derivative of our something and put things back where you found them and likewise for our other derivative. Now we have the derivative with respect to x of x and the derivative with respect to x of y. dx over dx is just 1 and dy over dx is dy over dx. And so this gives us our left-hand side. On the right-hand side, we have the derivative of a product, so we can use the product rule. So that'll be x times the derivative with respect to x of y, otherwise known as dy dx, plus y times the derivative with respect to x of x and simplifying. And so there's our right-hand side. Now remember, we want to solve for dy dx. So here's a couple of places where dy dx appears, so we'll try and solve our equation. So the first thing we'll do is we'll get all of our dy dx terms onto one side and, if in doubt, factor out. And now I can find dy dx by dividing through by 3y squared minus x and I get my final expression for dy dx. We can also use implicit differentiation to simplify some ordinary differentiation problems. This is especially useful when we have a function like cube root that corresponds to a simpler inverse function. In this case, if I have y equals the cube root of something, then I can eliminate the cube root through a simple algebraic manipulation. And then I can differentiate both sides. Over on the left-hand side, I need the derivative of y cubed. So I need the derivative of something cubed, which will be 3y squared dy dx. And I'll also need the derivative of x squared plus 5x plus 4, which will be 2x plus 5. And since I want to find dy dx, I'll solve by dividing by 3y squared. And while we could leave the derivative in the form 2x plus 5 over 3y squared, a good rule is to answer the question in the same dialect it was asked in. And what that means in this particular case is that since we're given the function as a function of x only, our derivative should also be expressed as a function of x only. And so we know what y is equal to, so we'll substitute that in to get our final expression.