 Let's take a look at an example of an implicit differentiation problem. Remember the idea behind implicit differentiation is that we're working with an implicit function, one in which y is not explicitly stated in terms of x. So as we take the derivative, we just need to treat it a little bit differently, but otherwise we're going to use the same rules that you've learned up to this point. Derivative of x cubed is simply 3x squared. Now for 3x squared y, that derivative is going to require the product rule because we have a product of two different variable terms, an x and a y. So I'm going to offset that in square brackets. So we keep the 3x squared. Now derivative of y is 1, but we need to include dy dx because this is where we need to think about the fact that it is only implied that y is a function of x and we are differentiating here with respect to x. So essentially any time you take the derivative of a y term, we're going to include that dy dx. So that's the end of the first part of the product rule. Then we have plus. Now we're going to keep the y and derivative of 3x squared simply is 6x. Now derivative of y cubed would be 3y squared, but again, it's only implied that y is a function of x here, so we need to include dy dx and on the right derivative of 1 of course is just 0. So now we're just going to rearrange things a little bit. The object here is to solve for dy dx because that's our derivative. So we're going to keep the dy dx terms, these two terms right here. We're going to keep those on the left side of the equation and we're going to move everything else over to the right side. So let me just rewrite what we would have then. Now the dy dx you can essentially treat as a greatest common factor and we're going to factor that out on the left. And then we simply solve for dy dx and we are going to simplify it a little bit when we're done. Because you'll notice that we could divide everything by 3 and so our answer in the end is negative x squared minus 2xy all over x squared plus y squared. Remember that with implicit differentiation, it's pretty common to end up with both x's and y's in your final answer for your derivative. Another thing that perhaps you've noticed is really the only calculus here is that first line in which you found the derivative. After that point, it's all algebra.