 Let's recap the major ideas of section 2.7 and act of calculus on derivatives of functions given implicitly. So the main idea of this section is to use the chain rule to find a derivative where complete information about the function involved is not known. We've actually seen something like this before back in screencast 257. When we found the derivative of a function that was the composite of two other functions, but neither of whose formulas we knew. What we did back then was use the chain rule to get as far as we could to set up a skeleton framework for a formula. And then we use the result of that formula to find the answer which involved estimating derivative values off of a graph. What we're doing in this section is fairly similar to that. We're given an expression. This is not necessarily a function of x, but rather just a mathematical expression that combines both x and y in the same place, usually in equation. And y here is taken to be a function of x that is unknown but which is known to be differentiable. So this situation comes up quite often in real life and we're given a curve that is not the graph of a function, but it's traced out by an expression. And we would like to find the slope of the tangent line to this curve at a particular point. Define dy over dx which is the slope of the tangent line for such an expression. The basic idea is to take the derivative with respect to x of both sides of the equation that I'm given. And then treat y as a function of x, again unknown but known to be differentiable. And whenever I encounter a y in the derivative taking process, I want to use the chain rule. For example here, the derivative of y squared with respect to x would be 2y times y prime. Because we would treat y squared as a composite function whose inner function is y and the outer function is x squared. And here the derivative of x times y would have to be done using the product rule. Once the derivative process is taken place, we simply solve for dy over dx as if it were a variable and we will have our derivative expression. This is the basic idea behind derivatives of implicit functions. But it makes more sense when we actually do it. So we're gonna now move on to a couple of videos where this idea is instantiated and you should make sure of course also to practice and analyze other examples as well.