 Let's look at a couple of examples of how to use the product rule for finding derivatives. Suppose we have the function h of x equals x squared plus 1 times the quantity x squared minus 2x. Now according to the product rule, you need to identify two different variable expressions. So let's consider this first one, the x squared plus 1 as the f function, and the other one as our g function. So as we go to find the derivative, remember you keep the first function, so we shall keep the x squared plus 1. And now we multiply by the derivative of the second function, the g expression. So that derivative would be 2x minus 2. You're simply using the power rule in order to get that derivative. Now it's always a plus. Now we keep the second function, the x squared minus 2x, and we need to multiply now by the derivative of the first function of the f function, that's the x squared plus 1, and according to the power rule that would be 2x. Now of course from here we could multiply it out, simplify it, we will of course get a polynomial when we're done, but otherwise just for the sake of showing you how the product rule works, you know, it's sufficient to simply stop here. Let's look at another example. Now this one we have once again two different quantities multiplied together, so this first one would be the f function, this one is our g function. So according to the product rule, we're going to keep the first function, now we need to multiply by the derivative of the second function. Of course that's x raised to the one-third power, so as we find our derivative that's going to be one-third x to the negative two-thirds, and according to the power rule, the rest of the derivative is simply 18x. Then we have plus. Now we're going to keep the second function, and we need to multiply by the derivative of the first function, so square root of x again is x to the one-half power, so the derivative of that is going to be one-half x to the negative one-half, and then we have minus three to finish out using the power rule. And once again, simply for the sake of demonstrating how the product rule goes, it's sufficient to just stop here. So let's look at one more example. In this one we're asked to find the equation of the tangent line to y equals 2x squared minus 6x times the quantity negative x minus one at the x value of negative two. So of course remember to write the equation of any line. We need to have the slope and where is the slope coming from but the derivative. So we're going to go ahead and find our derivative. So remember how it's going to go with the product rule. We're going to keep the first function, multiply by the derivative of the second, so that's simply negative one plus now we keep the second function and we multiply by the derivative of the first. Now the slope we're going to get by evaluating that derivative at negative two. So I shall leave that to you and try the number crunching, but hopefully when you give it a shot you get negative 34. So that's the slope we want to use. Now the other thing we need is the original point that would be on the curve. So we do need to evaluate the original function at negative two. So if you substitute negative two into that original function, once again I'll leave you to do the number crunching, but you do get 20. So we can utilize point slope form of the line and simply substitute in. So we'd have y minus 20 equals negative 34 times the quantity x plus two. Remember it's minus and negative two, so it turns into plus two. You can leave it like that. If you care to bring the 20 over, you are most welcome to do that as well. You can multiply it out and put it in slope intercept form if you wish, otherwise it's fine to stop there.