 While we can do many derivatives by using basic rules of algebra and a couple of simple derivative rules, if we know a few more derivative rules, we'll be able to handle a lot more derivatives. So we'll introduce the product and quotient rule. So suppose I have a function that is a product of two other things, f of x and g of x, then the derivative is going to be the first function times the derivative of the second, plus the second function times the derivative of the first. So suppose I want to find the derivative of 2x plus 5 times 3x plus 7. Again, the last thing that we do determines the type of function, and in this case this is a product, and we have a way of finding the derivative of a product. And because you found this way on the internet, it must be true. But you might be somewhat skeptical and want to verify that the derivative is actually correct, so let's see if we can find the derivative in two different ways. So first we'll use our product rule. So if our function looks like the product of two things, then the derivative is going to be the first one times the derivative of the second, plus the second one times the derivative of the first. And we'll actually write that down. Now we need to find the derivative of 3x plus 7 and the derivative of 2x plus 5. And at this point the calculus has done. Anything you do after this point falls into the category of algebra. But let's verify this by finding the derivative another way. And we actually found this derivative earlier. We found that the derivative was 12x plus 29. But this isn't what we found as the derivative using the product rule. Is the product rule false? Maybe we should abandon it. Maybe some things on the internet aren't true. Before we completely abandon our product rule we might do a little bit of algebra. And we find that we do actually get the same derivative just in completely different forms. And that's an important thing to remember. Depending on how we find the derivative, we may get two different expressions for the derivative. But as long as we've done everything correctly, those two expressions will be the same after some sort of algebraic or other type of simplification. What if we have a quotient of two functions? Again if you have a hammer every problem looks like a nail. In this case we just got the hammer of the product rule. So let's rewrite our quotient as a product. And here since h of x equals f of x over g of x then cross multiplying gives us h of x times g of x equals f of x. And over on the left hand side I have a product and I know what that will differentiate us. So f of x equals h of x times g of x. So my derivative of f of x will be h of x g prime of x plus g of x h prime of x. Now we actually want to find the derivative of h of x. So we'll solve for it. And while this messy expression is perfectly reasonable as a derivative, let's clean it up a little bit. This is a compound fraction. So let's multiply through by the denominator g of x and after all the dust settles we get this nice expression for the derivative. And so this gives us the quotient rule. If my function is a quotient then the derivative is going to be the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all over the square of the denominator. So let's find the derivative of x squared plus 3x plus 1 over x squared plus 7. And since it's on the internet we may want to verify it. So we will invoke our quotient rule and we'll write it down. Denominator times derivative numerator minus numerator times derivative denominator all over denominator squared. And now we need to find the derivative of x squared plus 3x plus 1 and x squared plus 7. And we get our derivative. So let's verify that this actually is the derivative. So if h of x is x squared plus 3x plus 1 over x squared plus 7, then cross multiplying gives me x squared plus 3x plus 1 equals h of x times x squared plus 7. And we know how to differentiate products. So the derivative of h of x times x squared plus 7 is h of x times derivative of the second plus the second times the derivative of the first. And over on the left hand side I have the derivative of x squared plus 3x plus 1. So I'll fill that in. And now I want to solve for h prime of x. So we'll do some algebra. And after all the dot settles we get this expression for h prime of x which we verify is the same as what we got using the quotient rule. Or let's try a more complicated function. So again the last thing that we do is this type of function we're dealing with. So here if we take a look at this, this is a quotient. And so we can find the derivative of a quotient denominator times derivative of numerator minus numerator times derivative of denominator all over denominator squared. So now we want to find the derivative of this expression which is a product. So we'll use the product rule. First times derivative second plus second times derivative first. But now we need to fill in the derivative of 2x minus 5 and x squared plus 7. And we get this as our derivative. We also need the derivative of x cubed plus 8 so we'll differentiate that. And we'll substitute in our values to get our final form of the derivative.