 Let's look at a few different other types of problems involving both use of the product and quotient rules. So in this first problem, we're asked to find f prime of 2 given different types of combinations of f of x, and you can see here the four conditions that were given that g of 2 is equal to 3, g prime of 2 equals negative 2, h of 2 equals negative 1, and h prime of 2 equals 4. So the first thing we're asked to do is in finding f prime of 2, given the fact that the f function is twice the g function plus the h function. So if we want to find, let's just find f prime of 2 for, or of f for x first. So 2 is simply a constant, so that's going to stay put. So we'll have two times the derivative of the g function plus the derivative of the h function. So now we can go ahead and evaluate specifically f prime of 2. So that's going to be 2 times g prime of 2, which you'll notice were given, plus h prime of 2, which we also are given. So we have 2 times g prime of 2 up there is a negative 2 plus h prime of 2 is 4, so in the end for this one we get 0. Let's look at a different scenario for what f of x is given to be. Now we're told f of x is 4 minus h of x. So in general the derivative of the f function then, of course derivative of 4 is 0, minus we would have h prime of x. So in evaluating specifically f prime of 2, we will have 0 minus, and go back to our conditions, h prime of 2 was given to be 4. So we have negative 4 for that answer. Now we have one involving use of the quotient rule. So if you remember how the quotient rule goes, f prime of x, I'm simply going to abbreviate here, is starting in the derivative, you keep the h function times the derivative of g minus keep the g function times derivative of h, and then we need to square the h in the denominator. So in evaluating f prime of 2 then, we need h of 2, which go back, that was negative 1, so we have negative 1, g prime of 2, which was given to be negative 2, minus now we need g of 2, which we were given to be 3, times h prime of 2, which we were given to be positive 4, and that's all over h of 2, that quantity squared, and h of 2 remember was negative 1, so we have negative 1 squared. So when we simplify that one, we come out with negative 10 in the very end. So let's try one now that's a product rule. So using the product rule in general, f prime is going to be g times h prime plus h times g prime. So we want f prime of 2, so we need g of 2, which if you go back, that was given to be 3, h prime of 2, which was given to be 4, plus now we need h prime of 2, which was given to be negative 1, and g prime of 2, which was given to be negative 2. So when you simplify that, you get 14. So really just using the properties and the actual statements of the product and the quotient rules. Now let's take a look at a few problems in which you're working from the graph of a function. If you take a look at this one, the blue function going down diagonally left to right is the g of x function. The upside down v, looks like an absolute value graph, is an f of x function, and notice at the top we're given that h of x is the product of those two functions f and g, and we want to first find h prime of 1. Now you have to remember it's a product rule, all right? So if we want to find h prime of 1, we're going to have f of 1 times g prime of 1 plus g of 1 times f prime of 1. And all of these values we're going to get from the graph. All right, f of 1, so that's just the function value itself right here. f of 1 would be looks like it's right about here, maybe at about one and a half. Now we need g prime of 1. Now we're lucky because the g function looks to be linear with a slope of negative 1, so really the derivative at any x value along that g function is going to be negative 1. So that kind of makes it a little bit easier. Plus now we need g of 1, which looks to be about 3 from the graph. And now we need the slope, the derivative on the f function at 1. Well now here's another case, you'll have to try to estimate the slope of that left side, of the upside down v. If you try to do that I thought it looked to be about 2 or so. So when we evaluate that and simplify it we get 4 and a half. Now we want to find h prime of 2. So really the same process with the product rule, but now we're using h prime of 2. Now if you take a look at the f function though, this could be really quick and easy for you, because notice that's the vertex of the upside down v where x equals 2. So right away we know we cannot take a derivative there. So really without having to do any work at all, we know that one's not going to exist. And that's really because of right there up there at the top of the vertex of the f function we have that point, we have that cusp. So right away we know that doesn't exist. So let's go ahead and try h prime of 3. So if we were to write it out it's going to be f of 3 times g prime of 3. Once again we'll have to get all these values from the graph, plus g of 3 times f prime of 3. So estimating f of 3 from the graph of f, I thought that looked to be about 2. g prime of 3, remember we were saying the derivative anywhere along that line is negative 1. So that remains negative 1, plus g of 3, that looks to be about 1. And f prime of 3, so now we have to estimate the slope over there on that other side of the v going down. So we thought the left side was a positive 2, so this really should be a negative 2 on the right side of the v. So we simplify that, we get negative 4. So let's try one similar to this, but using the quotient rule instead. Otherwise the same idea, same graph. So remember with the quotient rule, let's just write it out general, k prime. So we would keep the denominator times the derivative of the numerator, then it's minus, we would keep f and multiply by the derivative of the denominator, and it's all going to be over g square. So once again we'll be getting all of the values we need from the graph. So the first thing we need is g of 1. So we had thought before that was 3, so let's just keep it that way. f prime of 1, remember that was what we were estimating for the slope on that left side of the v, which we estimated in the earlier problem to be 2, minus, now we need f of 1. And that's what we thought to be about equal to 2. And then g prime of 1, remember the slope anywhere on that g function really looks to be negative 1. So that's all over g of 1 squared, so that would be about 3 squared. So if we simplify that, give it a shot, it should get 8 9ths. So now same thing at 2. Now once again, remember that vertex, so right away we know this doesn't exist. And it's really because of that vertex, you know, when we get to the point of doing f prime of 2, that's really where the problem is going to lie. So we got one more. We need g of 3, which looks to be about 1, times f prime of 3, which we estimated to be about negative 2 on that right side of the upside down v, minus f of 3, which we think is about 2, times the slope anywhere on the g function, which is negative 1. All over 1 squared, because g of 3 looks to be about 1. And this is one that comes out 0. So this is just another example of how you can use product and quotient rules coming right from the graph of given functions in order to figure out other combinations of them and really composite functions as such.