 This lesson is on product and quotient rules. We are looking at two functions, which make a product, which are f of x and g of x. And what we are going to do is look at what could possibly be f of x, what could be g of x, and what could be just a constant multiplier, which we will factor out when we actually do product and quotient rule for derivatives. So on this a, we have y is equal to x sine x. We can make the x our f of x and the sine x becomes our g of x, multiplying f of x times g of x. For the next problem, we have y is equal to e to the x, tan x over 2. The fraction one-half can be taken out of this as the constant multiplier, and we'll have e to the x tan x, and that makes e to the x the f of x and tan x the g of x. With the next one, we see that we have an x in a square root of x minus 1. Again, the x itself can be f of x and the square root of x minus 1, the g of x. On the next problem, we have 3 to the x and x, and we can see again that 3 to the x can be the f of x and the x can be the g of x. On the following one, we see that the 2 multiplies x and that times the absolute value. Again, the 2 becomes the constant multiplier. The x is our f of x and the absolute value of x is the g of x. On f, we have 2 cosine x as our function. On this one, 2 is the constant multiplier and cosine x is not multiplied by anything else, so that's just the function f of x. We do not have to use product rule on this. On this last one, we see x squared 2 to the x over 3. That means a 1 third can be factored out as the constant multiplier and we'll have x squared 2 to the x. And then the x squared becomes our f of x and the 2 to the x becomes the g of x. So now that we have looked at the different types of products, we need to learn how to do product rule. So what is this product rule? Well, if we have a function y is equal to some f of x function times some g of x function, then the product rule is y prime equals f times the prime of g, f times the derivative of g, plus g times the prime of f, or g times the derivative of f. So it's f g prime plus g f prime. How do we use the product rule? Well, let's do the problem. f of x is equal to x minus 3 times x plus 2. And if you notice, I made this f of x. So I have to change my letters a little bit. This will be now g of x and this will be h of x. So f prime of x will be equal to g of x times h prime of x plus h of x times g prime of x. So putting our functions in, we'll have g of x, which is x minus 3. Then we're going to take the derivative of x plus 2 with respect to x. And we're going to add to that h of x, which is x plus 2. And then we're going to take the derivative of our g of x, which is x minus 3. And that's with respect to x. So let's compute all this. The x minus 3 remains. The derivative of x plus 2 is, well, the derivative of x is 1. The derivative of 2 is 0. So it's times a 1 plus x plus 2 times the derivative of x minus 3. Well, the derivative of x is 1 and the derivative of negative 3 is 0. So that's times 1 also. And if we clean this up, we'll get 2x minus 1. And so now we have used our product rule to take the derivative of the product of 2 functions. Let's try a couple of more problems. Determine f prime of x when f of x is equal to 2 to the x times x to the 1 half. So f prime of x is equal to the 2 to the x times the derivative of x to the 1 half, which is 1 half x to the negative 1 half. Then we'll add to that the x to the 1 half and take the derivative of 2 to the x, which is 2 to the x natural log of 2. Now we clean up. And there are a couple of ways we can clean this up. One is to put the x to the negative 1 half in the denominator and then get a lowest common denominator with that. Another way is just to factor out the x to the negative 1 half. And that seems to be the simplest with this problem. We will also factor out 2 to the x because that is common to both terms. So we have 2 to the x and then x to the negative 1 half taken out of our problem as a factor. And in the first term it looks like we only have 1 half left. And in the next term we have x to the 1 half here. And if we factored out the negative 1 half, that means we have to put back an x in here in order to multiply out to the x to the 1 half there. So it would be x natural log of 2 because we factored out the 2 to the x. So that would be how you would simplify a problem as complicated as this one. For the next problem, if we want to find f prime of x, we have x to the fifth, which is our f of x, and then we take the prime on e to the x, which is e to the x. Add to that the e to the x times the derivative of x to the fifth, which is 5x to the fourth. Again, we will factor out to simplify this. We see there's an e to the x there, and we see an x to the fourth on this term and x to the fifth on that. Again, we take out the lowest one, which is x to the fourth, and we see what's left. Well, this says x to the fifth, so we need an x for this term, plus a 5 for the other term. And that gives us the answer for the derivative of x to the fifth times e to the x. Well, let's look at a proof of this. Suppose we have y equals a product of two functions, f of x and g of x, and we want to take the prime. So we use our definition of the derivative that says the limit as h approaches 0 of f of x plus h times g of x plus h minus f of x g of x. Okay, that's f of x plus h minus f of x all over h, our original way to look at definition of the derivative. What we want to do now is add in an f of x plus h times g of x, and then subtract it off. We have our original part of our definition, and we're going to just add and subtract something. Of course, adding and subtract the same thing means it adds up to 0. So we're going to add this in and then subtract it off. Then we're going to replace it. So now we have the limit as h approaches 0 of f of x plus h times g of x plus h minus g of x all over h, and add to that the limit as h approaches 0 of g of x factored out times f of x plus h minus f of x all over h. And as we begin to see what happens here, this is g prime and this is f prime. So we end up with the limit as h approaches 0 of f of x plus h times g prime of x times g of x f prime of x. Now when we take the limit as h approaches 0 of f of x plus h, this becomes 0 and we're left with f of x. So we have f of x times g prime of x plus g of x times f prime of x, which is the definition that we need for when we use product rule. Let's go on to quotient rule. Well, when we look at a problem that has a quotient, we have to use quotient rule to get that derivative. And of course, you can see very readily that the numerator becomes f of x and the denominator becomes g of x. And of course, when we use quotient rule, g of x cannot be equal to 0 or we just can't do it. So what is the quotient rule? Well, we can use the quotient rule to determine any quotient and it reads y prime is equal to g of x times f prime of x minus f of x times g prime of x all over g squared of x. Now the way I normally do these, I do the square first and I've started doing that by teaching students to make sure that they do that square first, the g squared first, and then do g of x times f prime minus f times g prime. So let's look at an example. Let's say f of x is equal to 12 e to the x over x minus 3. Well, in this problem, I put in a constant multiplier, which we are going to factor out and just put it out front as 12 times e to the x over x minus 3. So we take the prime of this, the first derivative, we just leave it completely out, and then we take the derivative on this quotient. So it's square that denominator, write that denominator up top, and then take the derivative on the numerator, which is e to the x minus we use the e to the x again, and then take the derivative on the denominator, which is the x minus 3, and we get the derivative that is 1. As usual, we have to clean things up. We can see we can factor out e to the x, so we have 12 e to the x times x minus 3 minus 1 over x minus 3 quantity squared, and a final clean up. We have 12 e to the x times x minus 4 over x minus 3 quantity squared, and that will give us the derivative using the quotient rule. Well, let's look at a proof of the quotient rule. Given y is equal to f of x over g of x, and of course g of x does not equal 0, then y prime equals f of x plus h over g of x plus h minus f of x over g of x all over h, and that is taken when the limit approaches 0. Once we get a lowest common denominator in the numerator, and then we can put that denominator into the denominator, we get limit as h approaches 0 of f of x plus h times g of x, that would be the cross multiplication here, minus f of x times g of x plus h, that's that multiplication, and then the denominator would be h times g of x times g of x plus h. Again, we're going to add and subtract something in. So now we're going to find that we have the limit as h approaches 0 of f of x plus h times g of x minus f of x times g of x plus h, which is what we see above, and we're going to add in f of x g of x and then subtract it out f of x times g of x, and this denominator is still the same. We again are going to rearrange things so that we can make derivatives. So once we have rearranged, we have the limit as h approaches 0 of g of x times f of x plus h minus f of x all over g of x times g of x plus h times h minus the limit as h approaches 0 of f of x times g of x plus h minus g of x over g of x times g of x plus h times h. And we can see we have here g of x times f prime of x, which would be the h in the denominator taken with the f of x plus h minus f of x. So that will give us f prime of x. And in the denominator, we are now left with g of x times g of x plus h. And for the second part of this, we have f of x times g prime of x, which is your g of x plus h minus g of x all over that h in the denominator. And in the denominator, we are left with g of x times g of x plus h. Meaning up a little bit more, we see that we have g of x times f prime of x minus f of x times g prime of x all over g squared of x. Remember, we've taken h to be 0. Put that in there so that both denominators become g of x and it becomes a square. This concludes your lesson on product rule and quotient rule for derivatives.