 So, what happens if we have somewhat more complicated functions? So for example, let's take a function that I can write as the product of two functions. So in general, the derivative of a product is absolutely not the product of the derivatives. If you apply this rule, I can guarantee that you will almost always be wrong. The derivative of a product is not the product of the derivatives, rather it involves something a little bit more complicated. If I have two differentiable functions, then the derivative of the product of the two functions is, this is somewhat complicated, the first times the derivative of the second, plus the second times the derivative of the other. Now, that's a mess of notation, so a somewhat easier way of writing that uses our prime notation. So if I want to write this using prime notation, there we have the derivative of that product f times g is the same as the first not differentiated times the derivative of the other, plus the second, again, not differentiated, times the derivative of the first. For example, suppose I know the function and derivative, function and derivative, and I want to find the derivative of the product of the two functions. So again, this fg'4 is the same as the derivative of the product of the two functions, and we're going to evaluate that derivative at x equal to 4. So the first thing we probably should do is just write down the product rule. Now our general product rule, if I want to find the derivative of the product, it's the first times derivative of second, plus the second times derivative of first, after we write this down about 10, 20, 30, 40, 500 times, you'll start to remember it. You don't have to commit it to memory, you just have to write it down and use it. So derivative of a product, I can write it down here, I want to find the derivative of the product at 4, and I'll just copy down the product rule, except I'll use the actual x value I'm interested in, and g, if only I knew what f of 4 was, how about that? If only I knew what g'4 was, there it is. And g of 4 up here, f' of 4 right here, so at this point I just spill in the blanks. I know what f of 4 is, g of 4, and the others, and part is part of the problem. We want to simplify this and get our final answer. Derivative of the product at 4 is equal to 17. Well, let's use an actual function. How about the derivative of 3x squared plus 5x plus 4 times 2x minus 8? Again, and importantly for later on, a little bit of analysis helps us decide what we need to do with this. If I wanted to evaluate this function for a given value of x, what would I have to do? Let's see, stuff in parentheses has to get done first, so powers have to get done before then, so let's see, take x squared times 3, take x times 5, and 3x squared 5x plus 4, stuff in parentheses has to get done first, x times 2, subtract 8, figure out what this is, figure out what this is. Finally, multiply the two things together. And the important part of that analysis is the last thing that I do before evaluating this function at any value of x, the last thing I do is a product. And what that means is the first thing I will have to worry about is how do you find the derivative of a product? And let's see, oh, that means I have to use the product rule. So let's go ahead and write that down. So again, derivative of a product is the first times the derivative of the other plus the second times the derivative of the first. And I just have to identify what I'm multiplying together. So I'm looking for the derivative of 3x squared plus 5x plus 4 times 2x minus 8. And let's see, so this is my f function. This is my g function, so following the directions for the product rule. I'm going to start out by just copying my f function over. Then I need my g prime function. Well, here's my g of x. So I need to write down the, well, I'll just write down that I should differentiate it. I need my g function again, 2x minus 8. And then I need the derivative of that first function. Now, it's worth noting what we've done here is we've actually set up what we need to do for the product rule. Don't do anything to this, keep it in parentheses. But I do need to differentiate 2x minus 8. Likewise, 2x minus 8, don't do anything to it. 3x squared plus 5x plus 4, I need to differentiate that. So I know what I need to do at this point and now it's just a matter of doing it. So that 3x squared plus 5x plus 4 doesn't change. I need the derivative of 2x minus 8, that's just going to be 2. The 2x minus 8 doesn't change and I need the derivative of 3x squared plus 5x plus 4, which is going to be 6x plus 5. And there's our derivative. Now, here's a useful check when you're applying the product rule or when we get to it later, the quotient rule or the chain rule. One of the useful things you can do to check your work here is that whenever you find the derivative of something more complicated than a sum or a difference or a constant multiple, you always end up with echoes of the original function. What's an echo? Well, it's not the same thing, but it's often a portion of what you started with. So here our original function had 3x squared plus 5x plus 4 and 2x minus 8. If I look at what I claim to be the derivative, here is an echo of the original function. Here's another echo of the original function. And then I get some extra pieces that are new, that are caused by the process of differentiation. But the general guideline is if you don't see these echoes of the original function, there's a pretty good chance you misapplied one of the derivative rules. One last note before we go on, we can leave it in this form. Going past this form, doing anything more with it, that's algebra and it's a good workout. And if you have nothing better to do for the rest of the day, it's probably worth doing the expanding and simplifying. But this is also a perfectly good form to leave the derivative in. We don't actually need to do the expansion and simplification. How about another function? So let's do a little bit of analysis on this again. This is square root of x times 5 plus 3 over x. And if I want to evaluate this for a given value of x, what do I need to do? Well, I find x, I take the square root of x, hold that thought. I take x, divide it into 3, add 5, figure out what that is. And then I'm going to take this value here, multiplied by the square root of x, and there's my function value. And again, the thing to note, the last thing that we do before we evaluate the function is to multiply. Which means that this function is a product and when I want to differentiate it, I need to invoke the product rule. So what's that going to be? The derivative of a product of two things is the first thing times the derivative of the second, plus the second thing times the derivative of the first. Now, in order to proceed again, this is one of the ones we have to remember that square root of x corresponds to a fractional exponent, and x is in the denominator correspond to a negative exponent. And that will help us simplify our expressions as a prelude to differentiating them. Now, notice here I've kept that square root of x because I'm not doing anything with it, it's just staying by itself. Again, I've kept this 3 over x because this is not being differentiated. But I do need to differentiate square root of x. I do need to differentiate 5 plus 3x to power minus 1. And I can differentiate that, that square root of x times derivative of 5 plus 3x minus 1. Derivative of the 5 is 0. This is constant times function, so the derivative is going to be constant 3 times derivative of x to power minus 1. Minus 1 comes out front, x to power is 1 more, negative 2. And that second term there, 5 plus 3x, don't do anything with that. Derivative of x to the n is nx to power n minus 1. And again, the hardest step in this problem, we would like to simplify this to the following extent. Because the original function was expressed in terms of a square root and an x in the denominator, we don't really want to carry around negative or rational exponents. We should simplify that as follows. And this is a purely algebraic step. Once we have gotten to here, we can leave the derivative in that form. It's just much more stylish to write the derivative in a somewhat simplified form. But again, not really worth going past this in our simplification.