 Hi there, and welcome to the screencast where we're going to do, as you can see from my big blue banner up on top here, some product rule examples. These are just going to be very basic examples here to give you the general gist there, more complicated examples coming up in the next few videos. So let's start by differentiating this function f of x equals x squared plus one times radical x minus one. This is going to be a product rule situation, or it can be a product rule situation because I can see that my f is defined as two functions multiplied by each other. Now we don't necessarily have to use the product rule on this, and we'll come back to that in a minute, but let's use the product rule just to get it under our fingers. So in the language of the product rule, we have two functions that are multiplied together. We'll just call them the first one and the second one. And the product rule itself tells us that the derivative of the function is the derivative of the first function, which I'll denote using my d over dx notations, the derivative of x squared plus one times the second function, kind of just left alone, plus the first function times the derivative of the second function, which I'll again denote using my d over dx notation, or radical x minus one. So what the product rule really does here is it doesn't so much take a derivative as it splits an existing complicated derivative up into little pieces that are easier to do. Let's just do these derivatives, these miniature derivatives, this one and this one here very quickly because they are quite simple. The derivative of x squared plus one is just 2x, and this was already being multiplied by a radical x minus one, and plus from here, and then I had x squared plus one, and the derivative of radical x minus one, just keep in mind that radical x is the same thing as x to the one half power, then I can use my rules for power functions, and that would give me one half x to the minus one half, and the minus one there would differentiate to zero. I'm actually not going to simplify that algebra because I don't want to complicate the issue here. You should definitely attempt to simplify the algebra and check your work using Wolfram Alpha, but I just want to point out what the derivative taking process is over with at this point, and we are left with this. Although I'm not going to simplify the algebra, I will point out that we can actually get around the product rule entirely by doing a little algebraic simplification on the front end of this. If you look at the original f of x definition here, I could just foil these two terms together and get something like x squared times radical x, that's x to the five halves, minus x squared plus radical x minus one. That's equal to the function I'm trying to differentiate, and it's actually a little easier to differentiate it this way if you do a little algebraic simplifying on the front end. So this derivative could be taken two different ways, one by using the product rule at the outset and then simplifying at the end, or by using algebraic simplification at the front end and then using some power function rules as we go. Your choice should end up with the same derivative either way. Now let's move on to another derivative where we're looking at the derivative of g of t equals t cubed minus t plus one times e to the t. There's not really much algebra we can do to simplify here, so we're just going to go ahead and jump straight into the product rule. The product rule, again, would say that g prime of t is, just to write it out, it would be the derivative with respect to t, that's the name of my variable now, t not x, derivative with respect to t of the first function, t cubed minus t plus one times e to the t plus, now here's the second wave of the product rule. It would be the first function, t cubed minus t plus one times the derivative with respect to t of the second function. So now we just have to take the derivatives that are in the two brackets here, but those are really simple. The first derivative is a polynomial, so this is going to give me three t cubed squared, sorry, three t squared minus one plus zero times e to the t plus, and I have t cubed minus t, that's a cube there, minus t plus one, and the derivative with respect to t of e to the t is e to the t itself. Now all the derivative taking is done, let's try to simplify the result a little bit. I see that there is a common factor of e to the t on both of these terms, so I will factor that e to the t completely out, and I'm left with three t squared minus one plus all this other stuff, t cubed minus t plus one, and there's just a tiny bit of simplifying that I can do here, namely I can add the one and the minus one together. So my final derivative that is fully simplified algebraically is e to the t is, or times t cubed plus three t squared minus t, again the minus one, the plus one, subtract off from each other. Finally, let's look at this one last example where we're going to take the derivative of h of x equals x times radical x. This is an example of where you should not use the product rule. Now I know what you're thinking, you're looking at this function and you're trained to think now that anytime I see two things multiplied together and need to take their derivative I'm going to use the product rule, like here's a thing and here's another thing multiplied together, but not so. We don't use the product rule on every single situation where I have one thing times another. This is much more simply done by doing a very quick algebraic simplification step at the beginning, namely getting this into one power of x. H of x is x times radical x, it's the same thing as x to the first power times x to the one half power. So I'm just going to simplify this by writing h of x equal to x to the three half power. So actually h of x is actually not strictly speaking two functions multiplied together despite all appearances, it's really just one single power of x. So that makes the derivative of one liner, h prime of x is now just going to be three halves times x to the one half and that's that. So this is a place where the product rule should definitely not be used. I mean you could, if you could go through and write out the product rule and all of its glory and apply it correctly and you're going to simplify everything down and it'll come out to be x three halves times x to the one half and you'll feel a little bit silly for having done that. So it's much easier if you can to simplify your function at the very beginning and then take it easy on the derivatives than to just pull out the most complicated derivative formula possible and apply it. So that's two examples of the product rule and one example where you should avoid the temptation to use the product rule. Thanks for watching.