 Hello and welcome to a screencast today about doing derivatives that involve your natural log of function. Okay, so the natural log function recall that if you want or for any positive real numbers x, that the derivative of the natural log of x is 1 over x. Okay, you can go back and watch a previous screencast to see where this came from or read about it in your book or look it up online or something. Here we're going to look at how we can actually use this function or use this new information to build a calculate more complicated derivatives. Okay, the first example I have here today is we want to differentiate y equals x squared natural log of x. And I kind of pause there because I didn't want to give it away, but what's really happening in between these two functions? Well really there is a time sign in here. So you could read this as x squared natural log of x, but even better you can read it as x squared times the natural log of x. Okay, let me get rid of that little mark there. It looked kind of funky. Okay, so we have a multiplication going on. We have to use the product rule. So let me write down product rule just to remind ourselves of what we're doing here. Okay, so the product rule says so y prime is going to be, we want to do the derivative of one function times the other plus and then kind of mix them up. So we'll do the derivative with respect to x of x squared. And that's an old friend times our function this ln of x. And I'm going to put it in cursive there so you can see when I'm writing. And then plus we're going to leave x squared alone. And then we're going to multiply by the derivative with respect to x of ln of x. Okay, so now we just go through and do the derivatives of the pieces that we have and throw everything together. So if we do the derivative of x squared, that if you recall is 2x times ln of x. And I'm going to get rid of all these parentheses in here too. Plus then x squared, that one comes along. We don't do anything with that one times. I'm going to underline these as I'm doing them too. So that one goes here. And that one is the one we're going to figure out now. So this is the one we need to practice. So that is 1 over x. Okay, you can certainly leave your answer like this or we can make it look a little bit nicer anyway. So y prime is going to be equal to 2x times the natural log of x. Plus now how can we pretty up this last term a little bit here? Well, x squared times 1 over x is simply just x. Okay, then if you want to do some factoring with us, I guess you could. But that's all algebra. Our calculus is done and for me this answer looks pretty darn nice. Okay, let's take a look at the next one then. Whoopsie, all those marks got on here. Okay, well I'm erasing, I'm going to talk as well. So here we've got differentiate and our function this time is the natural log of x squared plus 1. Okay, there is a very important difference for you to see between this function here. So natural log of x squared plus 1 and the one that we just did. So that was the natural log. Oops, let me bring this back up again here if it'll cooperate, here we go. x squared times the natural log of x. Okay, so it's definitely important to understand the difference between the two of these because obviously we're going to use something different. So is this a multiplication? Are we going to use the product rule for this one? No, this is a composition. This x squared plus 1 right here is inside of that natural log function. So what rule do we want to use then if we've got a composition? Well that's the chain rule. So remember f of x is our outside function, g of x is our inside function, and then we have to put our pieces together then, you know, how things work. So in this case our outside function is the ln of x. Our inside function is x squared plus 1, because that's what's inside of this natural log function. Okay, let's do our derivatives. So f prime of x, okay, ln of x, the derivative of that is 1 over x. That's what we were just practicing. And g prime, the derivative of x squared plus 1 is 2x. Again, that one should be an old friend, nice polynomial for you. Okay, now how do we put these pieces together again using the chain rule? Well if you recall we have to take the derivative of our outside function, evaluate it at our inside function, so we're going to take this and shovel it into there, and then we're going to multiply by the derivative of that inside function. Okay, so let me write it all out so we can see what that looks like. So we're going to do 1 over x squared plus 1, because that's our inside function times the derivative of that inside function, which is 2x. Okay, so again you can pretty this up just a little bit, so that would just be 2x over x squared plus 1. Okay, so in general what actually ends up happening then with this natural log function, how the chain rule works every single time, is if you have the ln of let's call the function g of x, then if you want to do the derivative of that, that's going to actually end up being the derivative of that inside function over the inside function. Okay, so I don't want to give you another rule to memorize, because all you have to do is just apply the chain rule, but in case you like rules and want to remember something like this, here's another one then to add to your toolbox. Thank you for watching.