Sal differentiates eˣcos(x) using the product rule.
- [Voiceover] So let's see if we can find the derivative with respect to x, with either x times the cosine of x. And like always, pause this video and give it a go on your own before we work through it. So when you look at this you might say, "well, I know how to find "the derivative with e to the x," that's infact just e to the x. And let me write this down. We know a few things. We know the derivative with respect to x of e to the x. E to the x is e to the x. We know how to find the derivative cosine of x. The derivative with respect to x of cosine of x is equal to negative sine of x. But, how do we find the derivative of their product? Well as you can imagine, this might involve the product rule. And let me just write down the product rule generally first. So if we take the derivative with respect to x of the first expression in terms of x, so this is, we could call this u of x times another expression that involves x. So u times v of x. This is going to be equal to, and I'm color-coding it so we can really keep track of things. This is going to be equal to the derivative of the first expression. So I could write that as u prime of x times just the second expression not the derivative of it, just the second expression. So times v of x and then we have plus the first expression, not its derivative, just the first expression. U of x times the derivative of the second expression. Times the derivative of the second expression. So the way you remember it is, you have these two things here, you're going to end up with two different terms. In each of them, you're going to take the derivative of one of them, but not the other one, and then the other one you'll take the derivative of the other one, but not the first one. So, the derivative of u times v is u prime times v, plus u times v prime. When you just look at it like that, it seems a little bit abstract and that might even be a little bit confusing, but that's why we have a tangible example here and I color-coded intentionally. We could say that u of x is equal to e to the x. And v of x is equal to the cosine of x. So v of x is equal to cosine of x. And if u of x is equal to e to the x, we know that the derivative of that with respect to x is still e to the x. That's one of the most magical things in mathematics. One of the things that makes e so special. So u prime of x is still equal to e to the x. And v prime of x, we know as negative sine of x. Negative sine of x, and so, what's this going to be equal to? This is going to be equal to the derivative of the first expression. So, the derivative of e to the x which is just, e to the x, times the second expression, not taking it's derivative, so times cosine of x. Plus the first expression, not taking its derivative, so e to the x, times the derivative of the second expression. So, times the derivative of cosine of x which is negative sine. Negative sine of x. And it might be a little bit confusing, because e to the x is its own derivative. This right over here, you can view this as this was the derivative as e to the x which happens to be e to the x. That's what's exciting about that expression, or that function. And then this is just e to the x without taking it's derivative - they are of course, the same thing. But anyway, now we can just simplify it. This is going to be equal to... We could write either as e to the x times cosine of x, times cosine of x minus e to the x. E to the x times sine of x. Times sine of x. Or, if you want, you could factor out an e to the x. This is the same thing as e to the x times cosine of x minus sine of x. Cosine of x minus sine of x. So hopefully this makes the product rule a little bit more tangible. And once you have this in your tool belt, there's a whole broader class of functions and expressions that we can start to differentiate.