Product rule example

So let's see if we can find the derivative with respect to ( x ) of ( F = e^x \cdot \cos(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 respect to ( e^x ). That's, in fact, just ( e^x )." And let me write this down. We know a few things. We know the derivative with respect to ( x ) of ( e^x ). ( e^x ) is ( e^x ). We know how to find the derivative of ( \cos(x) ). The derivative with respect to ( x ) of ( \cos(x) ) is equal to ( -\sin(x) ).

But how do we find the derivative of their product? Well, as you can imagine, this might involve the product rule. 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(x) )—times another expression that involves ( x ), so ( U \cdot V(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' ) of ( x ) times just the second expression—not the derivative of just the second expression—so times ( V(x) ), and then we have plus the first expression—not its derivative—just the first expression, ( U(x) ), times the derivative of the second expression. So the derivative of the second expression.

So what you remember is you have to do these two things here. You're going to end up with two different terms, and 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 ( U' ) derivative of ( U ) times ( V ) is ( U' \cdot V + U \cdot V' ).

Now when you just look at it like that, it seems a little bit abstract, and that might even be a little confusing. But that's why we have a tangible example here. I color-coded it intentionally so we can say that ( U(x) = e^x ) and ( V(x) = \cos(x) ). So ( V(x) = \cos(x) ), and if ( U(x) = e^x ), we know that the derivative of that with respect to ( x ) is still ( e^x ). That's one of the most magical things in mathematics—one of the things that makes it special.

So ( U' ) of ( x ) is still equal to ( e^x ), and ( V' ) of ( x ); ( V' ) of ( x ) we know is ( -\sin(x) ). 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^x ), which is just ( e^x ), times the second expression—not taking its derivative—so times ( \cos(x) ), plus the first expression—not taking its derivative—so ( e^x ) times the derivative of the second expression, so times the derivative of ( \cos(x) ), which is ( -\sin(x) ).

And it might be a little bit confusing because ( e^x ) is its own derivative, but this right over here you can view this as the derivative of ( e^x ), which happens to be ( e^x ). That's what's exciting about that expression or that function. And then this is just ( e^x ) without taking the derivative, of course the same thing.

But anyway, well, now we can just simplify it. This is going to be equal to—we could write this either as ( e^x \cdot \cos(x) - e^x \cdot \sin(x) ). Or if you want, you could factor out an ( e^x ). This is the same thing as ( e^x \cdot (\cos(x) - \sin(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.