Proof of the derivative of sin(x) | Derivatives introduction | AP Calculus AB | Khan Academy

What we have written here are two of the most useful derivatives to know in calculus. If you know that the derivative of sine of x with respect to x is cosine of x and the derivative of cosine of x with respect to x is negative sine of x, that can empower you to do many more, far more complicated derivatives.

But what we're going to do in this video is dig a little bit deeper and actually prove this first derivative. I'm not going to prove the second one; you can actually use it using the information we're going to do in this one. But it's just to make you feel good that someone's just not making this up, that there is a little bit of mathematical rigor behind it all.

So let's try to calculate it. The derivative with respect to x of sine of x, by definition, this is going to be the limit as delta x approaches 0 of sine of x plus delta x minus sine of x, all of that over delta x. This is really just the slope of the line between the point (x, sine of x) and (x plus delta x, sine of x plus delta x).

So how can we evaluate this? Well, we can rewrite sine of x plus delta x using our angle addition formulas that we learned during our trig identities. So this is going to be the same thing as the limit as delta x approaches zero.

I'll rewrite this using our trig identity as cosine of x times sine of delta x plus sine of x times cosine of delta x, and then we're going to subtract this sine of x up here minus sine of x, all of that over delta x. So this can be rewritten as being equal to the limit as delta x approaches 0 of—let me write this part in red—so that would be cosine of x sine of delta x, all of that over delta x.

And then that's going to be plus—I'll do all of this in orange—all I'm doing is I have the sum of things up here divided by delta x; I'm just breaking it up a little bit. Plus sine of x cosine of delta x minus sine of x, all of that over delta x.

And remember, I'm taking the limit of this entire expression. Well, the limit of a sum is equal to the sum of the limits, so this is going to be equal to—I'll do this first part in red—the limit as delta x approaches 0 of, let's see, I can rewrite this as cosine of x times sine of delta x over delta x, plus the limit as delta x approaches 0 of—and let's see, I can factor out a sine of x here, so it's times sine of x.

I factor that out, and I'll be left with a cosine of delta x minus 1, all of that over delta x. So that's this limit, and let's see if I can simplify this even more. Let me scroll down a bit.

So this left-hand expression I can rewrite. This cosine of x has nothing to do with the limit as delta x approaches 0, so we can actually take that outside of the limit. So we have cosine of x times the limit as delta x approaches 0 of sine of delta x over delta x.

And now we need to add this thing, and let's see how I could write this. So I have a sine of x here. Actually, let me rewrite this a little bit differently. Cosine of delta x minus 1, that's the same thing as 1 minus cosine of delta x times negative 1. And so you have a sine of x times a negative 1, and since the delta x has nothing to do with the sine of x, let me take that out—the negative and the sine of x.

So we have minus sine of x times the limit as delta x approaches 0 of what we have left over is 1 minus cosine of delta x over delta x. Now, I'm not going to do it in this video; in other videos, we will actually do the proof. But in other videos, we have shown using the squeeze theorem, or sometimes known as the sandwich theorem, that the limit as delta x approaches zero of sine of delta x over delta x is equal to one.

And we also showed in another video—and that's based on the idea that this limit is equal to one—that this limit right over here is equal to zero. And so what we are then left with—and I encourage you to watch the videos that prove this—and this, although these are really useful limits to know in general in calculus, that what you're going to be left with is just cosine of x times 1 minus sine of x times 0.

Well, all of this is just going to go away, and you're going to have that it's going to be equal to cosine of x, and you are done.