 In this video, we are going to determine the derivatives of sine and cosine, which turn out to be each other up to a change of sine. So let's consider first the derivative of sine. We claim that the derivative of sine of x is equal to cosine of x. Similarly, the derivative of cosine of x is going to equal negative sine of x. Let's see why the derivative of sine is equal to cosine. So if we take our function f to be sine of x, we want to compute f prime of x, which by the definition, f prime of x is going to be the limit as h approaches 0 of f of x plus h minus f of x all over h. So this is just the definition of the derivative. This just means that, oh, it's a limit of a difference quotient, right? If we plug in the specific value f, which is sine, into this, we end up with the limit as h approaches 0 of sine of x plus h minus sine of x over h. Now, if you try to approach this with just an algebraic perspective, you'd be really kind of stuck. You really can't foil out the numerator and so to speak, because we're not multiplying anything. And some of the other algebraic techniques we've tried would also not be crazy effective here. Instead, because we have sine and we have this x plus h inside of it, this makes us think about an appropriate trigonometric identity. That is to say, your trigonometric identity sense is tingling right now, telling you that there must be some identity that saves the day. And sure enough, we want to use the angle sum identity, which looks like sine of x plus h. This is equal to sine of x times cosine of h plus cosine of x times sine of h, for which if you didn't have that memorized, you can probably be OK. These are things we sometimes have to look up from time to time. But this is the appropriate identity that we're going to use that is sine of x plus h becomes sine of x cosine of h plus sine of h cosine of x, which we saw there. Yeah, I put those in slightly different words, but in terms of multiplication, it doesn't make much of a difference. This is how we can expand that trigonometric statement right there. So when we're in this setting, we have to, now that we've expanded the sine of x plus h, what I'm going to do is going to start grouping them according to the angle of the sine, because you'll notice that we have a sine of x, we have a sine of h, and we have a sine of x. So I want to bring the sine of x's together and segregate it from the sine of h, for which case if we bring those together, since everything's on top of this factor of h, we're going to get sine of x times cosine of h, just this term right here, minus sine of x, again, they're all over h, and then, like I said, we're going to segregate the sine of h times cosine of x all over h. The reason we want to do that as well, since we have a sine of x in common here, we can factor it out, we get a sine of x times cosine h minus 1 over h, and then we have also this cosine x, I'm going to put that to the side, sine of h, for which then, since we have a limit of two different terms right here, I'm going to write this as a sum of two different limits, we're going to take the limit as h approaches 0 of sine of x times cosine h minus 1 all over h. That's the first one. And then we add to that the second limit, which is the limit as h approaches 0 of cosine of x times sine of h over h, like so. Now you'll notice that with both of these limits, they have these factors of a trigonometric function involving the angle x. We have sine of x and we have cosine of x. But in both of these limits, we're looking at the limit as h approaches 0. So as h gets closer and closer and closer to 0, what does that mean for x? Nothing. x is actually independent of this choice of h here. x is a fixed point in this consideration. As such, with respect to h approaching 0, sine of x is a constant. So we can actually factor down the limit. Same thing with cosine of x. It's constant because the limit is asking what happens as h varies. No variation is on x right now. So we can factor those things out as constants. That gives the following situation right here. Now that we've pulled out the sine of x and cosine of x, you'll notice that we have some limits we need to calculate. But hopefully, this looks like a little bit of deja vu. These are limits that we've calculated already. We have previously shown that the limit as h approaches 0 of sine of h over h, that's going to converge towards the number 1. And we've also shown that the limit as h approaches 0 of cosine h minus 1 over h, that's going to converge to 0. So we get sine of x times 0. That's going to be a 0. We're going to get cosine of x times 1. That's going to be a cosine. So when you simplify that, we see that the derivative of sine turned out to be cosine, which is pretty phenomenal, right? The derivative of the trigonometric function sine is its complementary function, cosine. And so we see that the derivative of sine is equal to cosine. Now similarly, we could use a same type of argument that we saw on this proof right here. We could show that the derivative of cosine of x is equal to negative sine of x. There does need to be a negative sign right here. And where did that negative sign come from? Well, basically, the main difference in the argument, it's basically the same argument, but then the main difference comes from the following observation, the following trig identity. In order to expand the difference quotient, we need to use the angle sum identity for cosine of x plus h. This will look like cosine of x times cosine of h minus sine of x times sine of h. So if you make that one modification, you use that different angle sum identity, and then you go through the simplification of the difference quotient, basically everything else is gonna go the exact same. You take all the same steps, but because there was this negative inside the identity instead of the plus that sine had, that has a consequence of a negative sign popping out at the very end here. So the derivative of cosine of x is negative sine, not positive sine, it's negative sine. So let's put this to practical use right here. We can start computing derivatives that involve sines and cosines. Like what if we wanna compute the derivative of f of x equals three x cosine of x? Notice I can factor this as three x times cosine of x, so I can treat this as a product of two functions. So when I compute the derivative of f, I can use the product rule, which the product rule tells me I'm gonna take the derivative of three x times that by cosine of x, and then I'm gonna add that to three x times the derivative of cosine of x. For which the derivative of three x, as we've seen before, will just be three. So we get three cosine of x. And then the derivative of cosine of x, like we just learned, is negative sine of x. For which we could write this as three cosine of x minus three x sine of x, like so. So knowing the derivative of cosine was very useful in this situation as we combined it with the product rule and other derivative things we've seen. How about g of x equals x squared times sine of x? Same basic idea. By the product rule, we see that the derivative of g right here is gonna be x squared prime times sine of x. And then we're gonna add to that x squared times the derivative of sine, right? For which the derivative of x squared by the power rule is gonna be a two x, we times that by sine. And then the derivative of sine, we've already learned is cosine of x. And that would then give us the derivative of this product function right here. And so we see that calculating derivatives involving sines and cosines is pretty much a cinch if we know the derivative of sine is cosine and the derivative of cosine is a negative sine.