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  • - [Instructor] 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 gonna 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.

  • So the derivative with respect to x of sine of x,

  • by definition, this is going to be the limit

  • as delta x approaches zero

  • of sine

  • of x plus

  • delta x minus

  • sine of x,

  • all of that over delta,

  • all of that over delta x.

  • This is really just the slope of the line between the point

  • x comma sine of x and x plus delta x

  • comma 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 write, rewrite this using our trig identity as cosine,

  • 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,

  • let me see if I can draw a relatively straight line,

  • all of that over

  • delta x.

  • So this can be rewritten as being equal to the limit

  • as delta x approaches zero 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 zero 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 zero

  • 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 one,

  • 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 zero,

  • so we can actually take that outside of the limit.

  • So we have the cosine of x times the limit

  • as delta x approaches zero

  • 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.

  • But actually, let me rewrite this a little bit differently.

  • Cosine of delta x minus one,

  • that's the same thing as one minus cosine of delta x

  • times negative one.

  • And so you have a sine of x times a negative one.

  • 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 zero of,

  • what we have left over is one minus cosine of delta x

  • over

  • delta x.

  • Now, I'm not gonna 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 is delta x approaches zero

  • of sine of delta x over delta x,

  • that this is equal to one.

  • And we also show 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 gonna be left with is just cosine of x

  • times one minus sine of x times zero.

  • Well, all of this is just gonna go away,

  • and you're gonna have,

  • that is going to be equal to cosine of x.

  • And you are done.

- [Instructor] What we have written here are two

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