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Proof of the derivative of sin(x)

Proving that the derivative of sin(x) is cos(x).

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  • ohnoes default style avatar for user Darth Vader
    At , Sal finished writing a very long expression:
    lim ∆x->0 [(cos x sin∆x + sin x cos ∆x - sin x)/x]
    I tried evaluating and got a wrong answer that the whole limit =(sinx-sinx)/x= 0/x, but why can't I just evaluate the whole thing here instead of using the limit properties and go through a lot of steps to get the final answer? Why is it wrong to do that? Does it have something to do with the limit properties?
    (3 votes)
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    • mr pink red style avatar for user David Lee
      If you just evaluate it without using properties, you'll get 0/0 which is not possible. Therefore, you need to use the properties to get rid of this problem. Also, if you actually use a graphing tool to draw it, and find the limit, you'll get the value that you got when you evaluate it.
      (5 votes)
  • aqualine ultimate style avatar for user xyzPoKeFaNxyz
    how can you change sin(x+ delta x)into cosx*sin(delta x)+sinx*cos(delta x)
    how can i find the formula?
    (3 votes)
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  • male robot hal style avatar for user Joseph Arcila
    How does it make sense what Sal did on to take out the cosx like that
    (1 vote)
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    • starky ultimate style avatar for user KLaudano
      One of the properties of limits is that the limit of f(x)*g(x) = limit of f(x) * limit of g(x). Sal applied this rule to transform the original limit into the product of the limits of cos(x) and sin(Δx)/Δx. Since cos(x) does not change with respect to Δx, the limit of cos(x) is simply cos(x). This left us with cos(x) * limit of sin(Δx)/Δx.
      (2 votes)
  • blobby green style avatar for user bakhshnabi983
    Limit of sin(1÷x) as x approaches to 0
    (1 vote)
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  • blobby green style avatar for user mahektalreja012
    Why is lim (as ∆x ->0)(sin∆x/∆x) = 1?
    and why is lim (as ∆x ->0)((1-cos∆x)/∆x) = 1?
    (1 vote)
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  • aqualine ultimate style avatar for user vedantkhanna
    What is the derivative of delta x? I was trying to use L'Hopital's rule to find the limit of sin x/delta x but I hit a roadblock ;(
    (0 votes)
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    • leaf green style avatar for user kubleeka
      Δx is a variable. If you're trying to use l'Hôpital's rule, you need to differentiate with respect to Δx, and the derivative of a variable with respect to itself is 1.

      But using l'Hôpital's rule doesn't help here anyway, because that requires knowing the derivative of sine.
      (2 votes)

Video transcript

- [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.