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One popular proof is to take $\sin (y) = x$ and then differentiate on both sides. But how do you prove it from first principles? Help very much appreciated.

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Why isn't that proof from first principles? It's a general principle that if you know the derivative of a function, and it has an inverse, then you know the derivative of the inverse, just by the chain rule. –  Ben Millwood Jun 4 '12 at 15:16
    
you can find the answer in the book written by thapelo vincent sello on amazon.com –  Seputla Mar 8 at 9:41
    
@Seputla: there are three books by that author on amazon.com. Which one do you mean? –  robjohn Mar 8 at 11:14
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1 Answer

Is this cheating? We want the limit as $h$ approaches $0$ of $\frac{\arcsin h-0}{h}$. Let $w=\arcsin h$. So we are interested in the limit of $\frac{w}{\sin w}$ as $w$ approaches $0$. Upside down, but familiar!

Now we know the derivative at $0$. We can get the derivative at $x$ by using the $\arcsin$ version of the addition law for sines.

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+1) No that is not cheating, sine must come into play somehow. –  AD. Jun 4 '12 at 15:14
    
@AD.: One can imagine defining $\arcsin$ first, for example as an integral. Just like with exponential/logarithm, we can take one or the other as fundamental. –  André Nicolas Jun 4 '12 at 15:18
    
Ok, I see your point. Personally, I like to define $\arcsin$ as the inverse of $\sin$ around 0. –  AD. Jun 4 '12 at 15:21
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It's not cheating, but it only gives the derivative at $x=0$... –  Hans Lundmark Jun 4 '12 at 20:50
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