# First principle proof for derivatives of $\arcsin{x}$

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

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
It's not cheating, but it only gives the derivative at $x=0$... – Hans Lundmark Jun 4 '12 at 20:50

We may define $\displaystyle\sin^{-1} x$ as $\displaystyle\int^x_0 \frac{1}{\sqrt{1-t^2}}dt$. Then by differentiating both sides, we get that $\displaystyle(\sin^{-1}x)'=\frac{1}{\sqrt{1-x^2}}.$

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