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.
First principle proof for derivatives of $\arcsin{x}$
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$\begingroup$
calculus
derivatives
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5Why 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. – 2012-06-04
1 Answers
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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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0+1) No that is not cheating, sine must come into play somehow. – 2012-06-04
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0@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. – 2012-06-04
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0Ok, I see your point. Personally, I like to define $\arcsin$ as the inverse of $\sin$ around 0. – 2012-06-04
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2It's not cheating, but it only gives the derivative at $x=0$... – 2012-06-04