I was trying to solve the following exercise:
Show that $(\arcsin (x))'=\frac{1}{\sqrt{1-x^2}}$ for all $x\in (-1,1)$.
So far I got that $(\arcsin (x))'=\frac{1}{\cos(\arcsin (x))}$, but i don't know how to go from $\frac{1}{\cos(\arcsin (x))}$ to $\frac{1}{\sqrt{1-x^2}}$. Any help is appreciated.
Best Answer
Setting $g(x)= \arcsin(x) \in [ \frac{-\pi}{2}, \frac{\pi}{2} ]$
$\sin( g(x))=x \longrightarrow g'(x) \cos( g(x))= 1$
On the other hand, as $ g(x) \in [ \frac{-\pi}{2}, \frac{\pi}{2} ]$, we have:
$\cos( g(x))= \sqrt{1- \sin^2(g(x))}=\sqrt{1-x^2} $.
So we're done.