Derive $\frac{d}{dx} \left[\sin^{-1} x\right] = \frac{1}{\sqrt{1-x^2}}$ Derive $\frac{d}{dx} \left[\sin^{-1} x\right] = \frac{1}{\sqrt{1-x^2}}$ (Hint: set $x = \sin y$ and use implicit differentiation)
So, I tried to use the hint and I got:
$x = \sin y$
$\frac{d}{dx}\left[x\right] = \sin y\frac{d}{dx}$
$\frac{dx}{dx} = \cos y \frac{dy}{dx}$
$\frac{dy}{dx} = \frac{1}{\cos y}$
$\frac{dy}{dx} = \sec y$
From here I need a little help.


*

*Did I do the implicit
differentiation correctly?

*How do I use this to help with
the original question?


First, note that $\displaystyle \sin^{-1}: [-1,1] \to [-\frac{\pi}{2}, \frac{\pi}{2}]$
The range is important, as for this range, you have that if $y = \sin^{-1} x$ then $\cos y  = \sqrt{1 - x^2}$, as we have $\sin y = x$ and $\cos y \ge 0$ whenever $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. 
i.e. since $\sin^2 y + \cos^2 y = 1$, we get $\cos y = \pm \sqrt{1 - x^2}$  and since $\cos y \ge 0$, we can say $\cos y = \sqrt{1- x^2}$.
Exercise
Suppose we defined $\sin^{-1}x$ as the unique angle $\theta$ in $[\frac{\pi}{2}, \frac{3 \pi}{2}]$ such that $\sin \theta = x$, what is the derivative of $\sin^{-1} x$?
