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?


You are on the right track. All that's left is to write $\sec(y)$ in terms of $x$. To do this, recall that $y=\arcsin(x)$. That is, $y$ is some angle, sine of which yields $x$. Thus, there is a right triangle with angle $y$, the side opposite $y$ has length $x$ and the hypotenuse is $1$. You need to compute the secant of this angle $y$.
arcsin http://homepage.mac.com/shelleywalsh/MathArt/arcsin.gif
This approach generalizes to finding the derivatives of the other inverse trig functions, and is a good way to wrap your head around composing trig functions and inverse trig functions of any flavor.
