Use inverse function theorem to evaluate $d/dx \arcsin(x)$ Use inverse function theorem to evaluate $d/dx \arcsin(x)$.
How is this any different then finding it using implicit differentiation? Thanks!

For the moment I'll assume that by the "inverse function theorem" you mean
$$
\frac{d}{dx} f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}.
$$
Let's see:
$$
\begin{align}
y & = f^{-1}(x) \\  \\
f(y) & = x \\  \\
\frac{d}{dx} f(y) & = \frac{d}{dx} x = 1\tag{implicit differentiation} \\  \\
f'(y) \frac{dy}{dx} & = 1 \\  \\
\frac{dy}{dx} & = \frac{1}{f'(y)} = \frac{1}{f'(f^{-1}(x))}.
\end{align}
$$
This uses implicit differentiation to prove the implicit function theorem.
So one answer to "What's the difference?" is that if you're to use the theorem, you just use the conclusion without writing the derivation, whereas if you're to use implicit differentiation, then you're writing out the derivation.  But probably the latter means you're to apply it to a particular concrete case rather than writing something like what appears above, where $f$ could be any function at all rather than some specified particular one.
