# 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!

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Do you mean the theorem that gives the derivative of the inverse of $f$ in terms of $f$? It's different in that you are just using a formula to obtain the derivative. Of course, the second step when using implicit differentiation brings you that formula. – David Mitra Apr 12 '12 at 15:51

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}
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.