Derivatives of Inverse Functions

If $g = f^{-1}$, then:

$g'(x) = \frac{1}{f'(g(x))}$

$\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}[\arccos x] = \frac{-1}{\sqrt{1-x^2}}$

$\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}$

With chain rule: $\frac{d}{dx}[\arctan u] = \frac{u'}{1+u^2}$.

$f(x) = x^3 + x$. $f(2) = 10$. Find $(f^{-1})'(10)$.

$f'(x) = 3x^2 + 1$. $f'(2) = 13$.

$(f^{-1})'(10) = \frac{1}{f'(2)} = \frac{1}{13}$.