Derivatives of Inverse Functions

If $y$ is a differentiable function of $x$, then (where the inverse exists and $\tfrac{dy}{dx}\neq 0$) $x$ is a function of $y$, and their derivatives are reciprocals:

This inverse-function rule is invaluable when a relation is given as $x$ in terms of $y$ (rather than the usual $y$ in terms of $x$): differentiate to get $\dfrac{dx}{dy}$, then take the reciprocal to obtain $\dfrac{dy}{dx}$ — no need to solve for $y$ explicitly.

Derivatives of inverse trigonometric functions. Applying this idea to the inverse trig functions gives standard results, quoted freely henceforth (each valid on the natural domain of the inverse function):

Combined with the chain rule, e.g. $\dfrac{d}{dx}\tan^{-1}(g(x)) = \dfrac{1}{1+[g(x)]^2}\cdot g'(x)$.

How $\sin^{-1}x$ is obtained. Put $y=\sin^{-1}x$, so $x=\sin y$. Then $\dfrac{dx}{dy}=\cos y$, and by the reciprocal rule $\dfrac{dy}{dx}=\dfrac{1}{\cos y}$. Since $\cos y = \sqrt{1-\sin^2 y}=\sqrt{1-x^2}$ (taking the positive root on the principal range), $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}$. The other inverse-trig results follow the same pattern.