# What is the derivative of f(x)=sin^-1(x) ?

Sep 11, 2014

Most people remember this
$f ' \left(x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$
as one of derivative formulas; however, you can derive it by implicit differentiation.

Let us derive the derivative.
Let $y = {\sin}^{- 1} x$.

By rewriting in terms of sine,
$\sin y = x$

By implicitly differentiating with respect to $x$,
$\cos y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

By dividing by $\cos y$,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} y$

By $\cos y = \sqrt{1 - {\sin}^{2} y}$,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {\sin}^{2} y}}$

By $\sin y = x$,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}$