# How do you find the derivative of sin^-1(x)?

Apr 12, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}$

#### Explanation:

To find derivative of ${\sin}^{- 1} x$, we use the concept of function of a function.

Let $y = {\sin}^{- 1} x$, then $x = \sin y$

Taking derivatives of both sides, we get

$1 = \cos y . \frac{\mathrm{dy}}{\mathrm{dx}}$ or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} y$

But $\cos y = \sqrt{1 - {\sin}^{2} y} = \sqrt{1 - {x}^{2}}$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}$