# Dy/dx? Ln(sin^-1(x))

Apr 24, 2018

$\frac{d}{\mathrm{dx}} \ln \left({\sin}^{-} 1 x\right) = \frac{1}{{\sin}^{-} 1 x \sqrt{1 - {x}^{2}}}$

#### Explanation:

The Chain Rule, when applied to logarithms, tells us that if $u$ is some function in terms of $x$, then

$\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \cdot \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

For $\ln \left({\sin}^{-} 1 x\right) ,$ we see:

$u = {\sin}^{-} 1 x$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}$

So,

$\frac{d}{\mathrm{dx}} \ln \left({\sin}^{-} 1 x\right) = \frac{1}{\sin} ^ - 1 x \cdot \frac{1}{\sqrt{1 - {x}^{2}}}$

$\frac{d}{\mathrm{dx}} \ln \left({\sin}^{-} 1 x\right) = \frac{1}{{\sin}^{-} 1 x \sqrt{1 - {x}^{2}}}$