# How do you find the second derivative of y= ln(1-x^2)^(1/2) ?

Oct 30, 2016

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

#### Explanation:

$y = \ln {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}$ can be rewritten as ${y}^{2} = \ln \left(1 - {x}^{2}\right)$

Differentiating wrt $x$ gives:
$\frac{d}{\mathrm{dx}} \left({y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(\ln \left(1 - {x}^{2}\right)\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} \left({y}^{2}\right) = \frac{d}{\mathrm{dx}} \left\{\ln \left(1 - {x}^{2}\right)\right\}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y\right) = \frac{1}{1 - {x}^{2}} \left(- 2 x\right)$

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

$\therefore \ln {\left(1 - {x}^{2}\right)}^{\frac{1}{2}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{x}{{x}^{2} - 1}$

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