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

Feb 16, 2017

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

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{chain rule}}$

color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx[ln(f(x))]=1/(f(x))xx f'(x))color(white)(2/2)|))

$y = \ln \left(1 - {x}^{2}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 - {x}^{2}} \times \frac{d}{\mathrm{dx}} \left(1 - {x}^{2}\right)$

$\textcolor{w h i t e}{\times \times x} = - \frac{2 x}{1 - {x}^{2}}$