# What is the derivative of y= ln(1-x^2)^(1/2)?

Oct 29, 2016

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

${y}^{2} = \ln \left(1 - {x}^{2}\right)$ *To be continued...

Differentiate the left hand side of the equation using implicit differentiation. Differentiate the right hand side of the equation using the rule:

If $g \left(x\right) = \ln \left(f \left(x\right)\right)$, $g ' \left(x\right) = \frac{f ' \left(x\right)}{f} \left(x\right)$

*Continued...

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

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

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

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

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

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

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

To tidy things up a bit, you can multiply the fraction to the right by one, which is the same as $\frac{- 1}{- 1}$.

So what you end up with is...

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