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

May 24, 2018

Recall that $\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$.

So, by the chain rule, when we put a function inside the natural log function, we see that its derivative is

$\frac{d}{\mathrm{dx}} \ln \left(f \left(x\right)\right) = \frac{1}{f} \left(x\right) \cdot f ' \left(x\right)$

So here,

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

If you want, you can simplify the negative/minus signs:

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