Derivative of Composite Natural Logarithm

Question

What is the derivative of $\ln\left(\left(\frac{1-x}{1+x}\right)^2\right)$ with respect to $x$? I used chain rule and my answer was $-4\frac{1-x}{1+x}$. But the answer should be $\frac{-4}{1-x^2}$. Is that right?

Answer 1

I assume you meant $y=\ln\left(\left(\frac{1-x}{1+x}\right)^2\right)$?

Then $y = 2 \left[ \ln(1-x) - \ln(1+x) \right]$.

With chain-rule:

$$\frac{\mathrm dy}{\mathrm dx} = 2 \left[ \frac{-1}{1-x} - \frac{1}{1+x} \right] = -2\cdot\frac{2}{1-x^2} = \frac{-4}{1-x^2}$$