Question

Calculati : [ln(1-x/1+x)]'

Answer 1

$\left(ln\dfrac{1-x}{1+x}\right)'=\left[ln(1-x)-ln(1+x)\right]'=\dfrac1{1-x}\cdot(-1)-\dfrac1{1+x}\cdot 1=\\\\=\dfrac1{x-1}-\dfrac1{x+1}=\dfrac{2}{x^2-1}.$

Green eyes.