Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

Wat is de afgeleide van $https://latex.codecogs.com/png.image?\dpi{110}\ln\sqrt{\frac{1+x}{1-x}}$	A. \(\boldsymbol{\frac {1} {1\,-\,x^2} }\)
B. \(\boldsymbol{\frac {1\,-\,x} {1\,+\,x} }\)
C. \(\boldsymbol{\frac {1} {1\,-\,x} }\)
D. \(\boldsymbol{-\,\frac 12 }\)
E. \(\boldsymbol{\frac {x\,-\,1} {x\,+\,1} }\)
F. \(\boldsymbol{\frac {-x} {1\,-\,x^2} }\)

$\\D\ln\sqrt{\frac{1+x}{1-x}}=\frac{1}{\sqrt{\frac{1+x}{1-x}}}.D\sqrt{\frac{1+x}{1-x}}=\frac{1}{\sqrt{\frac{1+x}{1-x}}}.\frac{1}{2\sqrt{\frac{1+x}{1-x}}}.D\frac{1+x}{1-x}=\frac12.\frac{1}{\frac{1+x}{1-x}}\frac{1.(1-x)-(1+x)(-1)}{(1-x)^2}\\=\frac12.\frac{1-x}{1+x}.\frac{1-x+1+x}{(1-x)^2}=\frac{1}{2}.\frac{2}{(1+x)(1-x)}=\frac1{1-x^2}$
$=\frac12.\frac{1}{\frac{1+x}{1-x}} \frac{1.(1-x)-(1+x)(-1)}{(1-x)^2}\\ =\frac12.\frac{1-x}{1+x}.\frac{1-x+1+x}{(1-x)^2}=\frac{1}{2}$
$=\frac12.\frac{1-x}{1+x}.\frac{1-x+1+x}{(1-x)^2}=\frac{1}{2}.\frac{2}{(1+x)(1-x)}=\frac1{1-x^2}$