Answer
$y'=\dfrac{2}{1-x^2}$
Work Step by Step
In order to derivate this function you have to apply the chain rule
Let's make an «u» substitution to make it easier
$f(u) = \ln(u)$
$u = \dfrac{1+x}{1-x}$
Derivate the function:
$f'(u) = \dfrac{u'}{u}$
Now let's find u'
*Note: Here you have to apply quotient rule
$u' = \dfrac{2}{(1-x)^2}$
Then undo the substitution, simplify and get the answer:
$y' =\dfrac{1-x}{1+x} \times \dfrac{2}{(1-x)^2}$
$y'=\dfrac{2}{(1+x)(1-x)}$
$y'=\dfrac{2}{1-x^2}$
