## Calculus, 10th Edition (Anton)

$y'=\dfrac{2}{1-x^2}$
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: $y' =\dfrac{1-x}{1+x} \times \dfrac{2}{(1-x)^2}$ $y'=\dfrac{2}{(1+x)(1-x)}$ $y'=\dfrac{2}{1-x^2}$