Answer
$f'(x) = -\dfrac{8}{(x-1)^2} \bigg (\dfrac{x+1}{x-1} \bigg )^3$
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
$u = \dfrac{x+1}{x-1}$
$f(u) = u^4 $
Derivate the function:
$f'(u) = 4u^3u'$
Now let's find u'
$u' = -\dfrac{2}{(x-1)^2}$
Then undo the substitution, simplify and get the answer:
$f'(x) = 4 \bigg ( -\dfrac{2}{(x-1)^2}\bigg ) \bigg (\dfrac{x+1}{x-1} \bigg )^3$
$f'(x) = -\dfrac{8}{(x-1)^2} \bigg (\dfrac{x+1}{x-1} \bigg )^3$
