Answer
$S^{\prime}(x)=\dfrac{-2x \tan x}{ (x^2-1)^{2}} +\dfrac{ \sec^2x}{(x^2-1)}$
Work Step by Step
We have: $S(x)=\dfrac{\tan x}{(x^2-1)}=\tan x (x^2-1)^{-1}$
We need to differentiate both sides with respect to $x$.
$S^{\prime}(x)=\dfrac{d}{dx} [\tan x (x^2-1)^{-1}]\\=\tan x \times (-1)(x^2-1)^{-2} \dfrac{d}{dx}(x^2-1)+(x^2-1)^{-1} \times \sec^2x\\=-\tan x (x^2-1)^{-2} \times 2x+\dfrac{ \sec^2x}{(x^2-1)}$
Therefore, $S^{\prime}(x)=\dfrac{-2x \tan x}{ (x^2-1)^{2}} +\dfrac{ \sec^2x}{(x^2-1)}$
