Answer
$-\displaystyle \frac{4x}{(x^{2}-1)^{2}}$
Work Step by Step
$f(x)$ is a quotient, $f=\displaystyle \frac{u}{v}$
$u(x)=x^{2}+1,\quad u^{\prime}(x)=2x$
$v(x)=x^{2}-1,\quad v^{\prime}(x)=2x$
$f^{\prime}(x)=\displaystyle \frac{d}{dx}[\frac{u(x)}{v(x)}]=\frac{u^{\prime}(x)\cdot v(x)-u(x)\cdot v^{\prime}(x)}{[v(x)]^{2}}$
$=\displaystyle \frac{2x\cdot(x^{2}-1)-(x^{2}+1)\cdot 2x}{[x^{2}-1]^{2}}$
$=\displaystyle \frac{2x\cdot(x^{2}-1-x^{2}-1)}{[x^{2}-1]^{2}}$
$=\displaystyle \frac{2x\cdot(-2)}{[x^{2}-1]^{2}}$
$=-\displaystyle \frac{4x}{(x^{2}-1)^{2}}$
