#### Answer

$\frac{3x^{4}+4x^{3}-2x^{-1}-x^{-2}}{(x+1)^{2}}$

#### Work Step by Step

Using the quotient rule, we have
$f'(x)=\frac{(x+1)\frac{d}{dx}(x^{4}+x^{-1})-(x^{4}+x^{-1})\frac{d}{dx}(x+1)}{(x+1)^{2}}$
$=\frac{(x+1)(4x^{3}-\frac{1}{x^{2}})-(x^{4}+x^{-1})\times1}{(x+1)^{2}}$
$=\frac{3x^{4}+4x^{3}-2x^{-1}-x^{-2}}{(x+1)^{2}}$