## Calculus (3rd Edition)

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