derivative of (x+1/(x^3+x-2))

解

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

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

解答ステップ

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

商の法則を適用:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( x + 1 ) ( x ^{3} + x - 2 ) - \frac{d}{d x} ( x ^{3} + x - 2 ) ( x + 1 )}{( x ^{3} + x - 2 ) ^{2}}

\frac{d}{d x} (x + 1) = 1

\frac{d}{d x} (x^{3} + x - 2) = 3 x^{2} + 1

= \frac{1 \cdot ( x ^{3} + x - 2 ) - ( 3 x ^{2} + 1 ) ( x + 1 )}{( x ^{3} + x - 2 ) ^{2}}

簡素化

\frac{1 \cdot ( x ^{3} + x - 2 ) - ( 3 x ^{2} + 1 ) ( x + 1 )}{( x ^{3} + x - 2 ) ^{2}} : \frac{- 2 x ^{3} - 3 x ^{2} - 3}{( x ^{3} + x - 2 ) ^{2}}

= \frac{- 2 x ^{3} - 3 x ^{2} - 3}{( x ^{3} + x - 2 ) ^{2}}