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

解

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

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

解答ステップ

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

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

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

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

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

拡張

(2 x - 1) (x^{3} + x + 1) - (3 x^{2} + 1) (x^{2} - x) : - x^{4} + 2 x^{3} + x^{2} + 2 x - 1

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