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

解

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

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

解答ステップ

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

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

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

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

拡張

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

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