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

解

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

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

解答ステップ

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

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

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

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

拡張

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

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