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

解

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

\frac{x ^{4} - 6 x ^{2} - 4 x - 3}{( x ^{2} - 1 ) ^{2}}

解答ステップ

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

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

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

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

拡張

(3 x^{2} + 3) (x^{2} - 1) - 2 x (x^{3} + 3 x + 2) : x^{4} - 6 x^{2} - 4 x - 3

= \frac{x ^{4} - 6 x ^{2} - 4 x - 3}{( x ^{2} - 1 ) ^{2}}