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

解

導関数

\frac{2 x ^{2} - 1}{x + 1}

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

解答ステップ

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

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

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

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

簡素化

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

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