derivative of (x^2-4/(x+1))

解

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

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

解答ステップ

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

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

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

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

簡素化

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

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