derivative f(x)=(x^2+2x-5)/((x+1)^2)

解

導関数

f (x) = \frac{x ^{2} + 2 x - 5}{( x + 1 ) ^{2}}

\frac{12}{( x + 1 ) ^{3}}

解答ステップ

\frac{d}{d x} (\frac{x ^{2} + 2 x - 5}{( x + 1 ) ^{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 ^{2} + 2 x - 5 ) ( x + 1 ) ^{2} - \frac{d}{d x} ( ( x + 1 ) ^{2} ) ( x ^{2} + 2 x - 5 )}{( ( x + 1 ) ^{2} ) ^{2}}

\frac{d}{d x} (x^{2} + 2 x - 5) = 2 x + 2

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

= \frac{( 2 x + 2 ) ( x + 1 ) ^{2} - 2 ( x + 1 ) ( x ^{2} + 2 x - 5 )}{( ( x + 1 ) ^{2} ) ^{2}}

簡素化

\frac{( 2 x + 2 ) ( x + 1 ) ^{2} - 2 ( x + 1 ) ( x ^{2} + 2 x - 5 )}{( ( x + 1 ) ^{2} ) ^{2}} : \frac{12}{( x + 1 ) ^{3}}

= \frac{12}{( x + 1 ) ^{3}}