derivative f(x)=(x^2-3x+4)/(x+2)

解

導関数

f (x) = \frac{x ^{2} - 3 x + 4}{x + 2}

\frac{x ^{2} + 4 x - 10}{( x + 2 ) ^{2}}

解答ステップ

\frac{d}{d x} (\frac{x ^{2} - 3 x + 4}{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 ^{2} - 3 x + 4 ) ( x + 2 ) - \frac{d}{d x} ( x + 2 ) ( x ^{2} - 3 x + 4 )}{( x + 2 ) ^{2}}

\frac{d}{d x} (x^{2} - 3 x + 4) = 2 x - 3

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

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

簡素化

\frac{( 2 x - 3 ) ( x + 2 ) - 1 \cdot ( x ^{2} - 3 x + 4 )}{( x + 2 ) ^{2}} : \frac{x ^{2} + 4 x - 10}{( x + 2 ) ^{2}}

= \frac{x ^{2} + 4 x - 10}{( x + 2 ) ^{2}}