derivative of (1-3x/(6+x))

解

\frac{d}{d x} (\frac{1 - 3 x}{6 + x})

- \frac{19}{( 6 + x ) ^{2}}

解答ステップ

\frac{d}{d x} (\frac{1 - 3 x}{6 + x})

商の法則を適用:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (1 - 3 x) = - 3

\frac{d}{d x} (6 + x) = 1

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

簡素化

\frac{( - 3 ) ( 6 + x ) - 1 \cdot ( 1 - 3 x )}{( 6 + x ) ^{2}} : - \frac{19}{( 6 + x ) ^{2}}

= - \frac{19}{( 6 + x ) ^{2}}