derivative of (3(2x+1)/(sqrt(6x+6x^2)))

解

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

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

解答ステップ

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

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

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

商の法則を適用:

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

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

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

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

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

簡素化

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

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