d/(dt)((3t^2)/(1+t^3))

解

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

\frac{3 ( 2 t - t ^{4} )}{( 1 + t ^{3} ) ^{2}}

解答ステップ

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

定数を除去:

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

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

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 t} ( t ^{2} ) ( 1 + t ^{3} ) - \frac{d}{d t} ( 1 + t ^{3} ) t ^{2}}{( 1 + t ^{3} ) ^{2}}

\frac{d}{d t} (t^{2}) = 2 t

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

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

簡素化

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

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