d/(dt)((5t)/(1+2t^2))

解

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

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

解答ステップ

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

定数を除去:

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

= 5 \frac{d}{d t} (\frac{t}{1 + 2 t ^{2}})

商の法則を適用:

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

= 5 \cdot \frac{\frac{d t}{d t} ( 1 + 2 t ^{2} ) - \frac{d}{d t} ( 1 + 2 t ^{2} ) t}{( 1 + 2 t ^{2} ) ^{2}}

\frac{d t}{d t} = 1

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

= 5 \cdot \frac{1 \cdot ( 1 + 2 t ^{2} ) - 4 tt}{( 1 + 2 t ^{2} ) ^{2}}

簡素化

5 \cdot \frac{1 \cdot ( 1 + 2 t ^{2} ) - 4 tt}{( 1 + 2 t ^{2} ) ^{2}} : \frac{5 ( 1 - 2 t ^{2} )}{( 1 + 2 t ^{2} ) ^{2}}

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