d/(dt)(\sqrt[4]{(e^{5t})/(t^2+1)})

解

\frac{d}{d t} (4 \frac{e ^{5 t}}{t ^{2} + 1})

\frac{e ^{\frac{5 t}{4}} ( 5 ( t ^{2} + 1 ) - 2 t )}{4 ( t ^{2} + 1 ) ^{\frac{5}{4}}}

解答ステップ

\frac{d}{d t} (4 \frac{e ^{5 t}}{t ^{2} + 1})

連鎖法則を適用:

\frac{1}{4 ( \frac{e ^{5 t}}{t ^{2} + 1} ) ^{\frac{3}{4}}} \frac{d}{d t} (\frac{e ^{5 t}}{t ^{2} + 1})

= \frac{1}{4 ( \frac{e ^{5 t}}{t ^{2} + 1} ) ^{\frac{3}{4}}} \frac{d}{d t} (\frac{e ^{5 t}}{t ^{2} + 1})

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

= \frac{1}{4 ( \frac{e ^{5 t}}{t ^{2} + 1} ) ^{\frac{3}{4}}} \cdot \frac{e ^{5 t} \cdot 5 ( t ^{2} + 1 ) - 2 t e ^{5 t}}{( t ^{2} + 1 ) ^{2}}

簡素化

\frac{1}{4 ( \frac{e ^{5 t}}{t ^{2} + 1} ) ^{\frac{3}{4}}} \cdot \frac{e ^{5 t} \cdot 5 ( t ^{2} + 1 ) - 2 t e ^{5 t}}{( t ^{2} + 1 ) ^{2}} : \frac{e ^{\frac{5 t}{4}} ( 5 ( t ^{2} + 1 ) - 2 t )}{4 ( t ^{2} + 1 ) ^{\frac{5}{4}}}

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