derivative \sqrt[5]{t}+2sqrt(t^5)

解

導関数

5 t + 2 t^{5}

\frac{1 + 25 t ^{\frac{23}{10}}}{5 t ^{\frac{4}{5}}}

解答ステップ

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

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d t} (5 t) = \frac{1}{5 t ^{\frac{4}{5}}}

\frac{d}{d t} (2 t^{5}) = 5 t^{\frac{3}{2}}

= \frac{1}{5 t ^{\frac{4}{5}}} + 5 t^{\frac{3}{2}}

簡素化

\frac{1}{5 t ^{\frac{4}{5}}} + 5 t^{\frac{3}{2}} : \frac{1 + 25 t ^{\frac{23}{10}}}{5 t ^{\frac{4}{5}}}

= \frac{1 + 25 t ^{\frac{23}{10}}}{5 t ^{\frac{4}{5}}}