derivative y=\sqrt[3]{t}(t^2+4)

解

導関数

y = 3 t (t^{2} + 4)

\frac{t ^{2} + 4 + 6 3 t t ^{\frac{5}{3}}}{3 t ^{\frac{2}{3}}}

解答ステップ

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

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 3 t, g = t^{2} + 4

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

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

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

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

簡素化

\frac{1}{3 t ^{\frac{2}{3}}} (t^{2} + 4) + 2 t 3 t : \frac{t ^{2} + 4 + 6 3 t t ^{\frac{5}{3}}}{3 t ^{\frac{2}{3}}}

= \frac{t ^{2} + 4 + 6 3 t t ^{\frac{5}{3}}}{3 t ^{\frac{2}{3}}}