derivative (t+4)^{(2/3)}(2t^2-4)^3

解

導関数

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

\frac{2 ( 2 t ^{2} - 4 ) ^{3} + 36 t ( 2 t ^{2} - 4 ) ^{2} ( t + 4 )}{3 ( t + 4 ) ^{\frac{1}{3}}}

解答ステップ

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

積の法則を適用:

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

f = (t + 4)^{\frac{2}{3}}, g = (2 t^{2} - 4)^{3}

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

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

\frac{d}{d t} ((2 t^{2} - 4)^{3}) = 12 t (2 t^{2} - 4)^{2}

= \frac{2}{3 ( t + 4 ) ^{\frac{1}{3}}} (2 t^{2} - 4)^{3} + 12 t (2 t^{2} - 4)^{2} (t + 4)^{\frac{2}{3}}

簡素化

\frac{2}{3 ( t + 4 ) ^{\frac{1}{3}}} (2 t^{2} - 4)^{3} + 12 t (2 t^{2} - 4)^{2} (t + 4)^{\frac{2}{3}} : \frac{2 ( 2 t ^{2} - 4 ) ^{3} + 36 t ( 2 t ^{2} - 4 ) ^{2} ( t + 4 )}{3 ( t + 4 ) ^{\frac{1}{3}}}

= \frac{2 ( 2 t ^{2} - 4 ) ^{3} + 36 t ( 2 t ^{2} - 4 ) ^{2} ( t + 4 )}{3 ( t + 4 ) ^{\frac{1}{3}}}