derivative h=(t+1)^{2/3}(2t^2-1)^3

解

導関数

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

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

解答ステップ

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

積の法則を適用:

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

f = (t + 1)^{\frac{2}{3}}, g = (2 t^{2} - 1)^{3}

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

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

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

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

簡素化

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

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