derivative f(x)=(x+5)^{2/3}(2x^2-1)^3

解

導関数

f (x) = (x + 5)^{\frac{2}{3}} (2 x^{2} - 1)^{3}

\frac{2 ( 2 x ^{2} - 1 ) ^{3} + 36 x ( 2 x ^{2} - 1 ) ^{2} ( x + 5 )}{3 ( x + 5 ) ^{\frac{1}{3}}}

解答ステップ

\frac{d}{d x} ((x + 5)^{\frac{2}{3}} (2 x^{2} - 1)^{3})

積の法則を適用:

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

f = (x + 5)^{\frac{2}{3}}, g = (2 x^{2} - 1)^{3}

= \frac{d}{d x} ((x + 5)^{\frac{2}{3}}) (2 x^{2} - 1)^{3} + \frac{d}{d x} ((2 x^{2} - 1)^{3}) (x + 5)^{\frac{2}{3}}

\frac{d}{d x} ((x + 5)^{\frac{2}{3}}) = \frac{2}{3 ( x + 5 ) ^{\frac{1}{3}}}

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

= \frac{2}{3 ( x + 5 ) ^{\frac{1}{3}}} (2 x^{2} - 1)^{3} + 12 x (2 x^{2} - 1)^{2} (x + 5)^{\frac{2}{3}}

簡素化

\frac{2}{3 ( x + 5 ) ^{\frac{1}{3}}} (2 x^{2} - 1)^{3} + 12 x (2 x^{2} - 1)^{2} (x + 5)^{\frac{2}{3}} : \frac{2 ( 2 x ^{2} - 1 ) ^{3} + 36 x ( 2 x ^{2} - 1 ) ^{2} ( x + 5 )}{3 ( x + 5 ) ^{\frac{1}{3}}}

= \frac{2 ( 2 x ^{2} - 1 ) ^{3} + 36 x ( 2 x ^{2} - 1 ) ^{2} ( x + 5 )}{3 ( x + 5 ) ^{\frac{1}{3}}}