derivative f(x)=(x^2+2x)^3(7x+1)^4

解

導関数

f (x) = (x^{2} + 2 x)^{3} (7 x + 1)^{4}

3 (x^{2} + 2 x)^{2} (2 x + 2) (7 x + 1)^{4} + 28 (7 x + 1)^{3} (x^{2} + 2 x)^{3}

解答ステップ

\frac{d}{d x} ((x^{2} + 2 x)^{3} (7 x + 1)^{4})

積の法則を適用:

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

f = (x^{2} + 2 x)^{3}, g = (7 x + 1)^{4}

= \frac{d}{d x} ((x^{2} + 2 x)^{3}) (7 x + 1)^{4} + \frac{d}{d x} ((7 x + 1)^{4}) (x^{2} + 2 x)^{3}

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

\frac{d}{d x} ((7 x + 1)^{4}) = 28 (7 x + 1)^{3}

= 3 (x^{2} + 2 x)^{2} (2 x + 2) (7 x + 1)^{4} + 28 (7 x + 1)^{3} (x^{2} + 2 x)^{3}