derivative y=(2x+1)^3+6x(2x+1)^2

解

導関数

y = (2 x + 1)^{3} + 6 x (2 x + 1)^{2}

96 x^{2} + 72 x + 12

解答ステップ

\frac{d}{d x} ((2 x + 1)^{3} + 6 x (2 x + 1)^{2})

簡素化

(2 x + 1)^{3} + 6 x (2 x + 1)^{2} : 32 x^{3} + 36 x^{2} + 12 x + 1

= \frac{d}{d x} (32 x^{3} + 36 x^{2} + 12 x + 1)

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (32 x^{3}) + \frac{d}{d x} (36 x^{2}) + \frac{d}{d x} (12 x) + \frac{d}{d x} (1)

\frac{d}{d x} (32 x^{3}) = 96 x^{2}

\frac{d}{d x} (36 x^{2}) = 72 x

\frac{d}{d x} (12 x) = 12

\frac{d}{d x} (1) = 0

= 96 x^{2} + 72 x + 12 + 0

簡素化

= 96 x^{2} + 72 x + 12