derivative of x/3 (2x+1^3)

解

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

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

解答ステップ

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

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

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

積の法則を適用:

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

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

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

\frac{d x}{d x} = 1

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

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

簡素化

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

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