derivative of (t(t(2t+1)^2))

解

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

16 x^{3} + 12 x^{2} + 2 x

解答ステップ

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

簡素化

(x) (x (2 x + 1)^{2}) : x^{2} (2 x + 1)^{2}

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

積の法則を適用:

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

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

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

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

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

= 2 x (2 x + 1)^{2} + 4 (2 x + 1) x^{2}

簡素化

2 x (2 x + 1)^{2} + 4 (2 x + 1) x^{2} : 16 x^{3} + 12 x^{2} + 2 x

= 16 x^{3} + 12 x^{2} + 2 x