derivative of 0.001x^3-0.14x^2+7x+160

解

\frac{d}{d x} (0.001 x^{3} - 0.14 x^{2} + 7 x + 160)

0.003 x^{2} - 0.28 x + 7

解答ステップ

\frac{d}{d x} (0.001 x^{3} - 0.14 x^{2} + 7 x + 160)

和/差の法則を適用:

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

= \frac{d}{d x} (0.001 x^{3}) - \frac{d}{d x} (0.14 x^{2}) + \frac{d}{d x} (7 x) + \frac{d}{d x} (160)

\frac{d}{d x} (0.001 x^{3}) = 0.003 x^{2}

\frac{d}{d x} (0.14 x^{2}) = 0.28 x

\frac{d}{d x} (7 x) = 7

\frac{d}{d x} (160) = 0

= 0.003 x^{2} - 0.28 x + 7 + 0

簡素化

= 0.003 x^{2} - 0.28 x + 7