derivative of x^3+2x^2-x+4

解

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

3 x^{2} + 4 x - 1

解答ステップ

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

和/差の法則を適用:

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

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

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

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

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

\frac{d}{d x} (4) = 0

= 3 x^{2} + 4 x - 1 + 0

簡素化

= 3 x^{2} + 4 x - 1