derivative f(x)=-x^3+x^2+4x-2

解

導関数

f (x) = - x^{3} + x^{2} + 4 x - 2

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

解答ステップ

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

和/差の法則を適用:

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

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

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

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

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

\frac{d}{d x} (2) = 0

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

簡素化

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