derivative f(x)=(x-2)(x-1)(x+1)(x+2)

解

導関数

f (x) = (x - 2) (x - 1) (x + 1) (x + 2)

4 x^{3} - 10 x

解答ステップ

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

積の法則を適用:

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

f = x - 2, g = (x - 1) (x + 1) (x + 2)

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

\frac{d}{d x} (x - 2) = 1

\frac{d}{d x} ((x - 1) (x + 1) (x + 2)) = 3 x^{2} + 4 x - 1

= 1 \cdot (x - 1) (x + 1) (x + 2) + (3 x^{2} + 4 x - 1) (x - 2)

簡素化

1 \cdot (x - 1) (x + 1) (x + 2) + (3 x^{2} + 4 x - 1) (x - 2) : 4 x^{3} - 10 x

= 4 x^{3} - 10 x