derivative f(x)=(x^3-3x)(2x^2+3x+5)

解

導関数

f (x) = (x^{3} - 3 x) (2 x^{2} + 3 x + 5)

10 x^{4} + 12 x^{3} - 3 x^{2} - 18 x - 15

解答ステップ

\frac{d}{d x} ((x^{3} - 3 x) (2 x^{2} + 3 x + 5))

積の法則を適用:

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

f = x^{3} - 3 x, g = 2 x^{2} + 3 x + 5

= \frac{d}{d x} (x^{3} - 3 x) (2 x^{2} + 3 x + 5) + \frac{d}{d x} (2 x^{2} + 3 x + 5) (x^{3} - 3 x)

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

\frac{d}{d x} (2 x^{2} + 3 x + 5) = 4 x + 3

= (3 x^{2} - 3) (2 x^{2} + 3 x + 5) + (4 x + 3) (x^{3} - 3 x)

簡素化

(3 x^{2} - 3) (2 x^{2} + 3 x + 5) + (4 x + 3) (x^{3} - 3 x) : 10 x^{4} + 12 x^{3} - 3 x^{2} - 18 x - 15

= 10 x^{4} + 12 x^{3} - 3 x^{2} - 18 x - 15