derivative f(x)=(2x^2-x+1)*(x^2+2x+3)

解

導関数

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

8 x^{3} + 9 x^{2} + 10 x - 1

解答ステップ

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

積の法則を適用:

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

f = 2 x^{2} - x + 1, g = x^{2} + 2 x + 3

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

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

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

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

簡素化

(4 x - 1) (x^{2} + 2 x + 3) + (2 x + 2) (2 x^{2} - x + 1) : 8 x^{3} + 9 x^{2} + 10 x - 1

= 8 x^{3} + 9 x^{2} + 10 x - 1