derivative f(x)=(8x+9)(x^3-6)

解

導関数

f (x) = (8 x + 9) (x^{3} - 6)

32 x^{3} + 27 x^{2} - 48

解答ステップ

\frac{d}{d x} ((8 x + 9) (x^{3} - 6))

積の法則を適用:

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

f = 8 x + 9, g = x^{3} - 6

= \frac{d}{d x} (8 x + 9) (x^{3} - 6) + \frac{d}{d x} (x^{3} - 6) (8 x + 9)

\frac{d}{d x} (8 x + 9) = 8

\frac{d}{d x} (x^{3} - 6) = 3 x^{2}

= 8 (x^{3} - 6) + 3 x^{2} (8 x + 9)

簡素化

8 (x^{3} - 6) + 3 x^{2} (8 x + 9) : 32 x^{3} + 27 x^{2} - 48

= 32 x^{3} + 27 x^{2} - 48