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人気のある微分積分 >

derivative f(x)=(2x-8)4(x^2+x+1)5

解

導関数

f (x) = (2 x - 8) 4 (x^{2} + x + 1) 5

解

20 (6 x^{2} - 12 x - 6)

解答ステップ

\frac{d}{d x} ((2 x - 8) \cdot 4 (x^{2} + x + 1) \cdot 5)

簡素化

(2 x - 8) \cdot 4 (x^{2} + x + 1) \cdot 5 : 20 (2 x - 8) (x^{2} + x + 1)

= \frac{d}{d x} (20 (2 x - 8) (x^{2} + x + 1))

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 20 \frac{d}{d x} ((2 x - 8) (x^{2} + x + 1))

積の法則を適用:

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

f = 2 x - 8, g = x^{2} + x + 1

= 20 (\frac{d}{d x} (2 x - 8) (x^{2} + x + 1) + \frac{d}{d x} (x^{2} + x + 1) (2 x - 8))

\frac{d}{d x} (2 x - 8) = 2

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

= 20 (2 (x^{2} + x + 1) + (2 x + 1) (2 x - 8))

拡張

2 (x^{2} + x + 1) + (2 x + 1) (2 x - 8) : 6 x^{2} - 12 x - 6

= 20 (6 x^{2} - 12 x - 6)

グラフ

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