d/(dt)((x)(k(x^2-5x+4)))

解

\frac{d}{d t} ((x) (k (x^{2} - 5 x + 4)))

k (3 x^{2} \frac{d x}{d t} - 10 x \frac{d x}{d t} + 4 \frac{d x}{d t})

解答ステップ

\frac{d}{d t} ((x) (k (x^{2} - 5 x + 4)))

定数を除去:

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

= k \frac{d}{d t} (x (x^{2} - 5 x + 4))

積の法則を適用:

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

f = x, g = x^{2} - 5 x + 4

= k (\frac{d x}{d t} (x^{2} - 5 x + 4) + \frac{d}{d t} (x^{2} - 5 x + 4) x)

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

= k (\frac{d x}{d t} (x^{2} - 5 x + 4) + (2 x \frac{d x}{d t} - 5 \frac{d x}{d t}) x)

拡張

\frac{d x}{d t} (x^{2} - 5 x + 4) + (2 x \frac{d x}{d t} - 5 \frac{d x}{d t}) x : 3 x^{2} \frac{d x}{d t} - 10 x \frac{d x}{d t} + 4 \frac{d x}{d t}

= k (3 x^{2} \frac{d x}{d t} - 10 x \frac{d x}{d t} + 4 \frac{d x}{d t})