d/(dt)((x(t,C))(1-1/(ktC)))

解

\frac{d}{d t} ((x (t, C)) (1 - \frac{1}{k \cdot t \cdot C}))

\frac{d}{d t} (x (t, C)) (1 - \frac{1}{k tC}) + \frac{1}{C k t ^{2}} x (t, C)

解答ステップ

\frac{d}{d t} ((x (t, C)) (1 - \frac{1}{k tC}))

k, C

を定数として扱う

積の法則を適用:

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

f = x (t, C), g = 1 - \frac{1}{k tC}

= \frac{d}{d t} (x (t, C)) (1 - \frac{1}{k tC}) + \frac{d}{d t} (1 - \frac{1}{k tC}) x (t, C)

\frac{d}{d t} (1 - \frac{1}{k tC}) = \frac{1}{C k t ^{2}}

= \frac{d}{d t} (x (t, C)) (1 - \frac{1}{k tC}) + \frac{1}{C k t ^{2}} x (t, C)