d/(d{r)}(1/(2{r)(a+b+c)})

解

\frac{d}{d r} (\frac{1}{2 r ( a + b + c )})

- \frac{1}{2 ( a + b + c ) r ^{2}}

解答ステップ

\frac{d}{d r} (\frac{1}{2 r ( a + b + c )})

定数を除去:

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

= \frac{1}{2 ( a + b + c )} \frac{d}{d r} (\frac{1}{r})

指数の規則を適用する:

\frac{1}{a} = a^{- 1}

= \frac{1}{2 ( a + b + c )} \frac{d}{d r} (r^{- 1})

乗の法則を適用:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= \frac{1}{2 ( a + b + c )} (- 1 \cdot r^{- 1 - 1})

簡素化

\frac{1}{2 ( a + b + c )} (- 1 \cdot r^{- 1 - 1}) : - \frac{1}{2 ( a + b + c ) r ^{2}}

= - \frac{1}{2 ( a + b + c ) r ^{2}}