derivative of a/(1+\frac{b{x+c}})

解

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

\frac{ab}{( x + c + b ) ^{2}}

解答ステップ

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

定数を除去:

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

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

指数の規則を適用する:

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

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

連鎖法則を適用:

- \frac{1}{( 1 + \frac{b}{x + c} ) ^{2}} \frac{d}{d x} (1 + \frac{b}{x + c})

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

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

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

簡素化

a (- \frac{1}{( 1 + \frac{b}{x + c} ) ^{2}} (- \frac{b}{( x + c ) ^{2}})) : \frac{ab}{( x + c + b ) ^{2}}

= \frac{ab}{( x + c + b ) ^{2}}