derivative of A/(B+Ce^x)

解答

\frac{d}{d x} (\frac{A}{B + C e ^{x}})

- \frac{e ^{x} A C}{( B + e ^{x} C ) ^{2}}

求解步骤

\frac{d}{d x} (\frac{A}{B + C e ^{x}})

将

A, B, C

视为常数

将常数提出:

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

= A \frac{d}{d x} (\frac{1}{B + C e ^{x}})

使用指数法则:

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

= A \frac{d}{d x} ((B + C e^{x})^{- 1})

使用连锁律:

- \frac{1}{( B + C e ^{x} ) ^{2}} \frac{d}{d x} (B + C e^{x})

= - \frac{1}{( B + C e ^{x} ) ^{2}} \frac{d}{d x} (B + C e^{x})

\frac{d}{d x} (B + C e^{x}) = C e^{x}

= A (- \frac{1}{( B + C e ^{x} ) ^{2}} C e^{x})

化简

A (- \frac{1}{( B + C e ^{x} ) ^{2}} C e^{x}) : - \frac{e ^{x} A C}{( B + e ^{x} C ) ^{2}}

= - \frac{e ^{x} A C}{( B + e ^{x} C ) ^{2}}

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