derivative of 1/14 (e^{7x}+e^{-7x})

解答

\frac{d}{d x} (\frac{1}{14} (e^{7 x} + e^{- 7 x}))

\frac{e ^{- 7 x} ( e ^{14 x} - 1 )}{2}

求解步骤

\frac{d}{d x} (\frac{1}{14} (e^{7 x} + e^{- 7 x}))

将常数提出:

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

= \frac{1}{14} \frac{d}{d x} (e^{7 x} + e^{- 7 x})

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{1}{14} (\frac{d}{d x} (e^{7 x}) + \frac{d}{d x} (e^{- 7 x}))

\frac{d}{d x} (e^{7 x}) = e^{7 x} \cdot 7

\frac{d}{d x} (e^{- 7 x}) = - 7 e^{- 7 x}

= \frac{1}{14} (e^{7 x} \cdot 7 - 7 e^{- 7 x})

化简

\frac{1}{14} (e^{7 x} \cdot 7 - 7 e^{- 7 x}) : \frac{e ^{- 7 x} ( e ^{14 x} - 1 )}{2}

= \frac{e ^{- 7 x} ( e ^{14 x} - 1 )}{2}

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