derivative of (e^x+1/(e^x-1))

解答

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

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

求解步骤

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

使用微分除法定则:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

化简

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

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

d er i v a t i v e f (x) = 9 x^{8} e^{x} + e^{x} x^{9}

\int_{0}^{1} x^{2} ln (x) d x

\int_{0}^{N} e^{- s t} e^{- 3 t} d t

\frac{d}{d x} (\frac{1}{0})

12 x^{2} y + y^{^{'}} = 5 x^{2}