derivative (e^x+1)/(e^{2x)}

解答

导数

\frac{e ^{x} + 1}{e ^{2 x}}

\frac{- e ^{x} - 2}{e ^{2 x}}

求解步骤

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

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

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

\frac{d}{d x} (e^{2 x}) = e^{2 x} \cdot 2

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

化简

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

= \frac{- e ^{x} - 2}{e ^{2 x}}

t a y l or \frac{( e ^{2 x} - 1 )}{x}, - 0.5

\frac{d y}{d x} = (\frac{sec ( x )}{y})^{2}

\frac{\partial}{\partial x} (y x)

(2 x y^{- 1}) d x + (2 - 4 x^{2} y^{- 2}) d y = 0

t an g e n t y = 2 ln (x), at x = 4