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

解答

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

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

求解步骤

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

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

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

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

化简

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

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

\frac{d}{d y} (\frac{y}{y + \frac{c}{y}})

\int - \frac{7}{x ^{2} + 16} d x

t an g e n t f (x) = \frac{1}{2 x}, at x = 64

\frac{d}{d x} (cos (x + π))

\frac{d}{d x} (x^{2} + 4 x - 5)