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

解答

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

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

求解步骤

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

\frac{d}{d x} (x) = \frac{1}{2 x}

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

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

化简

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

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

\frac{\partial}{\partial x} (x e^{- 3 x y})

n = 1 \sum \infty \frac{n}{3}

\frac{d}{d x} (\frac{3 x - 4}{6 x ^{2} - 1})

d er i v a t i v e f (x) = ln (5 x - 7)

\int \frac{x ^{2} + 9}{x ^{2} + 4} d x