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

解答

导数

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

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

求解步骤

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

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

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

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

化简

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

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

t a y l or e^{- 2 x} + sin (3 x) + 3 x^{2} - 2 x + 5, 0

t an g e n t f (x) = \frac{6}{( 1 + e ^{- x} )}, at x = 0

\frac{d y}{d x} = \frac{y}{x} + 4 x + 1

\frac{\partial}{\partial x} (t e^{x t})

d er i v a t i v e y = e^{5 e^{x}}