What is the derivative of (2-xe^x/(x+e^x)) ?

The derivative of (2-xe^x/(x+e^x)) is (-e^xx^2-e^{2x}-2e^x-2)/((x+e^x)^2)

What is the first derivative of (2-xe^x/(x+e^x)) ?

The first derivative of (2-xe^x/(x+e^x)) is (-e^xx^2-e^{2x}-2e^x-2)/((x+e^x)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

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

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

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 4 e^{x}

\int sin (2 x - 1) d x

x^{2} y^{^{'}} - x y = y^{2}

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

\frac{4}{3} x y^{^{'}} - \frac{3}{8} y = 0

What is the derivative of (2-xe^x/(x+e^x)) ?
The derivative of (2-xe^x/(x+e^x)) is (-e^xx^2-e^{2x}-2e^x-2)/((x+e^x)^2)
What is the first derivative of (2-xe^x/(x+e^x)) ?
The first derivative of (2-xe^x/(x+e^x)) is (-e^xx^2-e^{2x}-2e^x-2)/((x+e^x)^2)