What is the derivative of (e^{-x}/((1+e^{-x))^2}) ?

The derivative of (e^{-x}/((1+e^{-x))^2}) is (e^{-2x}(-e^x+1))/((1+e^{-x))^3}

What is the first derivative of (e^{-x}/((1+e^{-x))^2}) ?

The first derivative of (e^{-x}/((1+e^{-x))^2}) is (e^{-2x}(-e^x+1))/((1+e^{-x))^3}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\frac{e ^{- 2 x} ( - e ^{x} + 1 )}{( 1 + e ^{- x} ) ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( - e ^{- x} ) ( 1 + e ^{- x} ) ^{2} - ( - 2 e ^{- x} ( 1 + e ^{- x} ) ) e ^{- x}}{( ( 1 + e ^{- x} ) ^{2} ) ^{2}} : \frac{e ^{- 2 x} ( - e ^{x} + 1 )}{( 1 + e ^{- x} ) ^{3}}

= \frac{e ^{- 2 x} ( - e ^{x} + 1 )}{( 1 + e ^{- x} ) ^{3}}

What is the derivative of (e^{-x}/((1+e^{-x))^2}) ?
The derivative of (e^{-x}/((1+e^{-x))^2}) is (e^{-2x}(-e^x+1))/((1+e^{-x))^3}
What is the first derivative of (e^{-x}/((1+e^{-x))^2}) ?
The first derivative of (e^{-x}/((1+e^{-x))^2}) is (e^{-2x}(-e^x+1))/((1+e^{-x))^3}