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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

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

Solution steps

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (2 (1 - sin (x)))

\int 8 x e^{4 x^{2}} d x

sin (x), \frac{π}{5}

f (x) = 4^{2}

\int ln (v) d v

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