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}

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\frac{d}{d x} (\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}}

y = ln (x^{2} + 1)

\frac{d r}{d s} = e^{r - 2 s}

\int \frac{16}{x ^{3} + 4 x} d x

\int 6 x^{5} y d x

ln (\frac{4 x + 1}{2 x - 7})

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}