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

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

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

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

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

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

Solution steps

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

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

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

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

Expand

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

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

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 3

x, 7, \frac{x ^{2}}{28}

\int - 2 x cos (4 x) d x

\int \frac{( 2 e ^{x} )}{x} d x

\int \frac{1}{16 x ^{2} - 25} d x

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