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

The derivative of (e^x-e^{-x}/(e^x+e^{-x)}) is 4/((e^x+e^{-x))^2}

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

The first derivative of (e^x-e^{-x}/(e^x+e^{-x)}) is 4/((e^x+e^{-x))^2}

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

\frac{4}{( e ^{x} + e ^{- x} ) ^{2}}

Solution steps

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

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

Simplify

\frac{( e ^{x} + e ^{- x} ) ( e ^{x} + e ^{- x} ) - ( e ^{x} - e ^{- x} ) ( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} ) ^{2}} : \frac{4}{( e ^{x} + e ^{- x} ) ^{2}}

= \frac{4}{( e ^{x} + e ^{- x} ) ^{2}}

What is the derivative of (e^x-e^{-x}/(e^x+e^{-x)}) ?
The derivative of (e^x-e^{-x}/(e^x+e^{-x)}) is 4/((e^x+e^{-x))^2}
What is the first derivative of (e^x-e^{-x}/(e^x+e^{-x)}) ?
The first derivative of (e^x-e^{-x}/(e^x+e^{-x)}) is 4/((e^x+e^{-x))^2}