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}