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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Expand

(a e^{a x} + a e^{- a x}) (e^{a x} + e^{- a x}) - (e^{a x} a - a e^{- a x}) (e^{a x} - e^{- a x}) : 4 a

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

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