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

The derivative of (e^x+e^{-x})/(e^{2x)} is (-e^{2x}-3)/(e^{3x)}

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

The first derivative of (e^x+e^{-x})/(e^{2x)} is (-e^{2x}-3)/(e^{3x)}

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derivative of

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

\frac{- e ^{2 x} - 3}{e ^{3 x}}

Solution steps

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

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

\frac{d}{d x} (e^{2 x}) = e^{2 x} \cdot 2

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

Simplify

\frac{( e ^{x} - e ^{- x} ) e ^{2 x} - e ^{2 x} \cdot 2 ( e ^{x} + e ^{- x} )}{( e ^{2 x} ) ^{2}} : \frac{- e ^{2 x} - 3}{e ^{3 x}}

= \frac{- e ^{2 x} - 3}{e ^{3 x}}

What is the derivative of (e^x+e^{-x})/(e^{2x)} ?
The derivative of (e^x+e^{-x})/(e^{2x)} is (-e^{2x}-3)/(e^{3x)}
What is the first derivative of (e^x+e^{-x})/(e^{2x)} ?
The first derivative of (e^x+e^{-x})/(e^{2x)} is (-e^{2x}-3)/(e^{3x)}