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

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

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

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

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derivative of

y = \frac{e ^{x} + e ^{- x}}{x}

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

Solution steps

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

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

\frac{d x}{d x} = 1

= \frac{( e ^{x} - e ^{- x} ) x - 1 \cdot ( e ^{x} + e ^{- x} )}{x ^{2}}

Simplify

\frac{( e ^{x} - e ^{- x} ) x - 1 \cdot ( e ^{x} + e ^{- x} )}{x ^{2}} : \frac{e ^{x} x - e ^{- x} x - e ^{x} - e ^{- x}}{x ^{2}}

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

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