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

The derivative of (e^x+e^{-x}/2) is 1/2 (e^x-e^{-x})

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

The first derivative of (e^x+e^{-x}/2) is 1/2 (e^x-e^{-x})

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

What is the derivative of (e^x+e^{-x}/2) ?
The derivative of (e^x+e^{-x}/2) is 1/2 (e^x-e^{-x})
What is the first derivative of (e^x+e^{-x}/2) ?
The first derivative of (e^x+e^{-x}/2) is 1/2 (e^x-e^{-x})