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Popular >

((e^x+e^{-x})/2)^'

Solution

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

Solution

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

Solution steps

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

(e^{x})^{'} = e^{x}

(e^{- x})^{'} = - e^{- x}

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

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Frequently Asked Questions (FAQ)

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