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

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

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

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

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derivative

\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}))

Simplify

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

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