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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derivative of

\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}))

Simplify

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

\frac{d}{d x} ((x)^{\frac{1}{3}})

\frac{d}{d x} (- \frac{ln ( 1 + 6 x )}{x})

\frac{d}{d x} (\frac{1}{x ( x - 1 )})

x \frac{d y}{d x} = 6 y

f (x) = 2 x^{2} + x - 3, at x = - 1

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})