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

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

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

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

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derivative of

y = \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})

\int x^{3} + 3 x - 2 d x

\frac{d}{d t} (\frac{1}{3} t^{3} + \frac{1}{2} t^{2})

\frac{d}{d x} (e^{- 8 x^{2}})

\frac{\partial}{\partial y} (e^{(x + y)} - 1)

\frac{\partial}{\partial z} (3 y cos (π x) + 2 x z)

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