What is (\partial)/(\partial x)(e^x-e^{-x}) ?

The answer to (\partial)/(\partial x)(e^x-e^{-x}) is e^x+e^{-x}

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(\partial)/(\partial x)(e^x-e^{-x})

\frac{\partial}{\partial x} (e^{x} - e^{- x})

e^{x} + e^{- x}

Solution steps

\frac{\partial}{\partial x} (e^{x} - e^{- x})

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

= e^{x} + e^{- x}

\frac{d}{d x} (1 - \frac{1}{2 x})

\frac{d y}{d x} = \frac{x - 1}{y}

\int \frac{x ^{2}}{x ^{2} - 16} d x

\int \frac{1}{( a + b x ) ^{3}} d x

\int_{0}^{0.5} (300 x) d x

What is (\partial)/(\partial x)(e^x-e^{-x}) ?
The answer to (\partial)/(\partial x)(e^x-e^{-x}) is e^x+e^{-x}