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}

f (x) = x^{3} - 8 x + 2

\frac{d}{d x} (3 x^{2} + x + 4)

\int_{1}^{4} x^{- \frac{3}{2}} d x

\int_{0}^{4} \frac{5 π y ^{4}}{4} d y

\frac{d y}{d x} + y cos (x) = 3 cos (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}