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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

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

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

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

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

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

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