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

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

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

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

e^{- y} - y e^{- x}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

e^{t} sin (6 t)

\frac{d}{d x} (sin (x) x)

f (x) = 7 e^{x} + 4 x, at x = 0

\int 5^{x} e^{- x} d x

\frac{\partial}{\partial x} ((x - a)^{- 1})

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