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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

\frac{\partial}{\partial x} (2 - x^{2} + y^{2})

t an g e n t \frac{1}{3 x}

a re a y = sin (2 x), y = cos (2 x), x = 0, x = \frac{π}{4}

l o n g d i v i s i o n \frac{4 x ^{2} - 3}{x - 1}

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