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 y}{d x} = 2 y - x y^{2}

\int - \frac{5}{12 - 12 x ^{2}} d x

\frac{d}{d x} (\frac{x ^{2} + 5}{x ^{2} + 6 x})

t an g e n t y = x^{2} - 2 x - 3, (2, - 3)

n = 0 \sum \infty c n (x - 8)^{n}

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