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

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

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(\partial)/(\partial x)(ye^x-xe^y)

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

x \to 0 + lim (\frac{1}{e ^{x} + 1})

[3 ln (x) - 12 sin (x)]

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

y^{^{''}} - 4 y^{^{'}} - 4 y = e^{2 x}

y = \frac{x ^{2}}{2} - 4 x + 6, y = \frac{x}{2} - 1, x = 0, x = 5

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