Partial Derivatives of x e^y + y e^x

Take the sum one term at a time

Differentiate term by term and, inside each term, ask which factor actually contains the moving variable. For $\frac{\partial f}{\partial x}$: in $x e^{y}$ the factor $e^{y}$ is a constant, so the term differentiates to $e^{y}$; in $y e^{x}$ the factor $y$ is a constant, so the term differentiates to $y e^{x}$.

Now move $y$ instead. In $x e^{y}$ the constant is $x$, leaving $x e^{y}$; in $y e^{x}$ the constant is $e^{x}$, leaving $e^{x}$.

At $(1,0)$ the slopes are $e^{0} + 0 = 1$ in $x$ and $1 + e \approx 3.718$ in $y$.

The mistake: a product rule with nothing to differentiate

• Writing $\frac{\partial}{\partial x}\left(x e^{y}\right) = e^{y} + x e^{y}$ runs the product rule on a factor that does not depend on $x$. The derivative of $e^{y}$ with respect to $x$ is $0$, so only the first term survives.
• Writing $\frac{\partial}{\partial x}\left(y e^{x}\right) = e^{x}$ drops the constant multiplier $y$.
• Reporting $e^{y} + e^{x}$ as the $x$ partial, which is really the mixed second partial $\frac{\partial^{2} f}{\partial x \partial y}$.

Symmetry gives a free check. Because $f(x,y) = f(y,x)$, the $y$ partial must be the $x$ partial with the letters exchanged. Compare $e^{y}+y e^{x}$ with $x e^{y}+e^{x}$: swap $x$ and $y$ in either and you get the other.

The one critical point

Set both partials to zero: $e^{y} + y\,e^{x} = 0$ and $x\,e^{y} + e^{x} = 0$. Multiply the first by $x$ and subtract the second. The $x\,e^{y}$ terms cancel and what is left factors.

Since $e^{x}$ is never zero, every critical point lies on the hyperbola $xy = 1$. Put $y = \frac{1}{x}$ into the second equation and it becomes $x\,e^{1/x} + e^{x} = 0$, which needs $x < 0$ because $e^{x}$ is positive. That left side has derivative $e^{1/x}\left(1 - \frac{1}{x}\right) + e^{x}$, and for $x < 0$ the bracket exceeds $1$, so the derivative is positive and the left side is strictly increasing. It can cross zero at most once, and it does cross at $x = -1$. So $(-1,-1)$ is the only critical point, and the value there is $-\frac{2}{e} \approx -0.736$.

The second derivatives there are $f_{xx} = y e^{x} = -e^{-1}$, $f_{yy} = x e^{y} = -e^{-1}$, and $f_{xy} = e^{y}+e^{x} = 2e^{-1}$, so the discriminant is negative.

A negative discriminant means $(-1,-1)$ is a saddle point, so this surface has no local maximum or minimum anywhere.