Gradient of e^(xy): Components and Saddle Point

The exponential survives, the exponent supplies the factor

With $u = xy$ and $f = e^{u}$, the chain rule gives $\frac{\partial f}{\partial x} = e^{u}\frac{\partial u}{\partial x}$. Differentiating the exponential changes nothing about it, so the entire difference between the two components is the derivative of the exponent.

The factor that appears is the other variable, which surprises people at first. Along a horizontal line, $f = e^{xy}$ is an exponential in $x$ with growth rate $y$, so the slope is proportional to $y$. That is precisely what $y\,e^{xy}$ says.

One critical point, and it is a saddle

Since $e^{xy}$ is never zero, $\nabla f = \left\langle 0,0\right\rangle$ requires $y = 0$ and $x = 0$ at the same time. The origin is the only candidate for a maximum or minimum, and $f(0,0) = 1$.

Walk away from it along two lines to see what kind of point it is.

• Along $y = x$: $f = e^{x^{2}} \geq 1$, so the origin looks like a minimum from this direction.
• Along $y = -x$: $f = e^{-x^{2}} \leq 1$, so it looks like a maximum from that one.

Higher one way and lower the other is the definition of a saddle. The level curves make the same point: $e^{xy} = c$ means $xy = \ln c$, a family of hyperbolas whose degenerate member is the pair of axes crossing at the origin.

Away from the origin the vector grows fast. At $(1,1)$, $\nabla f = \left\langle e, e\right\rangle \approx \left\langle 2.7183, 2.7183\right\rangle$, pointing outward along the line $y = x$, which is the direction in which $xy$ increases fastest.

The mistake: treating e^(xy) like e^x

The usual wrong answer is $\frac{\partial f}{\partial x} = e^{xy}$, from a reflex that the derivative of an exponential is itself. That rule applies to $e^{x}$; the moment the exponent is anything else, the chain rule attaches its derivative.

A second wrong answer, $xy\,e^{xy}$, multiplies by the whole exponent instead of by its derivative.

One test point will not catch both, so use two.

• At $(2,0)$ the function is constant in $x$, since $f(x,0) = e^{0} = 1$ for every $x$, so the true $x$ partial is $0$. The answer $e^{xy}$ gives $1$ there and is exposed, while $xy\,e^{xy}$ gives $0$ and slips through.
• At $(2,1)$ the true $x$ partial is $y\,e^{xy} = e^{2} \approx 7.389$, while $xy\,e^{xy}$ gives $2e^{2} \approx 14.778$. Avoid $(1,1)$ as a test point: there $x$ and $y$ are both $1$, every stray factor equals $1$, and the true answer and both wrong ones all return $e$.