Partial Derivatives of e^(xy): Method and Mistakes

Freeze one variable, then use the chain rule

A partial derivative measures how $f$ changes when one variable moves and the other is held fixed. Hold $y$ fixed in $f(x,y) = e^{xy}$. The exponent $xy$ is then a constant multiple of $x$, so you are differentiating $e^{cx}$ with $c = y$, and the chain rule brings that constant out front.

Now hold $x$ fixed instead. The exponent is a constant multiple of $y$, and the same chain rule runs with the roles swapped.

At the point $(1,2)$ this gives $\frac{\partial f}{\partial x} = 2e^{2} \approx 14.778$ and $\frac{\partial f}{\partial y} = e^{2} \approx 7.389$. From that point, a small step in $x$ changes $f$ twice as fast as the same step in $y$.

The mistake: forgetting the inside derivative

Two wrong answers turn up again and again on this function.

• $\frac{\partial f}{\partial x} = e^{xy}$ treats the exponential as if it differentiated to itself with no chain rule. The outer exponential is indeed unchanged, but the inside derivative $y$ still multiplies it.
• $\frac{\partial f}{\partial x} = xy\,e^{xy}$ differentiates the exponent as though both letters were moving. In a partial with respect to $x$, the variable $y$ is frozen, so $\frac{\partial}{\partial x}(xy) = y$, not $xy$.

A fast check catches the first error. On the line $y = 0$ the function is $f(x,0) = e^{0} = 1$ for every $x$, so the slope in $x$ has to be $0$ there. The correct formula gives $0 \cdot e^{0} = 0$, while $e^{xy}$ would give $1$.

What the two partials say about the surface

The exponential factor is positive everywhere, so both partials vanish only when $x = 0$ and $y = 0$. That single critical point is a saddle: along the line $y = x$ the function is $e^{x^2}$, which sits at or above $1$, and along $y = -x$ it is $e^{-x^2}$, which sits at or below $1$.

Notice the symmetry too. Swapping $x$ and $y$ leaves $f$ unchanged, and it swaps the two partials, which is a cheap way to confirm you did not lose a factor in one of them.