Gradient of xy: Why It Is (y, x) and Not (x, y)

Freeze one variable and the other becomes a coefficient

Hold $y$ fixed. Then $f = y \cdot x$ is a constant multiple of $x$, and the derivative of a constant multiple of $x$ is that constant.

Hold $x$ fixed instead. Now $f = x \cdot y$ is a constant multiple of $y$, and the same reasoning applies.

So $\nabla f(2,3) = \langle 3, 2\rangle$. It is worth saying out loud: the $x$-component is $y$.

The mistake: copying the point instead of swapping

Two errors show up on almost every set of papers. The first is pattern matching, writing $\nabla f = \langle x, y\rangle$ because the components should surely go in variable order. They do go in variable order, but the entry in the $x$ slot is $\frac{\partial f}{\partial x}$, and that partial happens to equal $y$.

The second is reaching for the product rule and differentiating both factors at once, producing $\frac{\partial f}{\partial x} = 1 \cdot y + x \cdot 1 = x + y$. The product rule applies only when both factors depend on the variable you are differentiating. Here $y$ does not depend on $x$, so its derivative with respect to $x$ is zero and the second term never appears.

• $\frac{\partial f}{\partial x} = y$, so the answer at a point uses the other coordinate.
• $\nabla f = \langle x + y, x + y \rangle$ is the product-rule error, and you can spot it because the two components come out equal.
• A quick sanity check: on the line $y = 5$, $f = 5x$ has slope $5$ in the $x$ direction, matching $\frac{\partial f}{\partial x} = y = 5$.

Level curves, and a perpendicularity check

For each nonzero $c$, setting $xy = c$ gives a hyperbola with the coordinate axes as asymptotes, sitting in the first and third quadrants when $c > 0$ and in the second and fourth when $c < 0$. The level set $c = 0$ is the degenerate case, the two axes themselves. The graph $z = xy$ is a hyperbolic paraboloid, the standard saddle used in every Calc 3 course.

Take the point $(2,3)$, which sits on the level curve $xy = 6$. Differentiating $xy = 6$ implicitly gives $y + x\,y' = 0$, so $y' = -y/x = -3/2$ and a tangent direction is $\langle 2, -3\rangle$. Dot it with the gradient:

The gradient vanishes only at the origin, where $\nabla f(0,0) = \langle 0,0\rangle$. That is the saddle point at the centre of the surface, and it is not a maximum or a minimum: $f$ is positive in the first quadrant and negative in the second.