Gradient of xy + yz + zx: Symmetric Quadratic Form

Only the terms containing your variable survive

Differentiate term by term. With respect to $x$, the term $xy$ gives $y$, the term $zx$ gives $z$, and the term $yz$ has no $x$ in it, so it contributes zero.

The function is completely symmetric in the three variables, so the other components follow by relabelling. Each one is the sum of the two variables you did not differentiate.

If you write $s = x + y + z$, there is a compact form worth remembering: $\nabla f = \left\langle s - x,\; s - y,\; s - z \right\rangle$. It makes the symmetry obvious and speeds up substitution.

The mistake: leaving the variable in its own component

The error to watch for is $\frac{\partial f}{\partial x} = x + y + z$, which keeps $x$ in the answer. It usually comes from differentiating the compact expression $s - x$ carelessly, or from pattern matching on the symmetry without checking the terms.

• Every component of this gradient must be missing its own variable. If $x$ appears in $\frac{\partial f}{\partial x}$, look again.
• A second slip is forgetting the $zx$ term when differentiating with respect to $x$, giving just $y$. Scan all three terms every time.
• Degree check: $f$ is degree 2, so each component must be degree 1. Both $y+z$ and $x+y+z$ pass that test, which is why the term by term scan matters more here.

Notice also that the gradient of a quadratic is linear, so the components should contain no products and no squares. Anything like $xy$ inside a component is wrong.

The origin is a saddle, not a minimum

At $(1, 2, 3)$ the value is $f = 2 + 6 + 3 = 11$ and the gradient falls out immediately.

For critical points, solve $y + z = 0$, $x + z = 0$ and $x + y = 0$ together. Subtracting the second from the third gives $y = z$, and combining with $y + z = 0$ forces $y = z = 0$ and then $x = 0$. The origin is the only critical point.

It is a saddle. Along the line $(t, t, t)$ the function is $f = 3t^2$, which climbs away from zero, while along $(t, -t, 0)$ it is $f = -t^2$, which falls. A single critical point with both behaviours nearby cannot be a maximum or a minimum.

One more identity that shows up in algebra and in optimisation problems with a fixed sum: $xy + yz + zx = \frac{(x+y+z)^2 - \left(x^2+y^2+z^2\right)}{2}$.