Gradient of x^2 + xy + y^2: Handling the Cross Term

Differentiate all three terms, including the cross term

Go term by term with $y$ frozen. The square $x^2$ gives $2x$. The cross term $xy$ is $y$ times $x$, so it gives $y$. The square $y^2$ is a pure constant and gives zero.

Repeat with $x$ frozen. Now $x^2$ is the constant, $xy$ is $x$ times $y$ and gives $x$, and $y^2$ gives $2y$.

At $(2,-1)$ this is $\langle 4 - 1,\ 2 - 2\rangle = \langle 3, 0\rangle$, pointing due east. The height there is $f(2,-1) = 4 - 2 + 1 = 3$.

The mistake: charging the cross term to only one component

A frequent answer is $\langle 2x + y, 2y\rangle$, where $xy$ has been used up in the $x$-component and forgotten in the $y$-component. Each partial derivative is computed from scratch on the whole function, so a term that contains both variables gets counted in both.

You can catch the error by slicing. Fix $x = 2$, so $g(y) = f(2,y) = 4 + 2y + y^2$ is a one-variable function with $g'(y) = 2 + 2y$. The correct formula gives $x + 2y = 2 + 2y$, which matches. The faulty formula gives $2y$, which does not.

• Ask of every term: does it contain the variable I am differentiating? If yes, it contributes.
• $xy$ contains both variables, so it contributes to both components.
• Slice-and-check with a fixed numeric value for the other variable is the fastest way to verify a suspect component.

A quadratic form, so the gradient is a linear map

Every component is linear in $x$ and $y$, which is what always happens for a quadratic form. You can write the gradient as a matrix acting on the position vector.

That matrix is invertible, so the only solution of $\nabla f = \mathbf{0}$ is $x = y = 0$. Since $f = \left(x + \tfrac{y}{2}\right)^2 + \tfrac{3}{4}y^2$ is a sum of squares, the origin is a global minimum with value $0$.

The level curves $x^2 + xy + y^2 = c$ are ellipses tilted at $45$ degrees, with the long axis along $y = -x$. Setting $y = -x$ gives $f = x^2$, which grows slowly, while $y = x$ gives $f = 3x^2$, which grows fast. The gradient is longest where the level curves crowd together, along $y = x$.