Partial Derivatives of 2x^2 - 3xy + 5y^2

Differentiating a quadratic form

Every term is degree two, so every partial is degree one. Holding $y$ fixed, the term $5y^2$ is a constant and contributes nothing, while $-3xy$ contributes its constant multiplier $-3y$.

At $(1,2)$ these evaluate to $4 - 6 = -2$ and $-3 + 20 = 17$. The function is climbing hard in the $y$ direction there and drifting down in $x$.

Quadratic forms like this one are the model every second derivative test is built on, because a smooth function near a critical point looks like a quadratic form plus a small error.

The mistake: losing the sign on the cross term

The $-3xy$ term is where marks are lost. Two errors show up repeatedly: dropping the minus sign, and keeping the wrong letter.

• $\frac{\partial}{\partial x}(-3xy) = -3y$, not $-3x$ and not $+3y$.
• $\frac{\partial}{\partial y}(-3xy) = -3x$.
• The coefficient $-3$ carries through untouched in both cases.

A quick check: the mixed partials must agree. Differentiating $4x - 3y$ with respect to $y$ gives $-3$, and differentiating $-3x + 10y$ with respect to $x$ also gives $-3$. If your two answers give different mixed partials, one of the cross-term signs is wrong.

What the constant Hessian tells you

Because the partials are linear, the second partials are numbers rather than functions.

The discriminant is $f_{xx}f_{yy} - f_{xy}^2 = 40 - 9 = 31 > 0$ with $f_{xx} > 0$, so the surface is a bowl everywhere, not just near one point. Setting both partials to zero gives the single solution $(0,0)$, the vertex of that bowl.

Since $f(0,0) = 0$ and the form is positive definite, $f(x,y) > 0$ at every other point in the plane. That is a global minimum, which quadratic forms hand you for free.