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

Differentiate term by term

A sum differentiates one piece at a time, so handle $x^3$, $y^3$, and $-3xy$ separately. With $y$ held fixed, $y^3$ is a constant and dies, while $-3xy$ is $-3y$ times $x$.

By the symmetry of the function under swapping $x$ and $y$, the other partial has the same shape with the letters exchanged.

At $(2,1)$ these give $3(4) - 3(1) = 9$ and $3(1) - 3(2) = -3$, so the surface rises steeply in the $x$ direction and falls in the $y$ direction from that point.

The mistake: dropping the cross term

The most common wrong answer is $\frac{\partial f}{\partial x} = 3x^2$, with the $-3xy$ term treated as if it were a constant. It is not: it contains $x$.

Use a two-part test on each term before you write anything down.

• Does the term contain the variable I am differentiating? If not, it contributes $0$.
• If it does, everything else in that term is a constant multiplier.
• So $-3xy$ contributes $-3y$ to $\frac{\partial f}{\partial x}$ and $-3x$ to $\frac{\partial f}{\partial y}$.

The mirror-image error is writing $\frac{\partial f}{\partial x} = 3x^2 - 3xy$, which keeps the $x$ in the cross term that should have been differentiated away.

Using the partials to locate critical points

Critical points are where both partials vanish at once, so solve the pair $3x^2 = 3y$ and $3y^2 = 3x$.

The real solutions of $x = x^4$ are $x = 0$ and $x = 1$, giving the critical points $(0,0)$ and $(1,1)$. Substituting back, $f(1,1) = 1 + 1 - 3 = -1$.

The second partials are $f_{xx} = 6x$, $f_{yy} = 6y$, and $f_{xy} = -3$. At $(1,1)$ the discriminant $f_{xx}f_{yy} - f_{xy}^2 = 36 - 9 = 27$ is positive with $f_{xx} = 6 > 0$, so $(1,1)$ is a local minimum. At $(0,0)$ the discriminant is $-9$, so the origin is a saddle.