Gradient of x^2 y^3: Exponent Bookkeeping Done Right

Only one exponent moves at a time

A monomial is the easiest gradient to compute once you see the pattern. Freeze $y$, and the factor $y^3$ becomes a coefficient sitting in front of $x^2$.

Freeze $x$, and now $x^2$ is the coefficient while $y^3$ is the live power.

At $(1,2)$: the first component is $2(1)(8) = 16$ and the second is $3(1)(4) = 12$, so $\nabla f(1,2) = \langle 16, 12\rangle$ with length $\sqrt{256 + 144} = 20$.

The mistake: bringing down both exponents at once

Because $x^2 y^3$ looks like a product of two powers, students often write $\frac{\partial f}{\partial x} = 2x \cdot 3y^2 = 6xy^2$, differentiating both factors in one step. That is the product rule applied where it does not belong.

The product rule is for factors that both depend on the differentiation variable. Under $\frac{\partial}{\partial x}$ the factor $y^3$ is frozen, so it behaves like the number $8$ would if $y$ happened to equal $2$. You would never differentiate the $8$ in $8x^2$.

• In $\frac{\partial f}{\partial x}$ the exponent on $y$ stays at $3$, not $2$.
• In $\frac{\partial f}{\partial y}$ the exponent on $x$ stays at $2$, not $1$.
• Degree check: $f$ has total degree $5$, so each component must have total degree $4$. Both $2xy^3$ and $3x^2y^2$ do, while the faulty $6xy^2$ has degree $3$ and fails.

Where the gradient vanishes, and a homogeneity check

Both components carry a factor of $x$ and a factor of $y$, so the gradient is the zero vector on both coordinate axes. If $x = 0$ then $2xy^3 = 0$ and $3x^2y^2 = 0$; the same happens if $y = 0$.

The two axes behave completely differently, which is worth the extra minute. On the $x$-axis with $x \ne 0$, moving to positive $y$ makes $f$ positive and moving to negative $y$ makes it negative, so $f$ takes both signs in every neighbourhood and the point is no kind of extreme value. The origin behaves the same way.

On the $y$-axis the sign of $y^3$ is locked. Near $(0,2)$ every nearby $y$ is still positive, so $f = x^2y^3 \ge 0$ with equality exactly on the axis, making $(0,2)$ a local minimum. Near $(0,-2)$ the cube is negative, so $f \le 0$ nearby and the point is a local maximum. These are not strict extremes, because the whole axis ties at the value $0$, but they are genuine extremes all the same.

Since $f$ is homogeneous of degree $5$, meaning $f(tx,ty) = t^5 f(x,y)$, Euler's relation gives a quick check on both components at once.

If your two components do not combine to $5f$ in this way, at least one of them is wrong.