Gradient of x^2 y z^3: Monomial in Three Variables

The power rule, applied to one variable at a time

A monomial makes the gradient mechanical. To differentiate with respect to $x$, treat $yz^3$ as a constant multiplier and apply the power rule to $x^2$.

For $y$ the exponent is 1, so it comes down as a 1 and $y$ disappears entirely. For $z$ the exponent 3 comes down and leaves $z^2$.

There is a shortcut worth knowing for monomials. Away from the coordinate planes you can factor $f$ out of every component, since $\frac{\partial f}{\partial x} = f \cdot \frac{2}{x}$ and similarly for the others.

The mistake: lowering every exponent at once

The most common wrong answer is $\frac{\partial f}{\partial x} = 2xz^2$ or $6xyz^2$, produced by differentiating two variables in the same step. A partial derivative changes exactly one exponent.

• Check the total degree. The function has degree $2 + 1 + 3 = 6$, so every component of the gradient must have degree 5.
• $2xyz^3$ has degree $1+1+3 = 5$, $x^2z^3$ has degree $2+3=5$, and $3x^2yz^2$ has degree $2+1+2=5$. All three pass.
• A wrong answer like $6xyz^2$ has degree 4, which flags the error before you substitute a single number.

The second slip is losing the $y$ component. Since $y$ appears to the first power, $\frac{\partial f}{\partial y} = x^2 z^3$ contains no $y$ at all, and students sometimes read that as a mistake and put the $y$ back.

A numerical check at a point

At $(2, 3, 1)$ the function value is $f = 4 \cdot 3 \cdot 1 = 12$.

The $z$ component is by far the largest, which the shortcut formula predicts: the exponent on $z$ is the biggest, and at $z=1$ the factor $\frac{3}{z}$ beats $\frac{2}{x} = 1$ and $\frac{1}{y} = \frac{1}{3}$. In relative terms the exponent is the whole story: raising $z$ by one percent raises $f$ by about three percent, raising $x$ by one percent raises it by two, and raising $y$ by one percent raises it by one.

Every component carries a factor of $x$ and a factor of $z$, so $\nabla f = \mathbf{0}$ exactly on the plane $x = 0$ and on the plane $z = 0$. The critical set is two whole planes. Setting $y = 0$ alone is not enough, since $\frac{\partial f}{\partial y} = x^2 z^3$ survives it.