Partial Derivatives of x^2 y^3: Method and Answers

Freeze one variable, then use one-variable rules

A partial derivative is an ordinary derivative in disguise. For $\frac{\partial f}{\partial x}$, treat $y$ as a number you happen not to know. Then $y^3$ is a constant sitting in front of $x^2$, and the power rule finishes the job.

Now swap roles. For $\frac{\partial f}{\partial y}$, the factor $x^2$ is the constant and $y^3$ is the live variable.

The pattern for any monomial $x^m y^n$ is the same: differentiate the exponent belonging to your variable and leave the other exponent untouched.

The mistake: reaching for the product rule

Because $x^2 y^3$ looks like a product, students often write $\frac{\partial f}{\partial x} = 2x \cdot y^3 + x^2 \cdot 3y^2$, differentiating both factors. That mixes two different derivatives into one answer.

The product rule applies when both factors depend on the variable you are differentiating. Here $y^3$ does not depend on $x$ at all, so its derivative with respect to $x$ is zero and the second term collapses.

• For $\frac{\partial f}{\partial x}$: $y$ is frozen, so $\frac{\partial}{\partial x}\left(y^3\right) = 0$.
• The product rule would give $2xy^3 + x^2 \cdot 0 = 2xy^3$, the same answer, only with a wasted term.
• Writing $3y^2$ instead of $0$ for that second factor is what produces the wrong answer.

You genuinely need the product rule only when both factors carry the differentiation variable, for example $\frac{\partial}{\partial x}\left(x^2 \sin x \cdot y^3\right)$.

Check the two answers at a point

Numbers catch algebra slips faster than rereading does. Take the point $(2, 1)$.

Sanity test the sizes against the surface. Moving in the $y$ direction from $(2,1)$ changes $f$ faster than moving in $x$, because the cubic exponent on $y$ combines with the large factor $x^2 = 4$.

The mixed second partials also agree, as Clairaut's theorem promises for a polynomial: differentiating $2xy^3$ with respect to $y$ and $3x^2y^2$ with respect to $x$ both give $6xy^2$.