Gradient of x^2/(y^2+1): When to Skip Quotient Rule

Split the function before you differentiate

Read the function as $f = x^2 \cdot \left(y^2+1\right)^{-1}$, a product of one factor in $x$ and one factor in $y$. Each partial then has exactly one live factor.

With $y$ frozen, $\frac{1}{y^2+1}$ is a constant multiplier and only $x^2$ moves.

With $x$ frozen, $x^2$ is the constant multiplier and the power rule handles $\left(y^2+1\right)^{-1}$.

At $(2,1)$: the first component is $\frac{4}{2} = 2$ and the second is $\frac{-2(4)(1)}{4} = -2$, so $\nabla f(2,1) = \langle 2, -2\rangle$.

The mistake: forcing the quotient rule on both components

Applied honestly to $\frac{\partial f}{\partial x}$, the quotient rule gives $\frac{\left(y^2+1\right)(2x) - x^2(0)}{\left(y^2+1\right)^2}$. The second term is zero because the denominator has no $x$, and cancelling one factor of $y^2+1$ returns $\frac{2x}{y^2+1}$. Correct, but three extra chances to slip.

Where it goes wrong is when students carry over a $2y$ from the other partial and write $\frac{\left(y^2+1\right)(2x) - x^2(2y)}{\left(y^2+1\right)^2}$. The derivative of the denominator with respect to $x$ is zero, not $2y$.

• If the denominator does not contain your variable, the whole fraction is a constant multiplier.
• Reserve the quotient rule for the component where both top and bottom actually move.
• Check by slicing: $f(x,1) = \frac{x^2}{2}$ has derivative $x$, and $\frac{2x}{y^2+1}$ at $y = 1$ gives $x$ as well.

A trough floor along the y-axis and a ridge along the x-axis

Both components vanish together only when $x = 0$, since $\frac{2x}{y^2+1} = 0$ forces $x = 0$ and then the second component is zero automatically. So the entire $y$-axis is a line of critical points, all with $f = 0$.

None of them is a strict minimum in the usual isolated sense, but each is a minimum in value, because $f \ge 0$ everywhere. The surface is a trough whose floor runs along the $y$-axis.

The signs read cleanly. The $x$-component has the sign of $x$, so the surface climbs away from the axis in both directions. The $y$-component has the sign opposite to $y$ whenever $x \ne 0$, so moving toward $y = 0$ always gains height. The ridge line sits along the $x$-axis, where $f(x,0) = x^2$ is as large as it gets for a given $x$.