Partial Derivatives of x/(y^2 + 1)

Look at the structure before choosing a rule

With $y$ held fixed, $\frac{1}{y^2+1}$ is a single number, and the function is that number times $x$. The derivative of $cx$ is $c$.

For the $y$ partial, rewrite as $f = x\left(y^2+1\right)^{-1}$ and use the chain rule with $x$ as a constant multiplier. The outer derivative is $-\left(y^2+1\right)^{-2}$ and the inner derivative is $2y$.

At $(3,1)$ the partials are $\frac{1}{2}$ and $-\frac{3}{2}$. The $x$ partial does not depend on $x$ at all, which says every slice taken at a fixed $y$ is a straight line.

The mistake: quotient rule where none is needed

Seeing a fraction, many students apply the quotient rule to $\frac{\partial f}{\partial x}$ and write the numerator as $1 \cdot \left(y^2+1\right) - x \cdot 2y$. The second term is wrong: during an $x$ derivative, the derivative of $y^2 + 1$ is $0$, not $2y$.

Applied correctly, the quotient rule still works; it is just slower.

• If the denominator has no $x$ in it, skip the quotient rule for $\frac{\partial f}{\partial x}$.
• If the numerator has no $y$ in it, factor it out before doing $\frac{\partial f}{\partial y}$.
• Choosing the lighter route is the difference between one line and four.

Reading the two partials

$\frac{\partial f}{\partial x}$ is always positive and at most $1$, with the maximum at $y = 0$. So the surface always rises as $x$ increases, most steeply along the $x$ axis and more gently for large $\lvert y \rvert$.

$\frac{\partial f}{\partial y}$ carries the sign of $-xy$, so the surface falls in the $y$ direction in the first and third quadrants and rises in the second and fourth. It is zero on both axes.

There are no critical points anywhere, because $\frac{\partial f}{\partial x} = \frac{1}{y^2+1}$ can never equal zero. The gradient is never the zero vector, so the surface has no flat spots at all.