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

Two quotient rules that collect differently

Write $D = x^{2}+y^{2}+1$ so the algebra stays short. Both partials are quotient rule problems, but they do not mirror each other, because $x$ appears squared in the numerator while $y$ appears only once.

Differentiating in $x$ with $y$ frozen, the numerator $x^{2}y$ has $x$-derivative $2xy$, and $D$ has $x$-derivative $2x$.

The bracket is where the simplification lives: $D - x^{2} = y^{2}+1$, so every $x^{2}$ cancels out of it.

Now freeze $x$. The numerator has $y$-derivative $x^{2}$ and $D$ has $y$-derivative $2y$, so this time the common factor pulled out front is $x^{2}$ and the bracket is $D - 2y^{2} = x^{2}-y^{2}+1$.

The mistake: reaching for a symmetry that is not there

When a function is unchanged by swapping $x$ and $y$, the second partial comes free: swap the letters in the first. The shortcut is so convenient that people reach for it here, where it fails. Swapping letters in $\frac{\partial f}{\partial x}$ produces $2xy\left(x^{2}+1\right)/D^{2}$, which is not the $y$-partial at all.

• Test the symmetry before you use it: $f(y,x) = \frac{xy^{2}}{x^{2}+y^{2}+1}$, which is not $f(x,y)$.
• $\frac{\partial f}{\partial x}$ carries a factor of $y$, so it vanishes along the whole $x$-axis.
• $\frac{\partial f}{\partial y}$ carries a factor of $x^{2}$, which is never negative, so its sign is decided entirely by the bracket $x^{2}-y^{2}+1$.

One line of checking catches a swapped answer. Put $y = 0$: the function is identically zero along the $x$-axis, so $\frac{\partial f}{\partial x}$ has to vanish there, and the factor $y$ makes it do so. On the same axis $\frac{\partial f}{\partial y} = \frac{x^{2}\left(x^{2}+1\right)}{\left(x^{2}+1\right)^{2}} = \frac{x^{2}}{x^{2}+1}$, which is positive away from the origin, and that is right: lifting $y$ off the axis makes $f$ positive.

Each vertical slice has a ridge

The sign of $\frac{\partial f}{\partial y}$ is the sign of $x^{2}-y^{2}+1$. Fix $x$ and walk upward in $y$: the surface climbs while $y^{2} < x^{2}+1$ and falls once $y$ passes that value, so the slice has a single ridge rather than growing forever.

At $(1,1)$ the denominator is 3, so $D^{2} = 9$. The $x$-partial is $\frac{2(1)(1)(2)}{9} = \frac{4}{9}$ and the $y$-partial is $\frac{1(1-1+1)}{9} = \frac{1}{9}$.

The $y$-component is still positive because $1 < \sqrt{2}$, so $(1,1)$ sits below the ridge of its slice. Move to $(1,2)$ and the same formula gives $-\frac{1}{18}$: past the ridge, on the way down.