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

Quotient rule with the other letter frozen

For $\frac{\partial f}{\partial x}$, hold $y$ fixed. The numerator $x + y$ is then $x$ plus a number, so its $x$ derivative is $1$. The denominator has $x$ derivative $2x$.

Expand the second product to $2x^2 + 2xy$ and subtract. The $x^2$ already sitting in the first bracket absorbs one of the two, which is why the simplified numerator carries $-x^2$ rather than $-2x^2$.

Swapping the two letters leaves $f$ unchanged, so the $y$ partial is the same expression with $x$ and $y$ exchanged. There is no second computation to do.

At the origin both partials equal $1$, so $\nabla f(0,0) = \left\langle 1, 1 \right\rangle$ and the surface rises fastest along the diagonal. At $(1,1)$ both equal $-\frac{1}{9}$, so somewhere between those points the climb has already turned into a gentle descent.

The mistake: differentiating the frozen letter

Both letters appear in the numerator and both appear in the denominator, which is what makes this one easy to get wrong. During $\frac{\partial f}{\partial x}$, every $y$ on the page is a fixed number.

• $\frac{\partial}{\partial x}(x + y) = 1$, not $1 + y$ and not $2$.
• $\frac{\partial}{\partial x}\left(x^2 + y^2 + 1\right) = 2x$, since $y^2 + 1$ contributes nothing.
• The quotient rule numerator is $u'v - uv'$ in that order, and reversing it flips the sign of the whole answer.

This function hands you a free check. Swap $x$ and $y$ in your answer for $\frac{\partial f}{\partial x}$ and you must land exactly on your answer for $\frac{\partial f}{\partial y}$, because the swap does nothing to $f$ itself. If the two are not mirror images, at least one of them is wrong.

Where the surface is flat

The denominator is at least $1$ everywhere, so both partials are defined on the whole plane and their signs are decided entirely by the numerators.

Set both numerators to zero. Subtracting one from the other kills the $1$ and the $-2xy$, leaving $2\left(y^2 - x^2\right) = 0$, so a critical point has to sit on $y = x$ or on $y = -x$.

• On $y = -x$ the first numerator becomes $1 + 2x^2$, which is never zero, so that diagonal carries no critical point at all.
• On $y = x$ it becomes $1 - 2x^2$, which vanishes at $x = \pm\frac{1}{\sqrt{2}}$.
• So there are exactly two critical points: $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$.

Those two points are the global maximum and the global minimum. Restricted to the diagonal $y = x$ the function is $\frac{2x}{2x^2 + 1}$, whose largest value is $\frac{\sqrt{2}}{2}$ at $x = \frac{1}{\sqrt{2}}$, and no point off the diagonal does better.