Gradient of x/(1 + y^2 + z^2): Quotient and Chain

Skip the quotient rule where you can

Let $D = 1 + y^2 + z^2$. For the $x$ partial, $D$ contains no $x$, so $f$ is just $\frac{1}{D}$ times $x$ and the derivative is that coefficient.

For $y$ and $z$ it is cleaner to write $f = xD^{-1}$ and use the power rule with the chain rule than to run the full quotient rule. With $x$ frozen, $\frac{\partial}{\partial y}\left(D^{-1}\right) = -D^{-2}\frac{\partial D}{\partial y}$ and $\frac{\partial D}{\partial y} = 2y$.

The $z$ component follows the same path with $2z$ in place of $2y$. The quotient rule gives the identical answer, since the numerator derivative term is zero, but it takes twice the writing and offers twice the chance of a sign error.

The mistake: a first power in the denominator

The most common wrong answer is $\frac{\partial f}{\partial y} = \frac{-2xy}{1 + y^2 + z^2}$, with the denominator left unsquared. Differentiating $D^{-1}$ produces $D^{-2}$, so the power in the denominator always goes up by one.

• Check the far field. As $y$ grows, $f$ flattens out toward zero, so $\frac{\partial f}{\partial y}$ must shrink faster than $f$ does. A squared denominator does that, an unsquared one does not.
• The minus sign is real: increasing $\left| y \right|$ makes the denominator bigger and pushes $f$ toward zero, so for positive $x$ and $y$ the function decreases.
• A third slip is treating $D$ as a constant for $y$ as well, which gives zero for two of the three components.

Substituting a point, and why there are no critical points

At $(2, 1, 1)$ the denominator is $D = 1 + 1 + 1 = 3$, so $f = \frac{2}{3}$ and $D^2 = 9$.

Reading the signs: pushing $x$ up increases $f$, while pushing $y$ or $z$ away from zero decreases it, because they only enlarge the denominator. The $y$ and $z$ components are equal here because the point is symmetric in those two coordinates.

The $x$ component is $\frac{1}{D}$, which is strictly positive everywhere. So $\nabla f$ is never the zero vector, and this function has no local maxima or minima anywhere in space. It has no global ones either: along the $x$-axis the denominator is 1 and $f = x$, which runs off to $+\infty$ one way and $-\infty$ the other.