Gradient of xy/(z^2+1): A Three Variable Gradient

Two easy partials, then one that needs the chain rule

Sort the variables by where they live. Both $x$ and $y$ appear only upstairs, and $z$ appears only downstairs. That tells you in advance which partial is the awkward one.

For $\frac{\partial f}{\partial x}$, everything except $x$ is a constant, so the whole function is $\left(\frac{y}{z^2+1}\right)x$.

For $\frac{\partial f}{\partial z}$, pull out the constant $xy$ and differentiate $\left(z^2+1\right)^{-1}$ with the power and chain rules.

At $(1,2,0)$ the denominator is $1$, and the third component carries a factor of $z = 0$, so $\nabla f(1,2,0) = \langle 2, 1, 0\rangle$. At $(1,1,1)$ the denominator is $2$ and the gradient is $\left\langle \tfrac{1}{2}, \tfrac{1}{2}, -\tfrac{1}{2}\right\rangle$.

The mistake: treating the denominator as inert in the z direction

Because $z$ never appears in the numerator, students often conclude that $\frac{\partial f}{\partial z} = 0$. That would say the value of $f$ does not depend on $z$ at all, which is false: at $z = 0$ the fraction is $xy$, and at $z = 3$ it is only $\frac{xy}{10}$.

The other slip is dropping the minus sign. Increasing $\left|z\right|$ grows the denominator, which shrinks $\left|f\right|$, so the $z$ partial must oppose the sign of $xyz$.

• A variable in the denominator still counts as a variable.
• $\frac{\partial}{\partial z}\left[\left(z^2+1\right)^{-1}\right] = -2z\left(z^2+1\right)^{-2}$, so the sign flips and the exponent drops to $-2$.
• Slice check with $x = y = 1$: $g(z) = \frac{1}{z^2+1}$ has $g'(1) = \frac{-2}{4} = -\tfrac{1}{2}$, matching the third component at $(1,1,1)$.

Reading the geometry from the three components

The factor $\frac{1}{z^2+1}$ never exceeds $1$ and never reaches zero, so it acts as a damping dial. Sliding away from the plane $z = 0$ in either direction flattens the saddle shape $xy$ without changing which way it tilts.

The plane $z = 0$ is exactly where the third component vanishes, because of the factor $z$. On that plane the gradient reduces to $\langle y, x, 0\rangle$, the gradient of $xy$ with a zero appended.

The gradient is the zero vector when $y = 0$, $x = 0$ and $xyz = 0$ all hold, which is the whole $z$-axis. Along that axis $f = 0$, and nearby values take both signs, so none of those points is a maximum or a minimum.