Multivariable calculus

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

Q: Does f_xx = 0 mean the graph is a plane?

No. It means each slice at fixed y is a straight line. The lines change slope as y changes, and the surface bends in the y direction, so the graph is a curved ruled surface rather than a plane.

Q: Where does f_yy change sign?

Where 6y^{2} = 2, that is y = \pm\frac{1}{\sqrt{3}} \approx \pm 0.577. The sign also depends on x, since f_{yy} carries a factor of x and flips as x crosses zero.

For f(x,y) = x/(1 + y^2) the second partials are f_xx = 0, f_yy = x(6y^2 - 2)/(1 + y^2)^3, and the mixed partial f_xy = -2y/(1 + y^2)^2. The pure x partial vanishes because the function is linear in x, but the surface is still curved: the mixed partial is nonzero whenever y is not zero.

f(x,y) = \frac{x}{1 + y^{2}}

\frac{\partial f}{\partial x} = \frac{1}{1 + y^{2}}

\frac{\partial f}{\partial y} = \frac{-2xy}{\left(1 + y^{2}\right)^{2}}

Second and mixed partials

\frac{\partial^{2} f}{\partial x^{2}} = 0

\frac{\partial^{2} f}{\partial y\,\partial x} = \frac{-2y}{\left(1 + y^{2}\right)^{2}}

\frac{\partial^{2} f}{\partial y^{2}} = \frac{x\left(6y^{2} - 2\right)}{\left(1 + y^{2}\right)^{3}}

Splitting the work between the two variables

In $x$ this function is linear: the whole factor $\frac{1}{1 + y^{2}}$ is a constant slope. In $y$ it is a rational function with a denominator that never vanishes, so both partials exist everywhere.

\frac{\partial f}{\partial x} = \frac{1}{1 + y^{2}}, \qquad \frac{\partial f}{\partial y} = \frac{-2xy}{\left(1 + y^{2}\right)^{2}}

Differentiating $f_x$ again in $x$ gives zero, since $\frac{1}{1 + y^{2}}$ has no $x$ in it. Differentiating $f_x$ in $y$ instead gives the mixed partial through the power rule on $\left(1 + y^{2}\right)^{-1}$ .

\frac{\partial^{2} f}{\partial x^{2}} = 0, \qquad \frac{\partial^{2} f}{\partial y\,\partial x} = \frac{-2y}{\left(1 + y^{2}\right)^{2}}

For $f_{yy}$ , differentiate $-2xy\left(1 + y^{2}\right)^{-2}$ using the product rule, then factor out one power of the denominator.

\frac{\partial^{2} f}{\partial y^{2}} = -2x\,\frac{\left(1 + y^{2}\right) - 4y^{2}}{\left(1 + y^{2}\right)^{3}} = \frac{x\left(6y^{2} - 2\right)}{\left(1 + y^{2}\right)^{3}}

The mistake students make

Seeing $f_{xx} = 0$ , students often conclude that the surface has no curvature at all, or that the Hessian is degenerate in a way that makes the second derivative test unnecessary. Neither follows. The Hessian determinant here is $D = 0 \cdot f_{yy} - \left(f_{xy}\right)^{2} = -\left(\frac{2y}{(1 + y^{2})^{2}}\right)^{2}$ , which is negative whenever $y \neq 0$ .

What $f_{xx} = 0$ really says is narrower: every cross section taken at fixed $y$ is a straight line, so the surface is ruled. Along other directions it bends, and the mixed partial is what records that bending.

The second slip is a chain rule miss on the way to $f_{yy}$ . Differentiating $\left(1 + y^{2}\right)^{-2}$ produces $-2\left(1 + y^{2}\right)^{-3} \cdot 2y$ , and dropping that inner $2y$ leaves a formula whose sign is wrong for large $y$ .

Clairaut symmetry with an asymmetric function

This function is not symmetric in $x$ and $y$ , which makes it a better test of Clairaut's theorem than a symmetric example. Start from $f_y$ and differentiate in $x$ .

\frac{\partial^{2} f}{\partial x\,\partial y} = \frac{\partial}{\partial x}\left(\frac{-2xy}{\left(1 + y^{2}\right)^{2}}\right) = \frac{-2y}{\left(1 + y^{2}\right)^{2}}

That matches the other order, even though the two computations look nothing alike: one was a power rule in $y$ , the other a constant multiple rule in $x$ . Continuity of both mixed partials is what forces the agreement, and here the denominator is at least 1 everywhere, so continuity holds on the whole plane.

The mixed partial is negative for $y > 0$ and positive for $y < 0$ , with extreme values at $y = \pm\frac{1}{\sqrt{3}}$ . That is the rate at which the slope in $x$ changes as you move in $y$ , which is exactly what a mixed partial measures.

Frequently asked questions

Does f_xx = 0 mean the graph is a plane?

No. It means each slice at fixed $y$ is a straight line. The lines change slope as $y$ changes, and the surface bends in the $y$ direction, so the graph is a curved ruled surface rather than a plane.

Where does f_yy change sign?

Where $6y^{2} = 2$ , that is $y = \pm\frac{1}{\sqrt{3}} \approx \pm 0.577$ . The sign also depends on $x$ , since $f_{yy}$ carries a factor of $x$ and flips as $x$ crosses zero.