Multivariable calculus

Second Partial Derivatives of arctan(xy)

Q: Why is arctan(xy) smooth everywhere when arctan of a quotient is not?

The denominator 1 + x^{2}y^{2} is at least 1 at every point, so nothing divides by zero. A function like \arctan(y/x) fails on the line x = 0, where the quotient itself is undefined.

Q: What is the Hessian determinant of arctan(xy)?

It is \frac{4x^{4}y^{4} - (1 - x^{2}y^{2})^{2}}{Q^{4}} with Q = 1 + x^{2}y^{2}. At the origin this equals -1, so the origin is a saddle point of the surface.

Write Q = 1 + x^2 y^2. For f(x,y) = arctan(xy) the second partials are f_xx = -2xy^3/Q^2, f_yy = -2x^3 y/Q^2, and the mixed partial f_xy = (1 - x^2 y^2)/Q^2. The mixed partial equals 1 at the origin and changes sign along the curves xy = 1 and xy = -1.

f(x,y) = \arctan(xy)

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

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

Second and mixed partials

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

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

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

From the arctangent rule to the second partials

The one variable rule is $\frac{d}{du}\arctan u = \frac{1}{1 + u^{2}}$ . With $u = xy$ the chain rule supplies a factor $y$ for an $x$ derivative and a factor $x$ for a $y$ derivative. Write $Q = 1 + x^{2}y^{2}$ , which is never zero.

\frac{\partial f}{\partial x} = \frac{y}{Q}, \qquad \frac{\partial f}{\partial y} = \frac{x}{Q}

For $f_{xx}$ the numerator $y$ is constant in $x$ , so differentiate $Q^{-1}$ and multiply. Since $\frac{\partial Q}{\partial x} = 2xy^{2}$ , the result carries three powers of $y$ .

\frac{\partial^{2} f}{\partial x^{2}} = y \cdot \left(-\frac{2xy^{2}}{Q^{2}}\right) = \frac{-2xy^{3}}{Q^{2}}

The mixed partial is a real quotient rule, because differentiating $\frac{y}{Q}$ in $y$ moves both the numerator and the denominator.

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

The mistake students make

The first error is forgetting the inner derivative, writing $\frac{\partial f}{\partial x} = \frac{1}{1 + x^{2}y^{2}}$ with no factor of $y$ . Every second partial then comes out wrong, and the symmetry check fails immediately.

The second error is treating $f_{xy}$ like $f_{xx}$ , holding the numerator fixed and differentiating only $Q^{-1}$ . That gives $-\frac{2x^{2}y^{2}}{Q^{2}}$ and loses the $Q$ term from the numerator, which is precisely the piece that makes $f_{xy}(0,0) = 1$ rather than $0$ .

Use the point $x = y = 1$ as a check. There $Q = 2$ , so the correct values are $f_{xx} = -\frac{2}{4} = -0.5$ and $f_{xy} = \frac{0}{4} = 0$ . A formula that does not give zero for the mixed partial on the hyperbola $xy = 1$ is wrong.

Symmetry and what the mixed partial shows

Because $\arctan(xy)$ is unchanged when $x$ and $y$ trade places, $f_{yy}$ is $f_{xx}$ with the letters swapped, and the mixed partial must be symmetric in $x$ and $y$ on its own. The expression $\frac{1 - x^{2}y^{2}}{Q^{2}}$ passes that test, since it depends only on the product $xy$ .

Clairaut's theorem confirms the same thing analytically. Differentiating $\frac{x}{Q}$ in $x$ gives $\frac{Q - x(2xy^{2})}{Q^{2}} = \frac{1 - x^{2}y^{2}}{Q^{2}}$ , and every derivative here is continuous because $Q \geq 1$ everywhere.

The sign of $f_{xy}$ tells you how the surface twists. It is positive inside the region $|xy| < 1$ , zero on the hyperbolas $xy = \pm 1$ , and negative beyond them, which is the arctangent flattening toward its horizontal asymptotes.

Frequently asked questions

Why is arctan(xy) smooth everywhere when arctan of a quotient is not?

The denominator $1 + x^{2}y^{2}$ is at least 1 at every point, so nothing divides by zero. A function like $\arctan(y/x)$ fails on the line $x = 0$ , where the quotient itself is undefined.

What is the Hessian determinant of arctan(xy)?

It is $\frac{4x^{4}y^{4} - (1 - x^{2}y^{2})^{2}}{Q^{4}}$ with $Q = 1 + x^{2}y^{2}$ . At the origin this equals $-1$ , so the origin is a saddle point of the surface.