Divergence of F = (sin(xy), cos(xy))

Chain rule, with the inner derivative matched to the variable

Both components wrap the same inside function $u = xy$, but the inner derivative is not the same, because you are differentiating with respect to a different variable each time.

First component, outer function $\sin$, differentiated with respect to $x$.

Second component, outer function $\cos$, differentiated with respect to $y$. The derivative of $\cos$ carries the minus sign.

Add them and you are done. As a spot check, at $(1,1)$ the divergence is $\cos 1 - \sin 1 \approx -0.3012$, so the field is a mild sink there.

The mistake: reusing the same inner factor twice

The wrong answer that shows up most often is $y\cos(xy) - y\sin(xy)$. The inner derivative $y$ from the first term has been copied into the second, but the second term is differentiated with respect to $y$, where the inner derivative of $xy$ is $x$.

A useful habit: before touching either component, write down both inner derivatives, $\frac{\partial u}{\partial x} = y$ and $\frac{\partial u}{\partial y} = x$, and then reach for the one that matches the variable you are working in.

The other frequent wrong answer, $y\cos(xy) + x\sin(xy)$, drops the minus sign that comes from differentiating cosine. Both errors survive a glance at the shape of the answer, so the only defence is doing the two steps separately and in order.

Every arrow has length 1, and the divergence still grows

This field is worth a second look because its arrows never change size. The two components are $\sin(xy)$ and $\cos(xy)$, so the length of every arrow is $\sqrt{\sin^{2}(xy)+\cos^{2}(xy)} = 1$. Only the direction varies from point to point.

Yet the divergence is not bounded. Collect the two terms with the standard combination of a sine and a cosine of the same angle and the size of the answer appears on its own.

So the divergence oscillates between $-r$ and $r$, where $r$ is the distance from the origin. Walk out along any ray that is not one of the axes and the angle $xy$ grows quadratically while the envelope $r$ grows only linearly, so the swings get both wider and faster. A field whose arrows all have length $1$ can still be a strong source at one point and a strong sink a short step away.

• On the whole $x$ axis, $y=0$, both terms vanish and the divergence is exactly $0$.
• On the $y$ axis, $x=0$, the divergence is just $y$, which is the envelope $r$ at full size.
• Everywhere else it sits somewhere between, at $\sqrt{x^{2}+y^{2}}\cos(xy+\varphi)$.

The lesson generalises past this example: divergence reads how the direction of a field turns as you move, not how long its arrows are. Fixing the length of every arrow puts no ceiling whatsoever on the divergence.