Is $\frac{\partial}{\partial x} \cos{\sqrt{x^2+y^2}}$ indeterminate or zero at (0,0)?

Solution 1:

Interestingly, this function is not only differentiable at $(0,0)$ (with the official definition), but in fact $C^1$. If one makes the standard mistake (which I've seen instructors make in teaching single-variable calculus) of blithely saying $f'(0)=\lim_{x\to 0} f'(x)$, it does work out nicely here. Note that \begin{align*} \lim_{(x,y)\to (0,0)} \frac{\partial f}{\partial x}(x,y) &= \lim_{(x,y)\to (0,0)} \left(\frac x{\sqrt{x^2+y^2}}\right)\left(-\sin\sqrt{x^2+y^2}\right) \\ &= \lim_{(x,y)\to (0,0)} -x\left(\frac{\sin\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\right) = 0\cdot 1 = 0. \end{align*} Nothing whatsoever indeterminate. I think @Dunkelheit's comment is on point. Since I do not use Wolfram ordinarily to answer such questions, I had not previously bumped into this phenomenon.