0
$\begingroup$

For the part highlighted in green I assume we are using Theorem 5.11 (below). However, from the the calculations of the partial derivatives we know that $\frac{\partial g}{\partial x}$ and $\frac{\partial g}{\partial y}$ exist and are equal to $0$ for $(x,y)=(0,0)$ and hence continuous at $(x,y)=(0,0)$.

But the above implies $g$ is differentiable at $(0,0)$. Where am I going wrong?

enter image description here

enter image description here

enter image description here

$\endgroup$

2 Answers 2

3
$\begingroup$

The partial derivatives of $g$ at $(0,0)$ exist (and are $=0$), but they are not continuous at $0$. (At any rate, you haven't checked their continuity at $(0,0)$, but wait).

If $g$ were differentiable at the origin its derivative would have to be $0$, because we have computed the partial derivatives, which turned out to be $0$. But this means that one should have $$\lim_{(x,y)\to(0,0)}{g(x,y)-g(0,0)\over \sqrt{x^2+y^2}}=0\ .$$ On the other hand, by definition of $g$, one has $${g(x,y)-g(0,0)\over \sqrt{x^2+y^2}}={r\cos\phi\cdot r\sin\phi\over r^2}=\cos\phi\sin\phi\ ,$$ which certainly does not go to $0$ when $r\to0$.

It follows that $g$ is not differentiable at $(0,0)$, which entails, without any calculation, that the partial derivatives $g_x$ and $g_y$ (as functions of two variables!) are not continuous at $(0,0)$.

$\endgroup$
0
$\begingroup$

Hint: For the partial derivatives, try to approach $(0,0)$ along the lines $x=0$ or $y=0$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy