 Laila Murphy

2022-11-17

Let $f:{\mathbb{R}}^{2}\to \mathbb{R}$ be defined as

then check whether its differentiable and also whether its partial derivatives ie ${f}_{x},{f}_{y}$ are continuous at $\left(0,0\right)$. I dont know how to check the differentiability of a multivariable function as I am just beginning to learn it. For continuity of partial derivative I just checked for ${f}_{x}$ as function is symmetric in $y$ and $x$. So ${f}_{x}$ turns out to be
${f}_{x}\left(x,y\right)=2x\mathrm{cos}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\mathrm{sin}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)$
which is definitely not $0$ as $\left(x,y\right)\to \left(0,0\right)$. Same can be stated for ${f}_{y}$. But how to proceed with the first part? kavdawg8w8

The partial derivatives at $\left(0,0\right)$ are ${\mathrm{\partial }}_{x}f\left(0,0\right)=0$ and ${\mathrm{\partial }}_{y}f\left(0,0\right)=0$.
The only candidate for the differential at $\left(0,0\right)$ is:
$Df\left(0,0\right)=\left[\begin{array}{cc}{\mathrm{\partial }}_{x}f\left(0,0\right)& {\mathrm{\partial }}_{y}f\left(0,0\right)\end{array}\right]=\left[\begin{array}{cc}0& 0\end{array}\right]$
which is simply the zero-functional.
We have:
$\begin{array}{rl}\underset{\left({h}_{1},{h}_{2}\right)\to \left(0,0\right)}{lim}\frac{|f\left({h}_{1},{h}_{2}\right)-f\left(0,0\right)-Df\left(0,0\right)\left({h}_{1},{h}_{2}\right)|}{‖\left({h}_{1},{h}_{2}\right)‖}& =\underset{\left({h}_{1},{h}_{2}\right)\to \left(0,0\right)}{lim}\frac{{h}_{1}^{2}+{h}_{2}^{2}}{\sqrt{{h}_{1}^{2}+{h}_{2}^{2}}}\cdot |\mathrm{cos}\frac{1}{\sqrt{{h}_{1}^{2}+{h}_{2}^{2}}}|\\ & =\underset{\left({h}_{1},{h}_{2}\right)\to \left(0,0\right)}{lim}\sqrt{{h}_{1}^{2}+{h}_{2}^{2}}\cdot \underset{\le 1}{\underset{⏟}{|\mathrm{cos}\frac{1}{\sqrt{{h}_{1}^{2}+{h}_{2}^{2}}}|}}\\ & =0\end{array}$
Hence, $f$ is differetiable at $\left(0,0\right)$ with the differential being 0, even though the partial derivatives are not continuous at $\left(0,0\right)$. reevelingw97

we have
$\frac{\mathrm{\partial }f\left(0,0\right)}{\mathrm{\partial }y}=\underset{h\to 0}{lim}h\mathrm{cos}\left(\frac{1}{h}\right)=0$
since $|h\mathrm{cos}\left(\frac{1}{h}\right)|\le |h|$ which tends to $0$

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