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MY ATTEMPT: \begin{equation} fx(0,0)= \lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h}=0 \end{equation} \begin{equation} fy(0,0)= \lim_{k\to 0}\frac{f(0,0+k)-f(0,0)}{k}=0 \end{equation} By Using the definition of differentiability for the functions of two variables: \begin{equation} \lim_{(h,k)\to (0,0)}\frac{f(a+h,b+k)-f(a,b)-hf_x(a,b)-kf_y(a,b))}{{\sqrt{h^2+k^2}}} \end{equation} \begin{equation} \lim_{(h,k)\to (0,0)}\frac{|hk|^p}{\sqrt{h^2+k^2}}=\text{0 if and only if p=2n where n=1,2,3..} \end{equation} so, $p=2n (n=1,2,3\ldots)$

PLEASE CHECK whether I attempted this question correctly or not and if there is any error please give suggestions to resolve it.

Thank you.

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  • $\begingroup$ why $p=2n$ and how did you proved that? If you show how you arrived at these values of $p$ it will be easier to help you. $\endgroup$
    – Marcos
    Commented Feb 23, 2022 at 10:54
  • $\begingroup$ actually, I got the answer. P>1. $\endgroup$
    – ashish
    Commented Feb 23, 2022 at 11:05
  • $\begingroup$ Yeah, that is the correct answer, but this needs to be proven. Why $p>1$ works and $p\leq1$ does not work? $\endgroup$
    – Marcos
    Commented Feb 23, 2022 at 11:08
  • $\begingroup$ You are right. Thank you for your advice. $\endgroup$
    – ashish
    Commented Feb 23, 2022 at 11:13

1 Answer 1

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We'll try to prove that your function is differentiable if and only if $p >1/2$

For the case $p > 1/2$:

We know that : $\sqrt {{h^2}+{k^2}}\ge \sqrt {2\lvert h \rvert \lvert k \rvert}$, thus : $$0\le \frac{\lvert hk \rvert ^p}{\sqrt {{h^2}+{k^2}}}\le \frac{\lvert hk \rvert ^p}{{\sqrt2 \lvert hk \rvert ^{1/2}}}=\frac{1}{\sqrt2}{(\lvert hk \rvert) ^{p-1/2}} $$ therefore if $p>1/2$, $$\lim\limits_{(h,k) \to (0,0)}\frac{\lvert hk \rvert ^p}{\sqrt {{h^2}+{k^2}}}=0$$ and the function is differentiable at (0,0)

Now if $p\le1/2$ : let's have a look at the limit of $\frac{\lvert hk \rvert ^p}{\sqrt {{h^2}+{k^2}}}$ along the curves : $k=mh , m\in \mathbb{R}^*$ :

$$\lim\limits_{h \to 0}\frac{\lvert mh^2\rvert ^p}{\sqrt {{h^2}+{m^2h^2}}}=\lim\limits_{h \to 0}\frac{{\lvert m\rvert}^p{\lvert h\rvert}^{2p-1}}{\sqrt {1+m^2}}$$ If $p=1/2$ the limit depends on m thus there's no limit and the function is not differentiable at $(0,0)$

If $p<1/2$ limit is $\infty$ so the function is also not differentiable at $(0,0)$

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