# For the function $f(x,y)=|xy|^{p}$ , find the value of $p$ for which $f$ is differentiable at $(0,0)$.

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

• 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. Commented Feb 23, 2022 at 10:54
• actually, I got the answer. P>1. Commented Feb 23, 2022 at 11:05
• Yeah, that is the correct answer, but this needs to be proven. Why $p>1$ works and $p\leq1$ does not work? Commented Feb 23, 2022 at 11:08
• You are right. Thank you for your advice. Commented Feb 23, 2022 at 11:13

## 1 Answer

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 $$(0,0)$$