Let $f(x,y)=|x|+|y|$ and $g(x,y)=|xy|^{1/2}$. $f$ and $g$ are differentiable in $(0,0)$?

Question

Let $f(x,y)=|x|+|y|$ and $g(x,y)=|xy|^{1/2}$. $f$ and $g$ are differentiable in $(0,0)$?

I'm having trouble calculating the limit after I apply the definition of differentiability. For example, in $g(x,y)=|xy|^{1/2}$, by the definition of differentiability, taking $h = (h_1,h_2)$,

$\dfrac{g(0+h)-g(0)-Bh}{||h||}= \dfrac{ \sqrt{|h_1.h_2|}}{||h||}= \dfrac{ \sqrt{|h_1.h_2|}}{\sqrt{h_1^2+h_2^2}}$.

but and now? I could for example take $h_1 = h_2$ and see that this limit goes to ${\sqrt{\dfrac{1}{2}}}$? Hence conclude that $g$ is not differentiable in $(0,0)$? or should I do something else?

For $f$, I have not been able to conclude the existence of the partial derivatives because the limit does not exist since by doing $t\rightarrow 0 ^ +$ and $t\rightarrow 0 ^- $ I get distinct limits. In this case I can already say that f is not differentiable? Thanks for any help!

Answer 1

U can check the differentiability by putting k=mh according to formula for dual variable (phi)(h, K) =$\frac{(hk)^1/2}{(h^2+k^2)^1/2}$ put k=mh and take limit as ${h\to∞}$ u get $\frac{m}{(1+m^2)^1/2}$ which is not differentiable

Answer 2

For the function $f$ being differentiable at $(0,0)$, then $\dfrac{\partial f}{\partial x}(0,0)$ would exist, then \begin{align*} \lim_{h\rightarrow 0^{+}}\dfrac{f(h,0)-f(0,0)}{h}=\lim_{h\rightarrow 0^{-}}\dfrac{f(h,0)-f(0,0)}{h}, \end{align*} but we can easily deduce a contradiction here.

Answer 3

Here is one way: Consider the function restricted to a line through the origin.

If the function is differentiable, then so will the restriction.

For $f$, consider the line along the $x$ axis. Then we have $\phi(t) = f(t,0) = |t|$. However, the absolute value is not differentiable at zero hence $f$ cannnot be differentiable.

Similarly, if we let $\gamma(t) = g(t,t) = |t|$ we see that $g$ is not differentiable.

To see that $a(t) = |t|$ is not differentiable at $t=0$, note that $\lim_{h \downarrow 0} {|t|-|0| \over t} = 1$, whereas $\lim_{h \uparrow 0} {|t|-|0| \over t} = -1$.