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> Check the differentiability of the function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ of two variables given by the formula $$f(x)=|x\_1|\cdot x\_2$$
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I still have problems with this. I started by trying to count partial derivatives:

$\displaystyle\frac{\partial f}{\partial x\_1}(x)=\lim\_{h\to 0}\frac{f(x\_1+h,x\_2)-f(x\_1,x\_2)}{h}=\lim\_{h\to 0}\frac{|x\_1+h|x\_2-|x\_1|x\_2}{h}$, so I think the problems can be at the points: $(0,x\_2)$, where $x\_2\neq 0$, because then we have that this limit is equal to $\displaystyle\lim\_{h\to 0}\frac{|h|x\_2}{h}$ which doesn't exist (left and right limits are not equal).

$\displaystyle\frac{\partial f}{\partial x\_2}(x)=\lim\_{h\to 0}\frac{f(x\_1,x\_2+h)-f(x\_1,x\_2)}{h}=\lim\_{h \to 0}\frac{|x\_1|(x\_2+h)-|x\_1|x\_2}{h}=|x\_1|$, so I think we haven't any problems here, this partial derivative always exists.

But what exactly can we deduce from these speculations about partial derivatives?

I've also tried to proudly find the differential of this function. I was taught that the function is differentiable at the point $x$ iff there exists (if there exists, there is only one) a linear mapping $L$ such that $(\*)\displaystyle\lim\_{h\to 0}\frac{f(x+h)-f(x)-L(h)}{\|h\|}=0$. Then we say that $Df(x)=L$ is a differential of function $f$ at the point $x$. For example consider function $g:\mathbb{R}^2\rightarrow \mathbb{R}, \ g(x)=x\_1\cdot x\_2$. We can find differential of this function by looking at the increment of this function: $g(x+h)-g(x)=(x\_1+h\_1)(x\_2+h\_2)-x\_1x\_2=x\_1h\_2+x\_2h\_1+h\_1h\_2$ . Then the candidate for $Df(x)$ is linear part of this increment: $L(h)=x\_1h\_2+x\_2h\_1$. When we check $(\*)$ it appears that indeed it is a desired differential.

But in my example: $f(x+h)-f(x)=|x\_1+h\_1|(x\_2+h\_2)-|x\_1|x\_2$ I'm confused, it seems hard. Do I have to consider a few cases depending on a sign of $x\_1, \ x\_2$ ?

Can anybody make it clear for me? It is really important to me to finally understand this topic.