I'll start by listing a definition:

"The function $f$ is differentiable in a $\in$ $D^0$ if there is a number $A_a \in R$ and a function $\omega_a$ continuous and null in $a$ so that, for any $x \in D$ we have:


where $\lim_{a\to\infty}\omega_a(x)=\omega_a(a)=0$ Immediately we see that $A_a=f'(a)$"

Now, I've found an example but I don't see very well how it's related to the formula above. The problem asks to prove that for a=1 $f$ is not differentiable in $(0,0)$ and for $a>1 \ f$ is differentiable in the origin.

$$f(x,y)=\frac{x^2y}{\sqrt{x^2+y^2}}, \ x^2+y^2\not=0$$ $$f(x,y)=0, \ x^2+y^2=0$$

$$f(x,y)=\frac{\partial f}{\partial x}(0,0)(x-0)+\frac{\partial f}{\partial y}(0,0)(y-0)+\omega(x,y)\sqrt{x^2+y^2}$$

I guess the last row applies the formula listed in the definition (right?). But in the last part in the definition formula says $\omega_a(x)(x-a)$, shouldn't then be? $$f(x,y)=\frac{\partial f}{\partial x}(0,0)(x-0)+\frac{\partial f}{\partial y}(0,0)(y-0)+\omega(x,y)(x-0)$$


Exactly what the equations involved look like depends on what dimension the domain and range of $f$ are, but you seem to have picked up on that.

The idea behind the definition is that, if the function $\omega_a$ exists and satisfies the conditions, then we can divide by $(x-a)$ to get that $$\frac{f(x)-f(a)}{x-a} = A_a + \omega_a(x)$$ When we take the limit as $x \to a$, the left side is the quotient definition of the derivative and the right side approaches $A_a$, since $\lim_{x\to a} \omega_a(x) = 0$.

If we can write $f$ in that form, then $f$ is locally linear (which is the same thing as differentiability), since we can write the difference between $f(x)$ and $f(a)$ as the sum of a linear term (the $A_a(x-a)$) and a term that tends to zero even when you divide it by $(x-a)$ (also referred to as tending to zero faster than $(x-a)$), namely $\omega_a(x)(x-a)$.

In the second problem, things are slightly different in that you have a function $f: \mathbb{R}^2 \to \mathbb{R} $, whereas the first function was just defined on $\mathbb{R}$. This makes the relevant equation slightly more complicated, but not really that much different. The analogous equation is that $f$ is differentiable at $(a,b) \in D^0$ if $$ f(x,y)-f(a,b) = A_{(a,b)}(x-a) + B_{(a,b)}(y-b) + \omega_{(a,b)}(x,y)\|(x-a,y-b)\|$$ for some constants $A_{(a,b)}, B_{(a,b)} \in \mathbb{R}$ and a continuous function $\omega_{(a,b)}(x,y)$ that is zero at $(a,b)$. Since we are working in two dimensions, not just one, the linear part now has one term for $x$ and one for $y$, and $\omega_{(a,b)}$ is a function of two variables, not just one. You wrote down the correct formula for the case $(a,b) = (0,0)$ as

$$f(x,y) = \frac{\partial f}{\partial x}(0,0)(x-0) + \frac{\partial f}{\partial y}(0,0)(y-0) + \omega(x,y)\sqrt{x^2 + y^2}$$

since the constants $A$ and $B$ have to be equal to the respective partial derivatives, if they exist near $(0,0)$, which they do. Showing that $f$ is or isn't differentiable at the origin reduces to showing that a continuous $\omega$ that vanishes at $(0,0)$ and satisfies the equation either exists or doesn't exist. The reason that the equation doesn't look exactly like it does in the definition is that you're working in more dimensions.

I'm not quite sure what you're saying about showing that $f$ is differentiable at the origin for $a>1$, since keeping the notation before would suggest that $a$ is the point $(0,0)$ itself.


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