Let $U \subset \mathbb{R}^{m}$ be an open set and $f: U \longrightarrow \mathbb{R}^{n}$. Then, $f$ is differentiable in $a \in U$ if only if for all $h \in \mathbb{R}^{m}$ with $a + h \in U$, theres exists a linear transformation $A(h): \mathbb{R}^{m} \longrightarrow \mathbb{R}^{n}$ such that $f(a+h) - f(a) = A(h)h$ and $h \longmapsto A(h)$ be continuous in $h=0$.

For "$\Longrightarrow$", I know that $$f(a+h)-f(a) = \mathrm{D}f(a)h + \frac{r(h)}{|h|}$$ where $\displaystyle \lim_{h\to 0}\frac{r(h)}{|h|} = 0$. But I cannot define the transformation $A(h)$. Someone can helps me? I like hints, no solutions.

  • $\begingroup$ I would guess the linear transformation $A(h)$ you are looking for would be the derivative of $f$ at $a+h$. $\endgroup$ – D_S Apr 3 '18 at 0:04
  • $\begingroup$ But and $\displaystyle \frac{r(h)}{|h|}$? $\endgroup$ – Lucas Corrêa Apr 3 '18 at 0:09

Take $A(h)=Df(a)$ for every $h$. For the converse part show that partial derivatives exist and are continuous at $a$.

  • $\begingroup$ For the first part, taking $A(h) = \mathrm{D}f(a)$, and $\displaystyle \frac{r(h)}{|h|}$? $\endgroup$ – Lucas Corrêa Apr 3 '18 at 12:59
  • $\begingroup$ $r(h)$ is just what you get from the equation for $f(a+h)-f(a)$. $\endgroup$ – Kavi Rama Murthy Apr 4 '18 at 10:52

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