Are all linear maps differentiable? Consider a linear map from $\Bbb R^n \rightarrow \Bbb R^m$. 
Question 1. How can we show that all such linear maps are differentiable without loss of generality? 
Question 2. Is it possible that even though a function is not continuous at a point, it can be differentiable at that point ? Say for example, consider a linear map $f: \Bbb R^2 \rightarrow \Bbb R$  
$f(x,y) =
\begin{cases}
\frac{xy}{x^2 + y^2}  & (x,y) \neq (0,0) \\[2ex]
0 & (x,y) = (0,0)
\end{cases}$

is not continuous at $(0,0)$ but is differentiable at $(0,0)$.
 A: A linear map $\phi$ between finite dimensional spaces is always differentiable, and its derivative at a given point is given by $\phi$.
This is immediate from the definition, since we have an identity
$$\phi(a+h) = \phi(a) + \phi(h)$$
so there is not even a lower order term.
Differentiability in a point $p$ implies continuity in that point. 
A: Question 1. Let $\ell$ be a linear map, for all $x\in\mathbb{R}^n$ and $h\in\mathbb{R}^n$, one has: $$\ell(x+h)=\ell(x)+\ell(h).$$
Therefore, one gets: $$\|\ell(x+h)-\ell(x)-\ell(h)\|=o(\|h\|).$$
Finally, $\ell$ is differentiable and for all $x\in\mathbb{R}^n$, one has: $\mathrm{d}_x\ell=\ell$.
Question 2. Let $f$ be a differentiable function, for all $x\in\mathbb{R}^n$ and $h\in\mathbb{R}^n$, one has: $$\|f(x+h)-f(x)-\mathrm{d}_xf(h)\|=o(\|h\|).$$
Therefore, $f$ is continuous at $x$, since $\mathrm{d}_xf$ is a linear form.
A: I'm going to give a sort of different answer that might help counter any pathological insidious thoughts you have to the second question.
Suppose that $f(x) = x^2$ whenever $x\neq 0$ and $f(0)=5$, clearly this isn't continuous . You might be tempted to write the following and conclude that $f$ is differentiable at $x=0$:
$$\lim_{h\to 0} \frac{f(x+h) -f(x)}{h} = \lim_{h\to 0} \frac{(x+h)^2 - x^2}{h}=  \lim_{h\to 0} \frac{ x^2 +2xh +h^2 - x^2}{h} =  2x$$
However this is faulty logic, remember that the term on the far left is computing $f'(x)$; therefore, $f(x) = 5$.  So this is more like
$$\lim_{h\to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h\to 0} \frac{h^2 - 5}{h} = \lim_{h\to 0} h - \frac{5}{h} = -\infty$$
and essentially you'll always run into a problem of this sort when you try to check if the derivative exists.
A: *

*Linear maps are totally differentiable, they are their own total derivative.

*If a function is totally differentiable at a point, it is continuous at that point. The existence of all partial derivatives at a point isn't sufficient but if they are all bounded and f is defined on an open subset S of $\mathbb{R^n}$ then f is continuous on S.
