Is it true that a multivariate function is differentiable if it's components are? For example, if I had $f : \mathbb{R}^n \to \mathbb{R} = x_1^2 + x_2^2 + ... + x_n^2$, would it follow that $f$ is differentiable from the fact that $g: \mathbb{R} \to \mathbb{R} = x^2$ is differentiable? Why?
 A: In your example you need three ingredients:


*

*The projection functions $p_i : \mathbb R^n \to \mathbb R, p_i(x_1,\ldots,x_n) = x_i$, are differentiable for $i = 1,\ldots, n$. This is very easy to verify.

*As you realized, $g : \mathbb R \to \mathbb R, g(x) = x^2$, is differentiable. The chain rule shows then that $f_i : \mathbb R^n \to \mathbb R, f_i(x_1,\ldots,x_n) = x_i^2$, is differentiable.

*The sum of differentiable functions is differentiable (see  Theoneandonly's comment). Now you see that $f = f_1^2 + \ldots + f_n^2$ is differentiable.
A: Note in the special case $f(x_1, x_2,...,x_n) = \sum_{i} f_i(x_i)$ for some differentiable $f_i:\mathbb{R} \to \mathbb{R}$, it is indeed true that the differentiability of $f$ follows from the differentiability of each $f_i$. Hence in your particular case, you're fine.
However, in the general case, you cannot say anything about the differentiability of a function $f:\mathbb{R}^n \to \mathbb{R}$ just because its "component functions" seem differentiable. It's possible to have functions such that, fixing any $n-1$ variables except say the $k$-th variable, the univariate function $g(x)=f(x_1, x_2, \cdots , \overbrace{x}^{k\text{th  index}}, \cdots, x_n)$ is always differentiable as a function $g:\mathbb{R} \to \mathbb{R}$, but $f$ still fails to be differentiable as a function $f:\mathbb{R}^n \to \mathbb{R}$. In other words, the existence of the partial derivatives does not imply differentiability. 
A: 
$f:U \subset \mathbb{R}^m \rightarrow \mathbb{R}^n$ then $\lim_{\textbf{x} \rightarrow \textbf{a}} f(\textbf{x}) = \textbf{k} \,\, \iff  \lim_{x_i \rightarrow a_i}  = k_i  \,\,\forall 1 \leq i \leq N$

From which that following result 

$f:U \subset \mathbb{R}^m \rightarrow \mathbb{R}^n$ is differentiable $\iff$ each component $f_i:U \rightarrow \mathbb{R}$ is differentiable 

we can write your function as composition of functions 
$h(x_1,x_2, \cdots ,x_n) = (x_1^2,x_2^2, \cdots ,x_n^2)$ and $g(x_1,x_2, \cdots ,x_n) = \sum_{i=1}^n x_i$ 
$$f = g \circ h$$

if $f$ and $g$ are differentiable and composition is defined then we have that $f \circ g $ is differentiable  

hence we have that $f$ is differentiable
