condition for differentiability Lets say you have a function:
$$f: \mathbb{R}^2 \rightarrow \mathbb{R^2}=((u(x,y),v(x,y)).$$
Does it follow directly from this definition:
http://en.wikipedia.org/wiki/Differentiable_function#Differentiability_in_higher_dimensions
That $f$ is differentiable if and only if both $u$ and $v$ are differentiable?
Update: A similar question is if $k:\mathbb{R}^2\rightarrow \mathbb{R}$ is differentiable, and $l:\mathbb{R}^2\rightarrow \mathbb{R}$ and $m:\mathbb{R}^2\rightarrow \mathbb{R}$ is differentiable, is then $k(l(x,y),m(x,y))$ differentiable? 
I am using this for something in complex analysis.
 A: Yes.
More generally, we have the following result :
Theorem : Let $U \subseteq \mathbb{R}^n$ be open, $f : U \to \mathbb{R}^m$ and $a \in U$. Then,
$$
f = \begin{pmatrix} f_1 \\ \vdots \\ f_m\end{pmatrix} \text{ is differentiable at $a$} \iff \forall i, f_i \text{ is differentiable at $a$}.
$$
Remark : Hence, in a sense, we don't really gain generality when working in $\mathbb{R}^m$ instead of $\mathbb{R}$.
Proof of the Theorem:
$[\Longrightarrow]$ Since $f$ is differentiable at $a$, there exists $A \in M_{m \times n}$ such that $\frac{\|r(h)\|}{\|h\|} \to 0$ when $h \to 0$ where $r(h) := f(a+h)-f(a)-Ah$. Denote by $A_i$ the $i$-th row of $A$. Then,
$$
\frac{\|f_i(a+h)-f_i(a)-A_i h\|}{\|h\|} = \frac{\|e_i^t(f(a+h)-f(a)-A h)\|}{\|h\|} \leq \frac{\|e_i\| \| r(h)\|}{\|h\|} \to 0
$$
when $h \to 0$. Hence $f_i$ is differentiable at $a$ (and its derivative is $A_i$).
$[\Longleftarrow]$ Let $A_i \in M_{1 \times n}$ be the derivative of $f_i$ at $a$. Let $A$ be the $m \times n$ matrix such that the $i$-th row of $A$ is $A_i$. Then,
\begin{align}
\frac{\|f(a+h)-f(a)-Ah\|}{\|h\|} &= \frac{\| \sum_{i = 1}^m e_i e_i^t(f(a+h)-f(a)-Ah)\|}{\|h\|} \\
&\leq \sum_{i = 1}^m \frac{\|e_i\| \|e_i^t(f(a+h)-f(a)-Ah)\|}{\|h\|} \\
&= \sum_{i = 1}^m \frac{\|f_i(a+h)-f_i(a)-A_ih\|}{\|h\|} \to 0
\end{align}
when $h \to 0$. Hence, $f$ is differentiable at $a$ (and its derivative is $A$). $\square$
The answer to your updated question is also yes, by the Theorem above and the so called Chain Rule :
Theorem (Chain Rule) : Let $U \subseteq \mathbb{R}^n, V \subseteq \mathbb{R}^m$ be open, $f : U \to V, g : V \to \mathbb{R}^l$ be two functions, $a \in U$ and $b := f(a) \in V$. If $f$ is differentiable at $a$ and $g$ is differentiable at $b$, then $g \circ f$ is differentiable at $a$ and $(g \circ f)'(a) = g'(b) f'(a)$.
The proof of the Chain Rule in this context is essentially the same as the proof of the single variable one.
