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:


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.


1 Answer 1



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.

  • $\begingroup$ This is the first time in my life I see $\sqcup$ used to denote an open set. $\endgroup$
    – Pedro
    Sep 6, 2014 at 23:52
  • $\begingroup$ @PedroTamaroff When writing by hand I prefer this because $U$ might interfer with the union operator. But it's true that on computer it doesn't really look alike. I'll change it. $\endgroup$
    – Amateur
    Sep 6, 2014 at 23:53
  • 1
    $\begingroup$ $\cup$ looks very different from $U$. At least in $\LaTeX$! In fact, $\sqcup$ denotes a disjoint union. $\endgroup$
    – Pedro
    Sep 6, 2014 at 23:54
  • $\begingroup$ @Amateur Thank you very much. This proof proves my first question? But is it correct that for my second question I can use your proof combined with the chain rule? For that I need both these things?, it does not follow from only one of them? $\endgroup$
    – user119615
    Sep 7, 2014 at 0:11
  • $\begingroup$ @user119615 The proof answers your first question, yes. As for your second question, I had in mind to define $f : \mathbb{R}^2 \to \mathbb{R}^2$ by $f(x,y) := (l(x,y), m(x,y))^t$. Then $k(l(x,y),m(x,y)) = k \circ f$. The proof I provided shows that $f$ is differentiable if $l$ and $m$ are. The result now follows from the Chain Rule. $\endgroup$
    – Amateur
    Sep 7, 2014 at 0:16

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