if derivative vanishes in a path connected set, then $f$ is constant on that set? Let $f: \mathbb{R}^d \to \mathbb{R}^m $ be a function sucht that $Df(a) = 0 $ for all $a \in U \subseteq \mathbb{R}^d$. Also, suppose $U$ is open and path-connected.
Question: IS $f$ constant on $U$ ??
 A: Here is little more precise answer: Let $a, b \in U$ be arbitrary. We are done if we can show $f(a) = f(b)$. Since $U$ is path-connected, we can find a $C^1$-path (obtained from a continuous path by a suitable regularization procedure) $\gamma :[0, 1] \to U$ from $a$ to $b$. Then
\begin{align*}
f(b) -f(a) &= f(\gamma(1))- f(\gamma (0)) = \int_0^1 \frac{\operatorname{d}}{\operatorname{d}t} f(\gamma(t)) \operatorname{d}t = \int_0^1 Df_{\gamma(t)} (\dot{\gamma}(t)) \operatorname{d}t = 0,
\end{align*}
where we have used the chain rule and $Df =0$. This argument also works if the path is only piecewise $C^1$ by applying it to the segments.
A: Although for the function of several variables there is no Mean Value Theorem, there is another similar result.

Suppose $f$ maps a convex open set $E \subset \mathbb{R}^d$ into
  $\mathbb{R}^m$, $f$ is differentiable in $E$, and there is a real
  number $M$ s.t. $\left\| {f'\left( x \right)} \right\| \leqslant M$
  for every $x \in E$. Then $\left| {f\left( a \right) - f\left( b\right)} \right|\leqslant M\left| {a - b} \right|$ for all $a,b \in E$.

In this case, we can put $M=0$ since $Df(a) = 0$.
A: Yes: pick any two points and join them with a piecewise-linear path. On each segment, $f$ is constant, and hence the values at the initial and final points coincide.
A: This answer is similar to that of Dragoslav, except it uses the mean value theorem.
We want to show that every coordinate function
$f_i : U \to \mathbb{R}$ is constant. Since $U$ is path-connected, it is enough to show that
$$
f_i(\gamma(0)) = f_i(\gamma(1))
$$
holds for every path $\gamma : [0,1] \to U$. The map $f_i \circ \gamma$ is a function of a single real variable, so we can apply the ordinary mean value theorem to obtain
$$
(f_i \circ \gamma)(1) - (f_i \circ \gamma)(0) = (f_i \circ \gamma)'(\xi)(1 - 0)
$$
for some $\xi \in [0,1]$. Furthermore, the chain rule for total derivatives yields
$$
(f_i \circ \gamma)'(\xi) = \underbrace{Df_i(\gamma(\xi))}_{= \, 0} \gamma'(\xi) = 0
$$
(note that the $f_i$ are totally differentiable iff $f$ is totally differentiable).
