Is it true the derivative of the length of a Lipschitz path equals its speed? (when they exist) $\newcommand{\ga}{\gamma}$
$\newcommand{\e}{\epsilon}$
I have the following situation:
$(X,d)$ is a metric space. $\gamma:[0,1] \to X$ is a Lipschitz path. The speed of $\gamma$ is defined as $$ \nu_{\ga}(s):=\lim_{\e \to 0} \frac{d\left( \ga(s),\ga(s+\e) \right)}{|\e|},
$$ if the limit exists.
Define $g(t)=L(\ga|_{[0,t]})$, and assume $g'(t_0),\nu_{\ga}(t_0)$ exist for a specific $t_0 \in [0,1]$.
Is it true that $g'(t_0)=\nu_{\ga}(t_0)$?
Note:
It is a known theorem, that for any Lipschitz path $\ga$, the speed $\nu_{\ga}(t)$ exists for almost all $t$, and $L(\ga)=\int \nu_{\ga}(t) dt$ (See "A course in metric geometry" by Burago,Burago and Ivanov, theorem 2.7.6). 
In fact, the proof actually shows that for almost all $t$, $g'(t),\nu_{\ga}(t)$ exist and are equal. I am asking about their equality at a single point (assuming existence).
Partial Results:
$$g'(t)=\frac{d}{dt}L(\ga|_{[0,t]})=\lim_{\Delta t \to 0+} \frac{L(\ga|_{[0,t+\Delta t]})-L(\ga|_{[0,t]})}{\Delta t}=\lim_{\Delta t \to 0+} \frac{L(\ga|_{[t,t+\Delta t]})}{\Delta t} $$
$$ \ge  \lim_{\Delta t \to 0+} \frac{d\left( \ga(t),\ga(t+\Delta t) \right)}{\Delta t} = \nu_{\ga}(t)$$
So, we established $g'(t) \ge \nu_{\ga}(t) $.
The question about the other direction still remains.
In the special case, where $X$ is a length space, and $\gamma$ is a geodesic (i.e locally a shortest path), we have that $d\left( \ga(t),\ga(t+\Delta t) \right)=L(\ga|_{[t,t+\Delta t]})$ (for small enough $\Delta t$), so the equality obviously holds.
Perhaps it will be easier to prove this for length spaces, but without the assumption that $\ga$ is a geodesic, I do not see how it helps.
 A: We can have the strict inequality $\nu_{\gamma}(t_0) < g'(t_0)$.
Let $(X,d)$ be the Euclidean plane, and let $\gamma \colon [0,1] \to \mathbb{R}^2$ be the path with
$$\gamma(n^{-1}) = \bigl(n^{-1}, (-1)^n\cdot n^{-2}\bigr)$$
for $n \in \mathbb{N}\setminus \{0\}$, linearly interpolating between $\gamma(n^{-1})$ and $\gamma\bigl((n+1)^{-1}\bigr)$, and by continuity $\gamma(0) = (0,0)$. One easily verifies that $\gamma$ is a Lipschitz path with Lipschitz constant $\frac{1}{2}\sqrt{29}$.
Further, for $0 < t \leqslant 1$, we have $t \leqslant d(\gamma(t),\gamma(0)) \leqslant t\sqrt{1 + t^2}$, so $\nu_{\gamma}(0)$ exists, and $\nu_{\gamma}(0) = 1$.
For $n \in \mathbb{N}\setminus \{0\}$, we have
$$\frac{\sqrt{5}}{(n+1)^2} < L(\gamma\lvert_{[(n+1)^{-1}, n^{-1}]}) = \sqrt{n^{-2}(n+1)^{-2} + \bigl(n^{-2} + (n+1)^{-2}\bigr)^2} < \frac{\sqrt{5}}{n^2},$$
and hence for $t \in \bigl((k+1)^{-1}, k^{-1}\bigr]$
$$\frac{\sqrt{5}}{k+2} < \sum_{n = k+1}^\infty \frac{\sqrt{5}}{(n+1)^2} < L(\gamma\lvert_{[0,t]}) < \sum_{n = k}^\infty \frac{\sqrt{5}}{n^2} < \frac{\sqrt{5}}{k-1},$$
whence
$$\sqrt{5}\frac{k}{k+2} < \frac{g(t)}{t} < \sqrt{5}\frac{k+1}{k-1},$$
which shows $g'(0) = \sqrt{5} > \nu_{\gamma}(0)$.
