I have the following question. Let $\gamma:[a,b]\rightarrow X$ be a rectifiable curve in a metric space $(X,d)$. If we consider the length function of $\gamma$, $L:[a,b]\rightarrow [0,L(\gamma)]$, we see that it is clearly a continuous, monotone increasing function. Therefore it is differentiable. In other words $L':[a,b]\rightarrow \mathbb{R}$ exists and $L(b)-L(a) = \int L'(t)dt$.
Now take $\delta>0$. Then we have that $d(\gamma(t),\gamma(t+\delta)) \leq L(t+\delta)-L(t)$. By dividing by $\delta$, we get:
$\frac{d(\gamma(t),\gamma(t+\delta))}{\delta} \leq \frac{L(t+\delta)-L(t)}{\delta}$.
Since $L$ is differentiable, the limit $\lim_{\delta\rightarrow 0}\frac{L(t+\delta)-L(t)}{\delta}$ exists. If I could bound $\frac{d(\gamma(t),\gamma(t+\delta))}{\delta}$ on the other side by the same limit I could define $|\gamma'(t)|$ as the limit
$\lim_{\delta \rightarrow 0}\frac{d(\gamma(t),\gamma(t+\delta))}{\delta}$.
I have tried to do this in many ways but haven't been able to. I know it should be possible since this construction is mentioned (without any details) in a survey about metric spaces of bounded curvature below.
Can you help me?
EDIT: I don't know if this would be of any use to anyone but I found a complete proof of this in the book The Geometry of Geodesics by Busemann.
"Therefore it is differentiable"?
Continuous monotone increasing functions need not be differentiable at every point. Consider the function $f : [0,1] \to \mathbb{R}$ defined by $$f(t) = \begin{cases} t & \text{if $t \in [0,\frac{1}{2}]$} \\ 2t - \frac{1}{2} &\text{if $t \in [\frac{1}{2},1]$} \end{cases} $$