$w(u)$ satisfies $$ w(u)=\frac{1}{u} \ \ \textrm{ for }1\leq u \leq 2, $$ $$ (uw(u))'=w(u-1) \ \ \textrm{ for }u>2. $$
I would like find numerical solutions for $w(u)'$. The derivative doesn't exist at 2 but does at all over values of $u\geq1$ I believe.
For $1 \leq u < 2$ we know $w(u)' = \frac{-1}{u^2}$.
How can I compute $w(u)'$ accurately for all other values?
Let us change a bit the notation and consider the following delay-differential equation
$$\dot{w}(t)=\dfrac{1}{t}(-w(t)+w(t-1)),t>2$$
with the initial condition $w(s)=1/s$, $s\in[1,2]$. This equivalent to the problem stated in the question where I have applied the chain rule and reorganized the terms to obtain an explicit delay-differential equation.
Now, the OP is right in saying that there is a discontinuity at for the derivative at $t=2$ since $$\lim_{t\to 2^-}\dot{w}(t)=-1/4$$ while $$\lim_{t\to 2^+}\dot{w}(t)=1/4.$$
In fact, this is a well-known phenomenon in delay-differential equations.
To solve for $\dot{w}(t)$, we need first to solve for $w(t)$ and this can be done by the propagation method, where we solve successively the delay-differential equation over the intervals, $[k, k+1]$ for $k\in\mathbb{Z}_{>0}$. The procedure is as follows:
It is also possible to obtain some analytical results. Over the interval $[k, k+1]$, the solution is expressed as
$$w(t)=\Psi(t,k)w(k)+\int_k^t\dfrac{1}{s}\Psi(t,s)w(s-1)ds,t\in[k,k+1]$$
where $$\Psi(t,s)=\exp\left(\int_s^{t}-\dfrac{1}{t}dt\right)=\dfrac{s}{t}.$$
This yields
$$w(t)=\dfrac{k}{t}w(k)+\dfrac{1}{t}\int_k^tw(s-1)ds,t\in[k,k+1].$$
Now, it is possible to numerically and successively integrate as
$$w(t)=\dfrac{2}{t}w(2)+\dfrac{1}{t}\log(t-1),\ t\in[2,3],$$
and this goes for all the next intervals. The issue here is that the integral may become more difficult to evaluate over time, but it can still be numerically evaluated.