Consider $x(t)$ and $y(t)$, both $C^1$. Then, their corresponding weak derivatives, with respect to $t$, are $$ \dot x(\phi) = \int_{t_0}^t \dot x(\tau) \phi(\tau) \,\mathrm d \tau \quad \text{ and } \quad \dot y(\phi) = \int_{t_0}^t \dot y(\tau) \phi(\tau) \,\mathrm d \tau, $$ with $\phi \in C_0^\infty (t_0, t)$. Suppose also, that $$ \lim_{a \rightarrow 0}\dot x(\phi) = \dot y(\phi), $$ where $a$ is parameter of $x(t)$. Would it not be correct to assume that this implies that $$ \lim_{a \rightarrow 0}\dot x(t) = \dot y(t), $$ for every $t \geq t_0$? With $\dot x(t),\ \dot y(t)$ of course being the classical time derivatives.
EDIT: Forgot to say that also $\lim_{a \rightarrow 0} x(t) = y(t),\ \forall t \geq t_0$.
Set $x(t)=a\sin(t/a)$ and $y(t)=0.$ Then $x$ and all its derivatives converge weakly to $y$ as $a\to 0,$ and $x$ converges to $y$ pointwise, but $\dot x(t)=\cos(t/a)$ does not converge to $y$ pointwise.