Suppose we have two continuous time stochastic processes $X_t$ and $Y_t$ defined on a fixed probability space $\Omega$ that satisfy $$ \forall t \in \mathbb{R}: \mathbb{P}\{X_t = Y_t\} = 1 $$ and for every choice of $ \omega \in \Omega$ the mapping $$ \mathbb{R} \to \mathbb{R}: t \mapsto X_t(\omega) $$ is differentiable.
It is possible to show that because of this the mapping $$ \mathbb{R} \to \mathbb{R}: t \mapsto Y_t(\omega) $$ is also differentiable for each $\omega$ and that furthermore holds that $$\forall t \in \mathbb{R}: \mathbb{P}\left\{\frac{dX_t}{dt} = \frac{dY_t}{dt}\right\} = 1 \ ? $$
No. Consider the probability space $[0,1]$ with its Borel $\sigma$-algebra and the Lebesgue measure and define the processes $X_t(\omega) = 0$ and $Y_t(\omega) = 1_{\{t = \omega\}}$ for a counter-example.