Let $X(t)$ be a Gaussian process with zero mean and covariance function $B(t,s) = 1/(1+(t-s)^2)$. Let $X'(t)$ be its $L^2$-derivative. I am looking for the joint distribution of $X(t)$ and $X'(t)$.
Since $X'(t)$ is a limit of Gaussian processes, it is itself Gaussian and since $X$ has zero mean, so has $X'$. I'm left to find the covariance.
How would I calculate this covariance?
For every $s\gt0$ and $t$, write $\dfrac{X(t+s)-X(t)}s=Y(t,s)+b(s)X(t)$ with $b(s)=\dfrac{B(0,s)-1}s$ and $Y(t,s)=\dfrac{X(t+s)-B(0,s)X(t)}s$. Consider the limit $s\to0$. By inspection $b(s)\to0$ hence $b(s)X(t)\to0$ in $L^2$. By hypothesis, $\dfrac{X(t+s)-X(t)}s\to X'(t)$ in $L^2$. Thus, $Y(t,s)\to X'(t)$ in $L^2$. A covariance computation shows that $Y(t,s)$ is independent of $X(t)$. This shows that, if $X'(t)$ exists, then $X'(t)$ is independent of $X(t)$.