If I wanted to differentiate a stochastic integral, is this logic correct? $$X_t = \int_0^t B^2_s dB_s\\ dX_t = d \int_0^t B^2_s dB_s \\ dX_t = B^2_tdB_t - B^2_0dB_0 \\ dX_t = B^2_tdB_t \quad (B_0 = 0) $$ Then would this follow by the same logic? $$X_t = \int_r^t B^2_s dB_s\\ dX_t = d \int_r^t B^2_s dB_s \\ dX_t = B^2_tdB_t - B^2_rdB_r $$
I am just taking the derivative, not $\frac{d}{dt}$ so I believe Leibniz rule does not apply (unless there is a rule for this that I am unaware of).
$X_{t}$ is not in general differentiable in the sense of ordinary calculus. The notation $$ dX_{t}=B_{t}^{2}dB_{t} $$ is equivalent to $$ X_{t_{1}}-X_{t_{0}}=\int_{t_{0}}^{t_{1}}B_{t}^{2}dB_{t}\qquad\text{for }t_{0}<t_{1}. $$
On the other hand, writing $$ dX_{t}=B_{t}^{2}dB_{t}-B_{r}^{2}dB_{r} $$ does not really make sense.