Is $d \langle X,Y \rangle = \langle dX,dY \rangle$.
I think the answer is yes because $ d \langle X,Y \rangle=\langle X,Y \rangle_t- \langle X,Y \rangle_s$ and $\langle dX,dY \rangle=\langle X_t-X_s,Y_t-Y_s \rangle= \langle X,Y \rangle_t- \langle X,Y \rangle_s$ after some basic computations. Am I correct?. Edit(IN response to Saz's comment ) In my stochastic analysis prof's terminology
$dX =\big( X_t-X_s\big)_{0\leq s \leq t}$ i.e it is the increment process of X Similarly we have
$dY =\big( Y_t-Y_s\big)_{0\leq s \leq t}$ i.e it is the increment process of Y and similarly we have for the quadratic covariation process
$ d \langle X,Y \rangle=\big(\langle X,Y \rangle_t- \langle X,Y \rangle_s \big)_{0 \leq s \leq t}$ i.e it is the increment process of $\langle X,Y \rangle$
So we have that $\langle dX,dY \rangle=\langle ( Y_t-Y_s)_{0\leq s \leq t} ( X_t-X_s)_{0\leq s \leq t} \rangle$ and then from the bilinearity of the quadratic variation I get that they are the same. I think I am a bit confused now .Can you help?