Suppose, we have a function $X(t)=\dfrac{exp[b(t)^Tc+b(t)^TAu]}{\int_a^bexp[b(s)^Tc+b(s)^TAu]ds}$
s or t is variable belong to (a,b). Practically we take finite points in this range. c and u are p* 1 vector, b(t) is p* n matrix and A is p* p matrix. Is it possible to differentiate $X(t)$ it with respect to c and find the critical values of c? I know basic rules about matrix differentiation, but I am wondering how to proceed with the integration extract c from that? After begining I end up with
$\dfrac{\partial X(t)}{\partial c} =0$
$=>\int_a^bexp[b(s)^Tc+b(s)^TAu]ds* [b(t)^T]=\int_a^bexp[b(s)^Tc+b(s)^TAu] b(s)^Tds$
How to go further? Can we proceed this way or we need to do something else?
$\def\v{{\rm vec}}\def\d{{\rm diag}}\def\D{{\rm Diag}}\def\o{{\tt1}}\def\p{{\partial}}\def\grad#1#2{\frac{\p #1}{\p #2}}\def\E{{\cal E}}\def\c#1{\color{red}{#1}}$Define the all-ones vector $\o$ and the variables $$\eqalign{ w &= Au + c \quad&\implies\quad dw = dc \\ p &= \exp\left(B^Tw\right) \quad&\implies\quad P = \D(p),\,p=P\o \\ h &= \int_a^b p\,ds \quad&\implies\quad H = \D(h),\,h=H\o \\ J &= \int_a^b PB^T\,ds \\ }$$ and their differentials $$\eqalign{ dp &= PB^Tdc \\ dh &= J\,dc \\ }$$ Now define the vector of interest and calculate its gradient. $$\eqalign{ x &= H^{-1}p \\ dx &= H^{-1}dp + \c{dH^{-1}}p \\ &= H^{-1}dp - \c{H^{-1}dH\,H^{-1}}P\o \\ &= H^{-1}PB^Tdc - H^{-2}P\,dh \\ &= \left(H^{-1}PB^T - H^{-2}PJ\right) dc \\ &= H^{-2}P\left(HB^T - J\right) dc \\ \grad{x}{c} &= H^{-2}P\left(HB^T - J\right) \\ }$$ NB: Since $\{H,dH,P,dP\}$ are diagonal matrices, they commute with each other.
Setting the gradient to zero yields $$\eqalign{ HB^T &= J \\ \left[\int_a^bP\,ds\right]\cdot B^T &= \int_a^bPB^Tds \\ B\cdot\left[\int_a^bp\,ds\right] &= \int_a^bBp\,ds \\ }$$ This confirms your result, and dividing through by $\left(\frac{1}{b-a}\right)$ it looks like some sort of mean-value theorem for the $p$ vector. $$\eqalign{ B\,\big\langle p\big\rangle &= \big\langle Bp\big\rangle \\ }$$
If we insert the explicit functional dependence on the time-like variables $$\eqalign{ B(t)\cdot\left[\int_a^bp(s)\,ds\right] &= \int_a^bB(s)p(s)\,ds \\ }$$ note that the LHS depends on $t$ while the RHS does not. This implies that $B$ is not a function of time, or that the RHS is zero and the integral on the LHS is orthogonal to $B(t)$ at all times $-$ which implies that the integral itself is zero $-$ which implies that both integrals are zero.