I would like to find the partial derivative of $f(y)$ with respect to c where $y$ follows multivariate normal/Gaussian density $N(x(t),\sigma^2I_n)$ i.e.
$f(y)=(2\pi)^{-n/2}|\sigma^2I_n|^{-1/2}exp[-1/2(y-x(t))^T|\sigma^2I_n|^{-1}(y-x(t))]$
Here, $x(t)=\dfrac{\exp(z(t))}{\int_a^p \exp(z(s))\,\text{d}s}$ and $z(t)=b(t)^Tc + b(t)^TAu$. I could do it if $x(t)$ is a function of $c$, but I am kind of lost at the very beginning since $x(t)$ is a composite function of $c$ i.e $x(t)=g(h(c))$.
N.B. Here the dimensions: $A$ is $p*p$, $b(t)$ is $p*n$, $c$ & $u$ is $p*1$, so $z(t)$ is $n*1$. And $\sigma^2I_n$ is the $n*n$ variance-covariance matrix.
Using the chain rule $\partial f(y)/\partial c= \partial f(y)/\partial x(t) *\partial x(t)/\partial z(t)*\partial z(t)/\partial c$ I have got ,
$\partial z(t)/\partial c= b(t)^T$
$\partial x(t)/\partial z(t) = x(t)b(t)^T - x(t) [\int_a^p exp[{z(s)}]b(s)^tds]/[\int_a^bexp[z(s)]ds$
$\partial f(y)/\partial c = N(x(t),\sigma^2I_n)|\sigma^2I_n|^{-1}(y - x(t)) $
Is it correct? It's looks too complicated, I thought it would be comparatively simple.