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.
$f(y)=(2\pi)^{-\dfrac{n}{2}}|\sigma^2I_n|^{-\dfrac{1}{2}}exp[-\dfrac{1}{2}(y-x(t))^T|\sigma^2I_n|^{-1}(y-x(t))]$
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 $\frac{\partial f(y)}{\partial c}= \frac{\partial f(y)}{\partial x(t)} *\frac{\partial x(t)}{\partial z(t)}*\frac{\partial z(t)}{\partial c}$ I have got ,
$\frac{\partial z(t)}{\partial c}= b(t)^T$
$\frac{\partial x(t)}{\partial z(t)} = x(t) - x(t) \frac{\int_a^p exp({z(s)})\frac{\partial z(s)}{ \partial z(t)}ds}{\int_a^bexp(z(s))ds}$
$\frac{\partial f(y)}{\partial x(t)} = N(x(t),\sigma^2I_n)|\sigma^2I_n|^{-1}(y - x(t)) $
Is using the chain rule is right path? If not, what else? If yes, is the partial derivatives are correct? It's looks too complicated, I thought the final result would be comparatively simple.