EDIT: This question has been solved!
Consider a $d$-dimensional Brownian motion $W$ and the stochastic differential equation(s)
$$ \begin{align} dX&=b(X)dt+\rho(X)dW\\ X_0&=x, \end{align} $$
where $\rho(X)\in\mathbb{R}^{d\times d}$ and $b(X)\in\mathbb{R}^d$. Moreover we have
$$ \begin{align} a(X)=\rho(X)\rho(X)^\top&=a+\sum_{i=1}^d \alpha_i X_i\\ b(X)&=b+\sum_{i=1}^d \beta_i X_i, \end{align}$$
where $a,\alpha_i$ are $d\times d$-matrices and $b, \beta_i$ $d$-vectors.
Further there is a $\mathbb{C}$-valued function $\phi$ and a $\mathbb{C}^d$-valued function $\psi=(\psi_1,\ldots,\psi_d)^\top$, which solve the Riccati equations
$$ \begin{align} \partial_t\phi(t,u)&=\frac{1}{2}\psi(t,u)^\top a\ \psi(t,u)+ b^\top \psi(t,u) \\ \phi(0,u)&=0\\ \partial_t\psi_i(t,u)&=\frac{1}{2}\psi(t,u)^\top \alpha_i\ \psi(t,u)+ \beta_i^\top \psi(t,u)\\ \psi(0,u)&=u \end{align} $$
For $T>0$ and $u \in \mathrm{i} \mathbb{R}^{d}$ define the complex-valued Itô process $$ M_t=\mathrm{e}^{\phi(T-t, u)+\psi(T-t, u)^{\top} X_t} $$
Now, the author states, that we find $$ \frac{d M_t}{M_t}=I_t d t+\psi(T-t, u)^{\top} \rho(X_t) d W_t $$ with $$ \begin{aligned} I_t=&-\partial_{T} \phi(T-t, u)-\partial_{T} \psi(T-t, u)^{\top} X_t \\ &+\psi(T-t, u)^{\top} b(X_t)+\frac{1}{2} \psi(T-t, u)^{\top} a(X_t) \psi(T-t, u) \end{aligned} $$
by applying Ito's formula to the real and complex part.
My question: How did the author come up with exactly this $I_t$? Unfortunately I cannot find precisely this $I_t$.
My attempt: Let's try this for $d=1$:
$$f(z):=\mathrm{e}^{\phi(T-t, u)+\psi(T-t, u) z}$$
Applying Ito's formula, I find
$$ \begin{align} M_t=\mathrm{e}^{\phi(T-t, u)+\psi(T-t, u) X_t}=f(X_t)&=f(X_0)+\int_0^t f'(X_s) dX_s+\frac{1}{2}\int_0^t f''(X_s) \langle X\rangle_s\\ &=f(X_0)+\int_0^t \psi(T-s, u)M_sd(b(X)dt+\rho(X)dW)_s\\ &\quad + \frac{1}{2}\int_0^t \psi(T-s, u)^2M_sd\langle b(X)dt+\rho(X)dW\rangle_s\\ &=f(X_0)+\int_0^t \psi(T-s, u)b(X_s)M_sds\\ &\quad + \int_0^t \psi(T-s, u)\rho(X_s)M_sdW_s\\ &\quad + \frac{1}{2}\int_0^t \psi(T-s, u)^2\rho(X_s)^2M_sds\\ \end{align}$$
In this case
$$ \begin{align} \frac{d M_t}{M_t}&= \Big(\psi(T-t, u)b(X_t) + \frac{1}{2} \psi(T-t, u)^2\rho(X_t)^2\Big)d t\\ &\quad +\psi(T-t, u) \rho(X_t) d W_t\\ &= \Big(\psi(T-t, u)b(X_t) + \frac{1}{2} \psi(T-t, u)\rho(X_t)\rho(X_t)^\top \psi(T-t, u)\Big)d t\\ &\quad +\psi(T-t, u) \rho(X_t) d W_t\\ &= \Big(\psi(T-t, u)b(X_t) + \frac{1}{2} \psi(T-t, u) a(X_t) \psi(T-t, u)\Big)d t\\ &\quad +\psi(T-t, u) \rho(X_t) d W_t \end{align}$$
And therefore
$$I_t = \psi(T-t, u)b(X_t) + \frac{1}{2} \psi(T-t, u) a(X_t) \psi(T-t, u)$$
Final thoughts: It seems, that I am missing these partial derivatives with respect to $T$. Was my definition of $f$ wrong? Should I have defined
$$f(t,z):=\mathrm{e}^{\phi(T-t, u)+\psi(T-t, u) z}$$
and apply the multi dimensional Ito formula? Maybe these derivates would come from there? Also, I would be curious how to do the calculation in the actual dimension $d\ne 1$.
I'd be very thankful for any hint or help on these questions. Thank you so much in advance!