Let $(X_t)_{t≥0}$ be an Ito process of the form $dX_t = µ(t)dt + σ(t)dW_t$ for some $µ ∈ \mathbb{L}^1(0, T)$ and $σ ∈ \mathbb{L}^2(0, T)$. I have been asked to apply Ito’s formula to $Y_t = g(t, X_t)$ for $g(t, x) = e^x + t \,sin(x)$ to write $Y_t$ as an Ito process.
So far, I have calculated the partial derivatives: $$g_t = sin(x)$$ $$g_x = e^x + t\,cos(x)$$ $$g_{xx}=e^x - t\, sin(x)$$ And I have plugged these into Ito's formula to give: $$dY_t = sin(X_t)\, dt\, + \, (e^{X_t} + t \, cos(X_t)) \, dX_t \, + \, \frac{1}{2}\sigma^2(e^{X_t} \, - \, t\, sin(X_t))\, dt$$
However I am told this is incorrect, and it also needs further steps. Can anyone help guide me in the right direction and inform me where I have went wrong?
EDIT:
I am told that the final solution is as follows:
$$dg(t, X_t) = (sin(X_t) + (e^{X_t} + t\,cos(X_t))\mu(t) + \frac{1}{2}(e^{X_t} - t\,sin(X_t)\sigma^2(t)))\, dt \, + (e^{X_t} + t\,cos(X_t))\sigma(t)\,dW_t$$
The martingale part of $dX_t$ is $dM_t = \sigma (t)dW_t$, which implies $d\langle M\rangle_t = \sigma(t)^2dt$. So all in all $$dY_t = g_t(t,X_t)dt + g_x(t,X_t)dX_t + \frac{1}{2}g_{xx}(t,X_t)\sigma(t)^2dt,$$ where $dX_t$ is given. I can't see anything wrong in your end result.