Ito formula - some doubts

How once can calculate stochastic differential of a process:

$$Y_t=e^{t^2+\int_0^ts \, dW_s}$$

There are two approaches, which one is correct (or both?).

1) $Z_t=t^2+\int_0^ts \, dW_s$ is an Ito process, so just apply Ito formula with respect to $Z_t$ to the function $f(Z_t)=e^{Z_t}$

2) Apply Ito formula with respect to $W_t$ and $t$ to the function $f(t,W_t)=e^{t^2+\int_0^ts \, dW_s}$

Thanks for any explanation. $W_t$ - standard Brownian motion (wiener process).

• In 2), your formula does not define a function $(t,W_t)\mapsto f(t,W_t)$. – Did Feb 10 '15 at 18:47
• I know, but my girlfriend argues that she can apply ito formula with respect to $W_t$, not $Z_t$ in order to calculate $dY_t$ – luka5z Feb 10 '15 at 18:48
• Yeah, and my grandfather disagrees. Sorry but I fail to see how this is even relevant. Do you have anything mathematical to share on the subject? – Did Feb 10 '15 at 18:50
• Since 2) does not exist at present, indeed 1) seems to be the only remaining option. – Did Feb 10 '15 at 18:56
• Why don't you write down your girlfriend's proposed solution? Otherwise this question is largely rhetorical. – encore Aug 9 '17 at 17:44