Ito's lemma: $$\text{if } dX_t = \sigma_tdW_t+\mu_t dt, \text{ where $W_t$ is a Wiener process, then } df(X_t)=\sigma_tf'(X_t)dW_t + (\mu_tf'(X_t)+ \frac{1}{2}\sigma_t^2f''(X_t))dt$$
now I want to apply this to $$Y_t=e^{\sigma W_t+\mu t}$$
Where $W$ is a Wiener process.
There are two ways I can think of to apply Ito's lemma to $Y_t$ but they give different results:
(1). We can of course set $X_t = \sigma W_t + \mu t$, and set $f(x)=e^x$. Then Ito's Lemma gives $df= f(X_t)(\sigma dW_t+(\mu + \frac{1}{2} \sigma ^2 )dt)$
(2). However, I am wondering why I get a different result if instead, I set $X_t=W_t$, and $f(x)=e^{\sigma x + \mu t}$.
Then Ito's lemma gives: $df=f(X_t)(\sigma dW_t + \frac{1}{2}\sigma^2dt)$.
Which of these applications is correct, and more importantly, why is the other incorrect?
The first one is correct.
In the second one, $f$ depends explicitly on $t$. This is not a problem per se, but it means that the correct form of Ito's lemma should include an additional $\frac{\partial f}{\partial t} dt$ term. This is an advantage of the otherwise misleading notation $\frac{dX_t}{dt}=\mu_t + \sigma_t \frac{dW_t}{dt}$: it reminds us what we are differentiating with respect to, so that we know not to forget about any explicit dependence on that variable.
To see the point, consider a more extreme example: $f(t,x)=t$. If I then define $Y_t=f(t,W_t)$, $dY_t$ can't possibly be $0$. Yet this is what your form of Ito's lemma without the $\frac{\partial f}{\partial t}$ term would imply.