Differential $dX_t$ of the process $X_t=e^{tW_t}$ In a quant interview, he asked me to compute the differential $dX_t$ of the Ito process $X_t=e^{tW_t}$. Is it correct to do the following?

$f(t,W_t)=e^{tW_t}$

We know that: 
$$d(f(t,W_t))=f_t(t,W_t)dt+f_w(t,W_t)dW_t+\dfrac{1}{2}f_{ww}(t,W_t)(dW_t)^2$$
we don't take $f_{tw}$ and $f_{tt}$, because: 
$$(dW_t)^2=dt$$
$$dW_t\,dt=0$$
$$dt^2=0$$

So we have: 

$$\begin{align}
d(e^{tW_t}) &amp;= W_te^{tW_t}dt+te^{tW_t}dW_t+\dfrac{1}{2}t^2e^{tW_t}dt\\
&amp;= \left(W_t+\dfrac{t^2}{2}\right)e^{tW_t}dt+te^{tW_t}dW_t
\end{align}$$
Naively $dX_t = X_{t+dt}-X_t$. Then we apply it to your formula

$$dX_t = X_{t+dt}-X_t = e ^{W_{t+dt}} -  e ^{W_{t}}$$

This is already looking good if I have an expression for the change of $W_t$.

$$W_{t+dt} = W_{t}+ dW_t$$

This gets places in your exponential and you expand:

$$ e^{x+\epsilon} =e^x \Big( 1+ \epsilon + \epsilon^2/2 + \dots\Big) $$

With $x= W_t$ and $ \epsilon= dW_t$.. This becomes something like Itô formula.
Naively $dX_t = X_{t+dt}-X_t$. Then we apply it to your formula

$$dX_t = X_{t+dt}-X_t = e ^{W_{t+dt}} -  e ^{W_{t}}$$

This is already looking good if I have an expression for the change of $W_t$.

$$W_{t+dt} = W_{t}+ dW_t$$

This gets places in your exponential and you expand:

$$ e^{x+\epsilon} =e^x \Big( 1+ \epsilon + \epsilon^2/2 + \dots\Big) $$

With $x= W_t$ and $ \epsilon= dW_t$.. This becomes something like Itô formula.