I am given $X = tW$ and $Y = e^{W}$ and am asked to compute $d(\frac{X}{Y})$. To me, this seems to be an application of Ito's product rule which I know to be:
$$d(X_1 X_2) = X_2 dX_1 + X_1dX_2 + G_1 G_2 dt$$ where $dX_1 = F_1dt + G_1dW$ and $dX_2 = F_2dt + G_2dW$
So following this, I write $$d(tW \cdot e^{-W}) = tW \cdot d(e^{-W}) + e^{-W}d(tW) +G_1G_2 dt$$
However, I am confused about computing $d(e^{-W})$ and $d(tW)$. The examples in my book give my a function $u(X,t)$, assume $X(\cdot) = W(\cdot) \implies dX = dW \implies F\equiv 0, G\equiv 1$, and apply Ito's chain rule:$$du(X,t) = (u_t + u_x F + \frac{1}{2}u_{xx}G^2)dt + u_x G dW$$ from $dX = Fdt + GdW$
Is $d(e^{-W}) = \frac{de^{-W}}{dt} + \frac{de^{-W}}{dW} = 0dt + -e^{-W}dW$? Or do we use the above chain rule somehow?
Two facts we can use are, first we have the product rule given by $$d(f(t)G_t) = f(t)dG_t + G_tdf(t) + df(t)dG_t.$$ In this case of the rule $f(t)$ is a deterministic function. Second from Ito's formula we know if $F_t = f(W_t)$ we have $$dF_t = \frac{1}{2}f''(W_t)dt + f'(W_t)dW_t.$$
Now given $X_t = tW_t$ and $Y_t = exp(W_t)$. We want to find $dX_t$ and $dY_t$.
Using the product rule for $X_t$ we have $$dX_t = W_tdt + tdW_t + dtdW_t = W_tdt + tdW_t.$$
Then using the fact from Ito's formula we have $$dY_t = exp(W_t)dW_t + \frac{1}{2}exp(W_t)dt.$$