I was given a process $X_t = 1 + 2t + e^{W_t}$ and I need to find its differential form, i.e. write as $dXt = u_t dt + v_t dW_t$
I just differentiated the given expression to find $$dX_t = 2 dt + e^{W_t}d W_t $$
However, the answer I am supposed to find is $dXt = 2dt + e^{W_t}dW_t + \frac 12 e^{W_t}d_t$.
I have no idea where the term $\frac 12 e^{W_t}d_t$ comes from.
Same thing happens with the process $X_t = W^2_t + \int_0^t W^2_s ds$ where I find $dX_t = 2 W_tdW_t + W_t^2dt $ while the answer says $dX_t = 2 W_tdW_t +dt + W_t^2dt $. No other explanation is given so a bit frustrating.
Informally speaking, $\mathrm{d}W_t$ is $O(t^{1/2})$, so $(\mathrm{d}W_t)^2$ is $O(t)$ and should not be neglected when applying Taylor's formula to order $1$ to a stochastic process to express its derivative. This is the content of the Itô's lemma stating that: $$\mathrm{d}Y_t=\frac{\partial f}{\partial t}(X_t,t)\mathrm{d}t+\frac{\partial f}{\partial x}(X_t,t)\mathrm{d}X_t+\frac{1}{2}\frac{\partial^2f}{\partial x^2}(X_t,t)(\mathrm{d}X_t)^2,$$ where $Y_t=f(X_t,t)$.
In your exercice, $f(x,t)=e^x$ and $X_t=W_t$, thus $$\mathrm{d}f(X_t,t)=e^{W_t}\mathrm{d}W_t+\frac{1}{2}e^{W_t}(\mathrm{d}W_t)^2,$$ but $(\mathrm{d}W_t)^2=\mathrm{d}t$, whence $\mathrm{d}e^{W_t}=e^{W_t}\mathrm{d}W_t+\frac{1}{2}e^{W_t}\mathrm{d}t$.