Suppose I have two continuous semimartingales $X$ and $Y$. By Ito's formula for semimartingales, we have
$$ X_tY_t=X_0Y_0+\int_0^t X_sdY_s+\int_0^tY_sdX_s+\langle X,Y\rangle_t $$ where $\langle X,Y\rangle_t$ is the predictable cross-variation between $X$ and $Y$.
The definition of $\langle X,Y\rangle_t$ that I am working with is the adapted process such that $X_tY_t-\langle X,Y\rangle_t$ is a martingale.
Suppose now $f,g\in C^2(\mathbb R)$ and $X_t:=f(W_t)$ and $Y_t:=g(W_t)$.
I want to know the explicit form of $\langle X,Y\rangle_t$.
From lecture notes, the professor wrote that, by Ito lemma,
$d(X_tY_t) =X_tdY_t+Y_tdX_t+d\langle X,Y\rangle_t= f(g'dW_t+\frac{1}{2}g''dt)+g(f'dW_t+\frac{1}{2}f''dt)+f'g'dt$.
My question is, why does the second equality hold?
I know, by Ito lemma, $X_tdY_t=f(g'dW_t+\frac{1}{2}g''dt)$, and $Y_tdX_t=g(f'dW_t+\frac{1}{2}f''dt)$. If the second equality above is correct, then it somehow implies that $d\langle X,Y\rangle_t=f'g'dt$. How do we know this is true? We are still on the way deriving the expression of $\langle X,Y\rangle_t$. Quite confusing. Thanks for help.
By Ito's lemma \begin{align} dX_t&=f'(W_t)\,dW_t+\tfrac12 f''(W_t)\,dt\,,&dY_t&=g'(W_t)\,dW_t+\tfrac12 g''(W_t)\,dt\,. \end{align} To $\langle X,Y\rangle_t$ only the two $dW_t$-terms contribute. Since $d\langle W,W\rangle_t=dt$ we get $$ d\langle X,Y\rangle_t=f'(W_t)g'(W_t)\,dt\,. $$