Using Ito calculus to show $d(\frac{1}{3} W_t^3) = W_t^2 dW_t + W_t dt$

Question

I am trying to self study some stochastic calculus and have come upon a result I do not understand. Can anyone show me how
\begin{align}
d(\frac{1}{3} W_t^3) = W_t^2 dW_t + W_t dt
\end{align}

where $W_t \in \mathbb{R}$ is Brownian motion. My notes (from my supervisor) say that the result is found using Ito's rule. I am not sure how this is applied in this case though?

Answer 1

Apply the Ito’s rule $$df(W)= f’(W_t)d W_t + \frac12f’’(W_t)dt$$ to $f(W_t)=\frac13 W_t^3$. Then, $f’(W_t)= W_t^2$, $f’’(W_t)=2 W_t $ and $$d(\frac{1}{3} W_t^3) = W_t^2 dW_t + W_t dt$$