# Compute dZ(t) : Ito's formula/lemma We need to find dZ(t). I know I have to use Ito's formula. But I am confused because in the Ito's formula we have f(y,t) is a twice differentiable function with two variables

But here Z(t) = 1/(2+x(t)), which just has one variable?

So, I am not sure how to proceed. Any tips will be appreciated!

It looks like $$Z(t)$$ is dependent on $$t$$ through $$X_t$$ and not directly on $$t$$, loosely speaking. Assume that $$Z_t = f(X_t)$$ and use Ito's formula with just one variable.
$$dZ_t = \frac{df}{dX} dX_t + \frac{1}{2} \frac{d^2f}{dX^2} d[X_t,X_t]$$
• $[X_t, X_t]$ is the quadratic variation of the Ito process $X_t$. By definition it is $[X_t, X_t] = \int_0^t \beta(s, X_s)^2 ds$, so $\frac{9t^2X_t^2}{2}dt$ is correct. The $\frac{1}{2}$ you get, because you use Taylor expansion to derive the Ito formula. – SmurfAcco Mar 28 '20 at 11:06
• The 1/2 comes before second derivative of function with respect to state variable $X_t$ due to Taylor expansion, as @SmurfAcco rightly said. The quadratic variation of $X_t$, $d[X_t,X_t]$ is given by $9t^2X_t^2$. It looks like you have missed $X_t^2$ term. – ForumWhiner Mar 28 '20 at 18:21