How to calculate $(dX_t)^2$ in Itô's Formula Let $T > 0$ and $g:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function where we denote $g'(x) = \frac{d}{dx}g(x)$. Let $(X_t)_{t\in[0,T]}$ be the solution to the following SDE,
$$X_{t}=  x_{0} +\int_{0}^{t} \frac{1}{2} g(X_s)ds + \int_{0}^{t} g(X_s) dW_{s},  t \in[0,T], x \in \mathbb{R}$$
Given the function g, define the additional function $h:\mathbb{R} \rightarrow \mathbb{R}$ as
$$ h(x) = \int_{0}^{x} \frac{1}{g(r)}dr, x\in \mathbb{R}$$
We assume that h is twice continuously differentiable and h has an inverse. Want to show that the solution X to the SDE can be written explicitly as
$$ X_{t} = h^{-1}\big(W_{t}+h(x_{0})\big)$$
My question and working:
I find that since h has an inverse we can write $h(X_{t}) = \big(W_{t}+h(x_{0})\big)$. We can then apply Ito's formula to $h(X_{t})$. I find the required derivatives that appear in Ito's formula to be $\partial_{t}f = 0$, $\partial_{x}f= \frac{1}{g(x)}$ and $\partial_{xx} = \frac{-g'(x)}{g^{2}(x)}$.
I am confused when applying this to Ito's formula, as we don't know what $U_{t}$ or $V_{t}$ are.
I obtain the following, where $(dX_t)^2$ is equal to the quadratic variation of x which would be the same as $V_{t}^{2}$.
$$dh(X_t) = 0dt + \frac{1}{g(X_t)} dX_t - \frac{1}{2}\frac{g'(X_t)}{g^2(X_t)} (dX_t)^2$$
The final answer is
$$dh(X_t) = \big[\frac{1}{g(X_t)}\frac{1}{2}g(X_s)g'(X_s) - \frac{1}{2}\frac{g'(X_t)}{g^2(X_t)}g^2(X_t) \big] dt + \big[ \frac{1 }{g(X_t)}g(X_t) \big] dW_t$$
I guess we used $(dXt)^{2}=V^{2}_{t}dt$ with $V_{t}=g(X_{t})$ to get the second term to match the SDE but then I am confused how we would know how to even arrive at the other two terms (and how we identified Vt if this was the case).
I thought to substitute the value of dXt given by the SDE provided for Xt, but this doesn't seem to work out for me.
Can someone please explain to me how to calculate $(dX_t)^2$?
Thank you in advance, any help is much appreciated.
 A: You need to understand that in terms of size, $dW_t\sim \sqrt{dt}$ is leading $dt$, $1\gg dW_t\gg dt$. Then it also should be clear that in
$$
(dX_t)^2=(b\,dt+a\,dW_t)^2
$$
the leading term is $a^2\,(dW_t)^2$. All the remaining terms have size order $(dt)^{3/2}$ or $(dt)^2$. In the "summation" of these "infinitesimals" towards the integral or solution of the SDE, only terms of size $dt$ or larger can give non-infinitesimal contributions. Thus the only part of the square playing a role is $a^2\,(dW_t)^2$. Assuming that the coefficients $a,b$ are smooth, they move slowly, so that the parts $(dW_t)$ get summed up. The average of $(dW_t)^2$ is $dt$. Thus indeed the Ito-Taylor formula gives
\begin{align}
dh(X_t)&=h(X_t+dX_t)-h(X_t)
\\
&=h'(X_t)dX_t+\tfrac12h''(X_t)(dX_t)^2~~~~\Bigl(+O(dt^{3/2})\Bigr)
\\
&=h'(X_t)(a\,dW_t+b\,dt)+\tfrac12h''(X_t)(a\,dW_t+b\,dt)^2
\\
&=(h'b+\tfrac12h''a^2)\,dt+h'a\,dW_t
\\
&=\tfrac12[h'g+h''g^2]\,dt+h'g\,dW_t=dW_t,
\end{align}
using $h'=1/g$, $h''=-g'/g^2$, $a=g$ and $b=\frac12g$ in the last line.
