# For an Ito Process, $d\ln{X} \neq \frac{dX}{X}$ and $(d\ln{X})^2 = (\frac{dX}{X})^2$, but $d\ln{X} \neq \pm \frac{dX}{X}$

In normal calculus we can write $$d\ln{x} = \frac{dx}{x}$$ since there is no quadratic variation to deal with. This isn't true for stochastic processes, and Ito's Lemma is used to calculate $$d\ln{X}$$. So when I was reading about volatility/realized volatility, I saw that there were two expressions used for realized variance $$\sigma ^2 dt$$, one is $$\frac{dX}{X}^2$$, while the other is $$d\ln{X}^2$$. So although these are equal I was wondering how come taking the square root doesn't lead to a valid relationship. I know that certain terms are 'negligible' thus leading the square operation to give the same results, but I don't see how to 'reverse' this using the square root.

Here's my work (it's just using Ito's Lemma and simplifying quadratic terms):

For an Ito Process of form: $$\frac{dX_t}{X_t} = \mu(t,X_t) dt + \sigma(t,X_t) dW_t$$

$$\begin{equation} (\frac{dX}{X})^2 = (\mu dt + \sigma dW)^2 = \sigma^2dt \end{equation}$$ while \begin{align} d\ln{X} = \frac{1}{X}dX - \frac{1}{2X^2}*dX^2 = \frac{dX}{X} - \frac{1}{2} \sigma ^2dt = (\mu - \frac{1}{2} \sigma ^2 ) dt + \sigma dW \\ (d\ln{X})^2 = ((\mu - \frac{1}{2} \sigma ^2 ) dt + \sigma dW)^2 = (\sigma dW) ^2 = \sigma ^2 dt \end{align}

so $$d\ln{X} \neq \frac{dX}{X}$$ and $$(d\ln{X})^2 = (\frac{dX}{X})^2$$.

In general, if $$x^2 = y^2$$, then $$x = \pm y$$, but $$d\ln{X} \neq \pm \frac{dX}{X}$$ since $$d\ln{X}$$ has the $$- \frac{1}{2} \sigma ^2 dt$$ term due to the quadratic variation of a Brownian motion.

So basically I am wondering why this is invalid, and what other sorts of operations on stochastic processes are invalid. Thanks!