Are the derivatives of the orthogonal polar factor locally integrable?

Let $$\mathbb{D}^2$$ be the closed unit disk, and let $$f:\mathbb{D}^2 \to \mathbb{R}^2$$ be a real-analytic map, satisfying $$\det df>0$$ everywhere except on a set of Hausdorff dimension not greater than $$1$$. Define $$U:= \{ p \in \mathbb{D}^2 \, | \, \det df>0\},$$

and set $$Q=Q(df)=df(\sqrt{df^Tdf})^{-1}$$ to be the orthogonal polar factor of $$df$$; $$\sqrt{}$$ is the unique symmetric positive-definite matrix square root. $$Q$$ is well-defined on $$U$$-in particular it is defined a.e. on $$\mathbb{D}^2$$. The restriction $$Q|_U$$ is smooth, even real-analytic. So, we can consider the derivatives $$Q_x,Q_y$$ of $$Q$$ on $$U$$.

Question: Are $$Q_x,Q_y$$ locally integrable on $$\mathbb D^2$$? i.e. is it true that $$Q_x,Q_y \in L^1_{loc}(\mathbb D^2)$$?

The easiest case is probably when $$f$$ is conformal a.e. (so it's actually holomorphic), as then $$Q(df)=\frac{\sqrt 2}{\|df\|}df$$ is obtained from $$df$$ simply by normalization. I asked on this case separately here.

Comment:

The only possible problem is when we approach points where $$df=0$$, as the polar factor map $$\text{GL}_2^+ \to \text{SO}_2$$ can be extended to a smooth map $$\text{GL}_2^+ \cup (\text{rank} = 1) \to \text{SO}_2$$.

However, the polar factor cannot be extended continuously on all $$\text{GL}_2^+ \cup \det^{-1}(0)$$, as singularities occur:

1. $$\lim_{t \to 0} Q(\left(\begin{matrix}t & 0 \\ 0 & t\end{matrix}\right))=\text{sgn}(t) \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$$, and

2. $$\lim_{t \to 0}Q(\left(\begin{matrix}0 & -t \\ t & 0\end{matrix}\right))=\text{sgn}(t) \left(\begin{matrix}0 & -1 \\ 1 & 0\end{matrix}\right)$$.

This question is a first step in understanding wether or not $$Q \in W^{1,p}(\mathbb{D}^2,\mathbb{R}^{4})$$ for some $$p \ge 1$$. (As you can see here).