If $f \in L^1 \cap L^2$ is $L^2$-differentiable, then $Df \in L^1 \cap L^2$ Working with the definition that $f \in L^2(\mathbb{R})$ is $L^2$-differentiable with $L^2$-derivative $Df$ if
$$ \frac{\|\tau_hf-f-hDf\|_2}{h} \to 0 \text{ as } h \to 0 $$
(where $\tau_h(x) = f(x+h)$), I want to try and show that

If $f \in L^1 \cap L^2$ is $L^2$-differentiable, then $Df \in L^1 \cap L^2$.

Showing that $Df \in L^2$ is immediate from the definition; however, I'm unsure on how to get that $Df \in L_1$. I've tried arguing from density of continuous compactly supported functions (for which $Df \in L^2 \implies Df \in L^1$ by Jensen's inequality), but to no avail.

I don't think that what you are trying to prove is true.
Consider e.g.
$$
f(x) = \frac{1}{x^2} \cdot \sin(x^2) \text{ for large } x,
$$
i.e. truncate $f$ somehow near the origin.
We then have $f \in L^1 \cap L^2$ with
$$
f'(x) = -\frac{1}{x^3} \cdot \sin(x^2) + \frac{2}{x} \cdot \cos(x^2).
$$
We have $f' \in L^2$, but (as shown below) $f' \notin L^1$.
Even with your definition of differentiability, I am convinced that $f$ is $L^2$-differentiable with $L^2$-derivative $f' \in L^2 \setminus L^1$.
EDIT: Ok, here is a proof that $f' \notin L^1$. First of all, we have $\frac{1}{x^3} \cdot \sin(x^2) \in L^1 ((1,\infty))$ (and we are considering everything away from the origin), so that $f' \notin L^1$ is equivalent to $\frac{2}{x} \cdot \cos(x^2) \notin L^1$. But the subsitution $\omega =x^2$ yields
$$
\int_1^\infty \bigg| \frac{2}{x} \cdot \cos(x^2) \bigg| \, dx = \int_1^\infty \bigg| \frac{\cos(\omega)}{\omega} \bigg| \, d\omega = \infty,
$$
because of $|\cos(\omega)| \geq 1/2$ on $(-\epsilon, \epsilon) + 2\pi \Bbb{Z}$, which easily implies that the integral diverges.
