Let $f$ be smooth with $f,f' \in C_0(\mathbb{R}) \cap L^1(\mathbb{R})$. Do the difference quotients converge in $L^1$? Let $f$ be a $C^1$-smooth map $\mathbb{R} \to \mathbb{R}$ and assume that $f,f' \in C_0(\mathbb{R}) \cap L^1(\mathbb{R})$. Define $g_h(t) = (1/h)(f(t+h)- f(t))$ for $h \neq 0$. 
I have been able to show $g_h \to f'$ uniformly as $h \to 0$, and I'm wondering whether $g_h \overset{L^1}{\to} f'$ as well. In case it helps, here is my proof of the former claim.

Proof: For each $h,t$ the mean value theorem implies the existence of an $s=s_{t,h}$ between $t$ and $h+t$ such that $g_h(t) = f'(s)$. Since $f'$ is in $C_0(\mathbb{R})$ it is uniformly continuous. It follows that $f'(s) \to f'(t)$ uniformly in $t$ as $h \to 0$ so we are done.

 A: For a function $g \colon \mathbb{R} \to \mathbb{R}$, let $\tau_tg (x) = g(x+t)$.
Then, for all $t \in \mathbb{R}$, the map $\tau_t$ is an isometry of $L^1(\mathbb{R})$. It is easily verified that for $f \in C_c(\mathbb{R})$, we have $\lim\limits_{t\to 0} \lVert \tau_t f - f\lVert_1 = 0$. Since the family $\{\tau_t : t\in \mathbb{R}\}$ is equicontinuous, we have $\lim\limits_{t\to 0} \lVert \tau_t f - f\lVert_1 = 0$ for all $f \in L^1(\mathbb{R})$ (even uniformly on compact subsets of $L^1$, but we don't need that).
For $h > 0$ and $g \in L^1$, let
$$\mu_h(g)(x) := \frac{1}{h}\int_0^h g(x+t)\,dt$$
the $h$-average of $g$. Then we have
$$\begin{align}
\lVert \mu_h(g) - g\rVert_1 &= \int_\mathbb{R} \lvert \mu_h(g)(x) - g(x)\rvert \,dx\\
&= \frac{1}{h}\int_\mathbb{R} \left\lvert \int_0^h g(x+t)-g(x)\,dt\right\rvert\,dx\\
&\leqslant \frac{1}{h} \int_\mathbb{R}\int_0^h \lvert g(x+t) - g(x)\rvert\,dt\,dx\\
&= \frac{1}{h} \int_0^h \int_\mathbb{R} \lvert g(x+t)-g(x)\rvert\,dx\,dt\\
&= \frac{1}{h}\int_0^h \lVert \tau_t g - g\rVert_1\,dt
\end{align}$$
and that converges to $0$ when $h\to 0$ by the above property of the $\tau_t$, in other words, $\lim\limits_{h \to 0} \lVert \mu_h(g) - g\rVert_1 = 0$.
Now, in your case, we have
$$\frac{f(x+h) - f(x)}{h} = \frac{1}{h} \int_0^h f'(x+t)\,dt = \mu_h(f')(x),$$
and hence we have indeed $L^1$-convergence of the difference quotients.
