We have 
$$d(t):=\int dx\Big(\frac{f(x+t)-f(x)}{t}-f'(x)\Big)
=\int dx\int_0^1 ds\,(f'(x+st)-f'(x)),$$
whence
$$|d(t)|\le\int_0^1 ds\, J_t(s),$$
where 
$$J_t(s):=\int dx\,|f'(x+st)-f'(x)|\underset{t\to0}\longrightarrow0 \tag{10}\label{10}$$
for each $s$, since $f'\in L^1$ (see e.g. [this answer][1]). Also, 
$$|J_t(s)|\le\int dx\,(|f'(x+st)+|f'(x)|)=2\|f'\|_1<\infty.$$

So, by dominated convergence, $d(t)\underset{t\to0}\longrightarrow0$.

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Instead of dominated convergence, one can use the fact that the convergence in \eqref{10} is uniform in $s\in[0,1]$.


  [1]: https://math.stackexchange.com/a/67674/96609