If $f \in W^{1, 1}(\mathbb{R}^n)$ then $\theta \mapsto f \circ (Id + \theta)$ is differentiable

Question

In a book I'm reading I have this statement:

If $f \in W^{1, 1}(\mathbb{R}^N)$, then the map defined by

\begin{align} W^{1, \infty}(\mathbb{R}^N; \mathbb{R}^N) &\rightarrow L^1(\mathbb{R}^N)\\ \theta &\mapsto f \circ (Id + \theta) \end{align}

is Fréchet differentiable.

I can't quite see how to prove this fact. The book uses the argument that:

$$ f(Id + \theta)(x) = f(x) + \nabla f(x) \cdot \theta(x) + o(\theta), \hspace{5mm} \dfrac{||o(\theta)||_{L^1}}{||\theta||_{W^{1, \infty}}} \xrightarrow{\theta \rightarrow 0} 0 . $$

Yet I fail to see how this equality holds, since $f$ is not necessarily differentiable at $x$.