How to prove the following claim?
Let $\Omega $ be an open subset of $ \mathbb{R}^n $, $ u \in C(\Omega) $ and $ \chi \in C_0^\infty (\Omega) $ with $ \text{supp} \: \chi \neq \emptyset $.
If $ \partial_j (\chi u) \in C(\Omega) $, then $ \partial_j u $ exists (and is continuous) on the interior of $ \text{supp} \: \chi $.
N.B. The claim in the title follows immediately from this.
On
The fundamental theorem of calculus allows us to write, for $ x_0 \in \Omega $,
\begin{equation} (\chi u) (x_0 + te_j) - (\chi u)(x_0) = \int_{0}^{t} \partial_j (\chi u) (x_0 + se_j) \: ds , \end{equation} where $e_j$ is the $i$-th standard unit vector in $ \mathbb{R}^n$ and $ t $ is a real parameter. Applying the fundamental theorem of calculus again along with the quotient rule, we obtain that for $ x_0 \in \text{int} (\text{supp} \: \chi) \subset \Omega $
\begin{align} \partial_j u (x_0) &= \left. \frac{d}{dt} u(x_0 + te_j) \right \vert_{t=0} \\ &= - \chi (x_0) \: u(x_0) \frac{\partial_j \chi (x_0)}{\lvert \chi (x_0) \rvert^2} + \partial_j (\chi u)(x_0) \\ &= - \frac{u(x_0)}{\chi (x_0)} \partial_j\chi(x_0) + \partial_j (\chi u)(x_0) . \end{align}
Hence $ \partial_j u $ exists and is clearly finite on the set $ \text{int} (\text{supp} \: \chi) $. q.e.d
In the interior of the support of $\chi$: The function $\frac 1\chi$ is smooth. Hence $u$ is $C^1$, being the product of two $C^1$ functions: $$u=(u\chi)\frac1\chi.$$