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By locality of weak differentiability, it is enough to work near the exceptional point; translate it to zero. Let $g_i$ denote the classical derivative off zero, assigning any value at zero. For $\varphi \in C_c^{\infty }(\Omega )$ choose a smooth cutoff ${\chi }_{\epsilon }$ which is zero on $B_{\epsilon }$, one outside $B_{2\epsilon }$, satisfies $|D{\chi }_{\epsilon }|\le C/\epsilon$, and has its transition region contained in $\Omega$. The product $\varphi {\chi }_{\epsilon }$ avoids the singular point, so classical integration by parts givesThe middle term has absolute value at mostwhen $n\ge 2$. The other terms tend to $\int u{D}_i\varphi$ and $-\int g_i\varphi$, since $u$ is locally integrable and $g_i\in L_{\mathrm{loc}}^1$. Therefore $\boxed{D_{i}u=g_i\text{ weakly on all of }\Omega }$. Its value at the single point is irrelevant. This is bounded isolated-point removability for weak first derivatives.

For $n=1$, take $u(x)=0$ for $x<0$ and $u(x)=1$ for $x>0$ on $(-1,1)$, with any value at zero. It is bounded and $C^1$ off zero, with classical derivative zero there. Butso its distributional derivative is a Dirac delta. If a locally integrable weak derivative existed, it would be zero off zero and hence zero almost everywhere, contradicting this identity. The step function is the required counterexample; the cutoff error in dimension one need not vanish.