Let $\Omega=B_1(0)=\{x\in\mathbb{R}^N:|x|<1\}$ and let $u\in L^1(\Omega)$ such that $u(x)=|x|^{-a}$, with $0<a<N$. Determine $\nabla u$ as distributional derivative. Then determine $a$ and $p$ such that $u\in W^{1,p}(\Omega)$.
Regarding the first part of the exercise, given $\phi\in\mathcal{D}(\Omega)$, we have: $$\begin{aligned} \langle\nabla u,\phi\rangle &= -\langle u,\nabla\phi\rangle = -\lim_{\varepsilon\to0^+}\int_{\Omega\setminus B_\varepsilon(0)}\frac1{|x|^a}\nabla\phi\,dx \\ &= \lim_{\varepsilon\to0^+}\left[\int_{\Omega\setminus B_\varepsilon(0)}\nabla\left(|x|^{-a}\right)\phi\,dx - \int_{\partial B_\varepsilon(0)}\frac1{|x|^a}\phi\nu\,d\sigma\right] \\ &= \lim_{\varepsilon\to0^+}\left[\int_{\Omega\setminus B_\varepsilon(0)} -a|x|^{-a-1}\frac x{|x|} \phi\,dx - \frac1{\varepsilon^a}\int_{\partial B_\varepsilon(0)}\phi\nu\,d\sigma\right] \\ &= \lim_{\varepsilon\to0^+}\left[\int_{\Omega\setminus B_\varepsilon(0)} -a|x|^{-a-2}x\phi\,dx - \frac1{\varepsilon^a}\int_{\partial B_\varepsilon(0)}\phi\nu\,d\sigma\right] \\ \end{aligned}$$
Regarding the last part of the exercise, we need $\nabla u\in L^p(\Omega)$ and that happens if $p(a+1)<N$, i.e. $a<\displaystyle\frac Np-1$.
For $a<N-1$ we also have that the second term in the limit vanishes for $\varepsilon\to0^+$, since: $$\left|\frac1{\varepsilon^a}\int_{\partial B_\varepsilon(0)}\phi\nu\,d\sigma\right| \le \frac{\lVert\phi\rVert_\infty}{\varepsilon^a}\mathcal{O}\left(\varepsilon^{N-1}\right) \to 0 \quad\text{if}\quad N-1-a>0\text{, i.e. }a<N-1.$$
So, my answer would be $a<\displaystyle\frac Np-1$. Is that correct?
Yes that is correct, another approach is using polar coordinates \begin{align} \int_{B_1(0)} |\nabla u(x)|^pdx &= \int_{B_1(0)} \left|-a\frac{x}{|x|^{a+2}}\right|^p dx \\ &= a^p \int_{B_1(0)} \frac{1}{|x|^{pa+p}} dx\\ &=\int_{0}^{1}\int_{S^{N-1}} \frac{1}{r^{pa+p}} r^{N-1}dSdr \\ &= \int_{0}^1 r^{N-1-pa-p}dr \int_{S^{N-1}} dS \\ &= Area(S^{N-1}) \int_{0}^1 r^{N-1-pa-p}dr \end{align} Hence, the integrand $\|\nabla u\|_{L^p(\Omega)}< \infty $ if only if $N-1-pa-p>-1$ in other words $u \in W^{1,p}(\Omega)$ if only if $\frac{N}{p}-1>a.$