I am reading chapter 3.6.1 in the 3rd edition of Sakurai's Modern Quantum Mechanics. On page 191 the connection between the $\mathbf{L^2}$ operator and the angular part of the Laplacian is shown. The line $(3.227)$ reads:

$$\langle\mathbf{x'}|\mathbf{x \cdot p}|\alpha\rangle = \mathbf{x'} \cdot \left(- i \hbar \nabla' \langle\mathbf{x'}|\alpha\rangle \right)$$ $$=- i \hbar r \frac{\partial}{\partial r} \langle\mathbf{x'}|\alpha\rangle$$

The whole further derivation is based on this relation.

I cannot understand why we take only the radial part of the gradient. Why are we allowed to neglect the angular part?

I am reading chapter 3.6.1 in the 3rd edition of Sakurai's Modern Quantum Mechanics. On page 191 the connection between the $\mathbf{L^2}$ operator and the angular part of the Laplacian is shown. The line $(3.227)$ reads:

$$\langle\mathbf{x'}|\mathbf{x \cdot p}|\alpha\rangle = \mathbf{x'} \cdot \left(- i \hbar \nabla' \langle\mathbf{x'}|\alpha\rangle \right)$$ $$=- i \hbar r' \frac{\partial}{\partial r'} \langle\mathbf{x'}|\alpha\rangle\tag{3.227}$$

The whole further derivation is based on this relation.

I cannot understand why we take only the radial part of the gradient. Why are we allowed to neglect the angular part?