I am currently studying Optics, fifth edition, by Hecht. In chapter 2.9 Spherical Waves, when discussing the spherical coordinates $x = r \sin(\theta) \sin(\phi)$, $y = r \sin(\theta)\sin(\phi)$, $z = r \cos(\theta)$, the author says that the Laplacian operator is
$$\nabla^2 = \dfrac{1}{r^2} \dfrac{\partial}{\partial{r}} \left( r^2 \dfrac{\partial}{\partial{r}} \right) + \dfrac{1}{r^2 \sin(\theta)} \dfrac{\partial}{\partial{\theta}} \left( \sin(\theta) \dfrac{\partial}{\partial \theta} \right) + \dfrac{1}{r^2 \sin^2 \theta} \dfrac{\partial^2}{\partial \phi^2}. \tag{2.67}$$
The author then continues as follows:
We can obtain this result without being familiar with Eq. (2.67). Start with the Cartesian form of the Laplacian, Eq. (2.61); operate on the spherically symmetrical wavefunction $\psi(r)$; and convert each term to polar coordinates. Examining only the $x$-dependence, we have $$\dfrac{\partial{\psi}}{\partial{x}} = \dfrac{\partial{\psi}}{\partial{r}} \dfrac{\partial{r}}{\partial{x}}$$ and $$\dfrac{\partial^2{\psi}}{\partial{x}^2} = \dfrac{\partial^2{\psi}}{\partial{r^2}} \left( \dfrac{\partial{r}}{\partial{x}} \right)^2 + \dfrac{\partial{\psi}}{\partial{r}} \dfrac{\partial^2{\psi}}{\partial{x}^2}$$ since $$\psi(\vec{\mathbf{r}}) = \psi(r)$$ Using $$x^2 + y^2 + z^2 = r^2$$ we have $$\dfrac{\partial{r}}{\partial{x}} = \dfrac{x}{r}$$ $$\dfrac{\partial^2{r}}{\partial{x}^2} = \dfrac{1}{r} \dfrac{\partial}{\partial{x}}x + x \dfrac{\partial}{\partial{x}} \left( \dfrac{1}{r} \right) = \dfrac{1}{r} \left( 1 - \dfrac{x^2}{r^2} \right)$$ and so $$\dfrac{\partial^2{\psi}}{\partial{x}^2} = \dfrac{x^2}{r^2} \dfrac{\partial^2{\psi}}{\partial{r}^2} + \dfrac{1}{r} \left( 1 - \dfrac{x^2}{r^2} \right) \dfrac{\partial{\psi}}{\partial{r}}$$ Now having $\partial^2{\psi}/\partial{x}^2$, we form $\partial^2{\psi}/\partial{y}^2$ and $\partial^2{\psi}/\partial{z}^2$, and on adding get $$\nabla^2 \psi(r) = \dfrac{\partial^2{\psi}}{\partial{r}^2} + \dfrac{2}{r} \dfrac{\partial{\psi}}{\partial{r}}$$ which is equivalent to Eq. (2.68). This result can be expressed in a slightly different form: $$\nabla^2 \psi = \dfrac{1}{r} \dfrac{\partial^2}{\partial{r}^2}(r \psi) \tag {2.69}$$ The differential wave equation can then be written as $$\dfrac{1}{r} \dfrac{\partial^2}{\partial{r}^2}(r \psi) = \dfrac{1}{v^2} \dfrac{\partial^2{\psi}}{\partial{t}^2} \tag{2.70}$$ Multiplying both sides by $r$ yields $$\dfrac{\partial^2}{\partial{r}^2}(r \psi) = \dfrac{1}{v^2} \dfrac{\partial^2}{\partial{t}^2}(r \psi) \tag{2.71}$$ Notice that this expression is now just the one-dimensional differential wave equation, Eq. (2.11), where the space variable is $r$ and the wavefunction is the product $(r \psi)$. The solution of Eq. (2.71) is then simply $$r\psi(r, t) = f(r - vt)$$ or $$\psi(r, t) = \dfrac{f(r - vt)}{r} \tag{2.72}$$ This represents a spherical wave progressing radially outward from the origin, at a constant speed $v$, and having an arbitrary functional form $f$.
Equation 2.61 is
$$\nabla^2 \equiv \dfrac{\partial^2}{\partial{x}^2} + \dfrac{\partial^2}{\partial{y}^2} + \dfrac{\partial^2}{\partial{z}^2} \tag{2.61}$$
Equation 2.68 is
$$\nabla^2 \psi(r) = \dfrac{1}{r^2} \dfrac{\partial}{\partial{r}} \left( r^2 \dfrac{\partial{\psi}}{\partial{r}} \right) \tag{2.68}$$
It seems that the author is attempting to connect the Laplacian with the differential wave equation. But how did the author get 2.70 from 2.69? I would greatly appreciate it if people would please take the time to clarify this.
The wave equation is $$\nabla^2 A=\frac{1}{c^2}\frac{\partial^2 A}{\partial t^2}$$ The author uses the letter $v$ instead of the letter $c$ and simply plugs in our expression for $\nabla^2 \psi$ in equation 2.69 into the wave equation to reach $$\nabla^2 \psi=\frac{1}{r}\frac{\partial^2}{\partial r^2}(r\psi)=\frac{1}{v^2}\frac{\partial^2 \psi}{\partial t^2}.$$ Equation 2.70.