Spherical symmetry math For spherical symmetry how  the last  four equations  calculations is done? ccan you explain please?

For reference see the equations 44 
 A: We have $\rho(x) = \bigl(\sum_i x_i^2\bigr)^{1/2}$, hence 
\begin{align*}
  \partial_i \rho(x) &= \frac 1{2(\sum_j x_j^2)^{1/2}}\cdot 2x_i = \frac{x_i}{\rho(x)}\\
  \partial_i^2\rho(x) &= \frac{\rho^2(x) - x_i^2}{\rho^3(x)}
\end{align*}
As $S$ is spherically symmetric, the values of $S$ only depend on $x$'s distance to the origin, that is we have $S(x) = \tilde S(\rho(x))$ for some function $\tilde S$. $\tilde S$ is denoted by $S$ again in your paper, but I will write $\tilde S$ here. By the chain rule, we have
\begin{align*}
  \Delta S &= \sum_i \partial_i^2(\tilde S \circ \rho)\\
              &= \sum_i \partial_i(\partial_\rho\tilde S \circ \rho \cdot \partial_i S)\\
   &= \sum_i \partial_\rho^2 \tilde S \circ \rho \cdot (\partial_i \rho)^2 + \partial_\rho \tilde S \circ \rho \cdot \partial^2_i \rho\\
   &= \partial_\rho^2 \tilde S \circ \rho \cdot \sum_i (\partial_i\rho)^2 +
  \partial_\rho\tilde S \circ \rho \cdot \sum_i \partial^2_i \rho
\end{align*}
Using the above, we have
\begin{align*}
  \sum_i (\partial_i \rho)^2(x) &= \sum_i \frac{x_i^2}{\rho^2(x)}\\
         &= 1\\
  \sum_i \partial_i^2\rho(x) &= \sum_i \frac{\rho^2(x) - x_i^2}{\rho^3(x)}\\
       &= \frac{d\rho^2(x) - \sum_i x_i^2}{\rho^3(x)}\\
       &= \frac{d\rho^2(x) - \rho^2(x)}{\rho^3(x)}\\
       &= \frac{d-1}{\rho(x)}
\end{align*}
Using this, we get
$$
  \Delta S(x) = \partial_\rho^2 \tilde S(\rho(x)) + \frac{d-1}{\rho(x)} \cdot \partial_\rho \tilde S(\rho(x))
$$
or in the notation of the paper
$$ \Delta S = \partial_\rho^2 S + \frac{d-1}{\rho} \cdot \partial_\rho S
$$
