In a rigorous setting, how do I do this derivative using Ricci Calculus?

Question

Let $R = \sqrt{x_k x_k}$. I want to calculate ${\partial R \over \partial x_i}$ and ${\partial^2 R \over \partial x_i \partial x_j}$. The solution was given as such: $$ {\partial R \over \partial x_j} = {\partial (x_i x_i)^{1/2} \over \partial x_j} = {1 \over 2R} (\delta_{ij} x_i + x_i \delta_{ij} ) = {x_i \over R}$$ In this problem, the derivative of the outer function (per chain rule) was simply given as $1 \over 2R$. However, I want to maintain the notation for $R$ in terms of $x$. What is the rule for changing the dummy variable when taking the derivative? The function can be neither: $$ {1 \over 2(x_i x_i)^{1/2}} (\delta_{ij} x_i + x_i \delta_{ij} )$$ nor: $$ {1 \over 2(x_j x_j)^{1/2}} (\delta_{ij} x_i + x_i \delta_{ij} )$$ Because both would violate the rule of the dummy variable being repeated more than twice. Do I just randomly choose some new variable to put in the expression for the denominator, i.e. $$ {1 \over 2(x_k x_k)^{1/2}} (\delta_{ij} x_i + x_i \delta_{ij} )$$ The same issue arises without loss of generality for taking the second derivative: $$ {\partial^2 R \over \partial x_i \partial x_j} = {\delta_{ij} \over R} + x_i {\partial \over \partial x_j} {1 \over R} = {\delta_{ij} \over R} - {x_i x_j \over R^3}$$