Show how this equation

${}_{bc}^{\text{'}a}=\frac{del{x}^{\text{'}a}}{del{x}^d}\frac{del{x}^f}{del{x}^{\text{'}b}}\frac{del{x}^g}{del{x}^{\text{'}c}}{}_{fg}^d+\frac{del{x}^{\text{'}a}}{del{x}^d}\frac{de{l}^2{x}^d}{del{x}^{\text{'}c}del{x}^{\text{'}b}}$

becomes this

${}_{bc}^{\text{'}a}=\frac{del{x}^{\text{'}a}}{del{x}^d}\frac{del{x}^f}{del{x}^{\text{'}b}}\frac{del{x}^g}{del{x}^{\text{'}c}}{}_{fg}^d-\frac{del{x}^d}{del{x}^{\text{'}b}}\frac{del{x}^f}{del{x}^{\text{'}c}}\frac{de{l}^2{x}^{\text{'}a}}{del{x}^{d}del{x}^f}$

if ${}^{la}$ be are Christoffel symbols of the second kind in new coordinates, and ${}^{a}be$ are Christoffel symbols in old $($previous$)$ coordinates

I need step by step solution with detail explanation