There is no "G1" in what you post. Did you mean g'?

We are given that \(\displaystyle g(x)= \sqrt{x}= x^{1/2}\) so that \(\displaystyle g'(x)= (1/2)x^{-1/2}. Further\(\displaystyle g(x)= \sqrt{x}= y\) so that \(\displaystyle x= y^2\) and \(\displaystyle x^{-1/2}= (y^2)^{-1/2}= 1/y\) so \(\displaystyle (1/2)x^{-1/2}= 1/(2y)\).\)