Answer: see proof below
Step-by-step explanation:
Use the Quotient rule for derivatives:
[tex]\text{If}\ y=\dfrac{a}{b}\quad \text{then}\ y'=\dfrac{a'b-ab'}{b^2}[/tex]
Given: [tex]y=\dfrac{2\sqrtx}{1-x}[/tex]
[tex]\sqrtx[/tex][tex]a=2\sqrt x\qquad \rightarrow \qquad a'=\dfrac{1}{\sqrt x}\\\\b=1-x\qquad \rightarrow \qquad b'=-1[/tex]
[tex]y'=\dfrac{\dfrac{1-x}{\sqrt x}-(-2\sqrt x)}{(1-x)^2}\\\\\\.\quad =\dfrac{\dfrac{1-x}{\sqrt x}-(-2\sqrt x)\bigg(\dfrac{\sqrt x}{\sqrt x}\bigg)}{(1-x)^2}\\\\\\.\quad =\dfrac{1-x+2x}{\sqrt x(1-x)^2}\\\\\\.\quad =\dfrac{x+1}{\sqrt x(1-x)^2}[/tex]
LHS = RHS: [tex]\dfrac{x+1}{\sqrt x(1-x)^2}=\dfrac{x+1}{\sqrt x(1-x)^2}\qquad \checkmark[/tex]
