Answer
a) $y=\sqrt x+C$
b) $y=2 \sqrt x+C$
c) $y=2x^2-2\sqrt x+C$
Work Step by Step
a) When $(\sqrt x)'=\dfrac{1}{2 \sqrt x}$
Thus, $y'=\dfrac{1}{2 \sqrt x}\implies y=\sqrt x+C$
b) When $( 2\sqrt x)'=\dfrac{1}{ \sqrt x}$
Thus, $y'=\dfrac{1}{\sqrt x}\implies y=2 \sqrt x+C$
c) When $( 2x^2-2\sqrt x)'=4x-\dfrac{1}{ \sqrt x}$
Thus, $y'=4x-\dfrac{1}{ \sqrt x} \implies y=2x^2-2\sqrt x+C$