If f(x)=√(1+√x) for x > 0, then f'(x) is :

$f'(x)=\frac{1}{2 \sqrt{1+\sqrt{x}}}\left(0+\frac{1}{2 \sqrt{x}}\right)$

$=\frac{1}{4 \sqrt{x}(\sqrt{1+\sqrt{x}})}=\frac{1}{4 \sqrt{x} . f(x)}$

Hence (2) is correct answer.