$f(x)=\dfrac{2x+\sqrt{3x}}{\sqrt{x}}=\dfrac{2x}{\sqrt{x}}+\dfrac{\sqrt3.\sqrt{x}}{\sqrt{x}}=2\sqrt{x}+\sqrt3$
$\to f'(x)=2.\dfrac{1}{2\sqrt{x}}=\dfrac{1}{\sqrt{x}}$
$f(z)=\dfrac{2z^3-3z+\sqrt{z}-1}{z}=2z^2-3+\dfrac{1}{\sqrt{z}}-\dfrac{1}{z}$
$\to f'(z)=2(z^2)'-\dfrac{(\sqrt{z})'}{(\sqrt{z})^2}+\dfrac{(z)'}{z^2}$
$=4z-\dfrac{1}{2z\sqrt{z}}+\dfrac{1}{z^2}$
