[tex]\it f(x) = x\sqrt{x} -3x
\\\;\\ \\\;\\
f'(x) = (x\sqrt{x} )' - (3x)' = \sqrt{x} +\dfrac{x}{2\sqrt{x}} -3 =
\\\;\\ \\\;\\
= \sqrt{x} + \dfrac{ \sqrt{x} \cdot \sqrt{x}}{2\sqrt{x}} - 3 = \sqrt{x} + \dfrac{ \sqrt{x}}{2} - 3 = \dfrac{2\sqrt{x}+\sqrt{x}-6}{2} =[/tex]
[tex]\it = \dfrac{3\sqrt{x}-6}{2}[/tex]
