Answer
$ p'(x) = \dfrac{3}{2\sqrt {3x}}$ and $p'(1) =\dfrac{\sqrt 3}{2} \\ p'(3) = \dfrac{1}{2} \\ p'(\dfrac{2}{3}) = \dfrac{3}{2\sqrt 2}$
Work Step by Step
Consider $p(x) =\sqrt {(3x)}$
We will use the derivative definition, such that
$p'(x) = \lim\limits_{h \to 0}\dfrac{\sqrt {(3(x+h))} - \sqrt {(3x)}}{h}$
$\implies \lim\limits_{h \to 0}\dfrac{3(x+h) - 3x}{h(\sqrt {(3x+h)} + \sqrt {(3x)})}=\dfrac{3}{2\sqrt {(3x)}}$
Now, we have
$ p'(x) = \dfrac{3}{2\sqrt {3x}}$ and $p'(1) =\dfrac{\sqrt 3}{2} \\ p'(3) =\dfrac{3}{2\sqrt {3(3)}}= \dfrac{1}{2} \\ p'(\dfrac{2}{3}) = \dfrac{3}{2\sqrt {3(\dfrac{2}{3})}}= \dfrac{3}{2\sqrt 2}$
