Answer
$\displaystyle p'(x) = \frac{3}{2\sqrt {3x}}$
$\displaystyle p'(1) = \frac{3}{2\sqrt{3}}=\frac{\sqrt 3}{2}$
$\displaystyle p'(3) = \frac{1}{2}$
$\displaystyle p'(\frac{2}{3}) = \frac{3}{2\sqrt 2}$
Work Step by Step
$p(x) =\sqrt (3x)$
Using the derivative definition:
$p'(x) = \lim\limits_{h \to 0}\frac{\sqrt (3(x+h)) - \sqrt (3x)}{h}$
$p'(x) = \lim\limits_{h \to 0}\frac{3(x+h) - 3x}{h(\sqrt (3(x+h)) + \sqrt (3x))}$
$p'(x) = \frac{3}{2\sqrt (3x)}$
Now,
$p'(1) = \frac{\sqrt 3}{2}$
$p'(3) = \frac{1}{2}$
$p'(\frac{2}{3}) = \frac{3}{2\sqrt 2}$
