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