## University Calculus: Early Transcendentals (3rd Edition)

$r'(s) = \frac{1}{\sqrt {2s+1}}$ $r'(0) = 1$ $r'(1) = \frac{1}{\sqrt 3}$ $r'(\frac{1}{2}) = \frac{1}{\sqrt 2}$
$r(s) = \sqrt (2s+1)$ 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, $r'(0) = 1$ $r'(1) = \frac{1}{\sqrt 3}$ $r'(\frac{1}{2}) = \frac{1}{\sqrt 2}$