## Thomas' Calculus 13th Edition

$r'(0) = 1 \\ r'(1) = \dfrac{1}{\sqrt 3} \\ r'(\dfrac{1}{2}) = \dfrac{1}{\sqrt 2}$ and $r'(s)= \dfrac{1}{\sqrt {(2s+1)}}$
Consider $r(s) = \sqrt {(2s+1)}$ We will use the derivative definition, such that $r'(s) = \lim\limits_{h \to 0}\dfrac{\sqrt (\sqrt {2(s+h)+1)} - \sqrt {(2s+1)}}{h}= \lim\limits_{h \to 0}\dfrac{(2(s+h)+1) - (2s+1)}{h(\sqrt {(2(s+h)}+1) + \sqrt {(2s+1))}}= \dfrac{1}{\sqrt {(2s+1)}}$ Now, we have $r'(0) = 1 \\ r'(1) =\dfrac{1}{\sqrt {(2(1)+1)}}= \dfrac{1}{\sqrt 3} \\ r'(\dfrac{1}{2}) =\dfrac{1}{\sqrt {(2(\dfrac{1}{2}) +1)}}= \dfrac{1}{\sqrt 2}$