## Calculus: Early Transcendentals (2nd Edition)

$y' =\frac{5}{\sqrt {10x+1}}$
$y = \sqrt {10x+1}$ $y' = (\sqrt {10x+1})'$ The inner function is $g(x) = 10x+1$ and the outer function is $f(u) = \sqrt u$ $f'(u) = \frac{1}{2\sqrt u}$ Therefore by THEOREM 3.14 Version 2 $y' = (\sqrt {10x+1})' = \frac{1}{2\sqrt {g(x)}}g'(x) = \frac{1}{2\sqrt {10x+1}} \times 10 =\frac{5}{\sqrt {10x+1}}$