Answer
$y' =\frac{x}{\sqrt {x^2+9}} $
Work Step by Step
$y = \sqrt {x^2+9}$
$y' = (\sqrt {x^2+9})'$
The inner function is $g(x) = x^2+9$
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 {x^2+9}}(2x) =\frac{x}{\sqrt {x^2+9}} $
