#### Answer

$dS=\pi {\frac{2r^2+h^2}{ (\sqrt{(r^2+h^2)})}}dr$

#### Work Step by Step

We are given: $S= \pi r \sqrt{(r^2+h^2)}$
on differentiating both sides and applying the product rule:
$dS= \pi r (\sqrt{(r^2+h^2)})'+\pi (\sqrt{(r^2+h^2)}){(r)'}$
$dS=\pi r{\frac{1}{2}\frac{1}{ (\sqrt{(r^2+h^2)})}(r^2+h^2)'}+\pi (\sqrt{(r^2+h^2)})dr$
$dS=\pi r{\frac{1}{2}\frac{1}{ (\sqrt{(r^2+h^2)})}(2rdr+2hdh)}+\pi (\sqrt{(r^2+h^2)})dr$
$dh=0$
$dS=\pi r{\frac{1}{2}\frac{1}{ (\sqrt{(r^2+h^2)})}2rdr}+\pi (\sqrt{(r^2+h^2)})dr$
$dS=\pi r^2{\frac{1}{ (\sqrt{(r^2+h^2)})}dr}+\pi (\sqrt{(r^2+h^2)})dr$
$dS=\pi {\frac{2r^2+h^2}{ (\sqrt{(r^2+h^2)})}}dr$