#### Answer

The Derivative is:
$y'=\frac{\sec x}{2\sqrt{x}}+\sqrt{x}\sec x\tan x$

#### Work Step by Step

$y=\sqrt{x}\sec x+3$
Applying Derivative Rules
$y'=f'(x)+g'(x)$ and $f'(x)=h'(x)\cdot r(x)+h(x)\cdot r'(x)$
$y'=((\frac{1}{2\sqrt{x}})(\sec x)+(\sqrt{x})(\sec x\tan x))+(0)$
$y'=\frac{\sec x}{2\sqrt{x}}+\sqrt{x}\sec x\tan x$