#### Answer

The Derivative is:
$y'=\sec^2 x$

#### Work Step by Step

$y=(\sin x +\cos x)\sec x$
Applying the distributive property of mathematics
$y=\sin x \sec x + \cos x\sec x$
$y=\sin x\frac{1}{\cos x}+\cos x\frac{1}{\cos x}$
$y=\tan x +1$
Applying Derivative Rules:
$y'=f'(x) + g'(x)$
$y'=\frac{d}{dx}(\tan x) + \frac{d}{dx}(1)$
$y'=(sec^2 x) + 0$
$y'=sec^2 x$