#### Answer

The Derivative is:
$y'=\sec x(1+x\tan x)-\frac{1}{x^2}$

#### Work Step by Step

$y=x\sec x +\frac{1}{x}$
Applying Derivative Rules
$y'=f'(x)+g'(x)$ and $f'(x)=h'(x)\cdot v(x) + h(x)\cdot v'(x)$
$y'=((x^{1-1})(\sec x)+(x)(\sec x\tan x))+(-1)x^{-1-1}$
$y'=(\sec x + x\sec x \tan x)-\frac{1}{x^2}$
$y'=\sec x(1+x\tan x)-\frac{1}{x^2}$