Answer
The Derivative is:
$y'=\frac{-x\sin x-\cos x}{x^2}+\sec x+x\sec x\tan x$
Work Step by Step
$y=\frac{\cos x}{x}+\frac{x}{\cos x}$
$y'=\frac{(-\sin x)(x)-(\cos x)(x^{1-1})}{(x)^2}+\frac{(x^{1-1})(\cos x)-(x)(-\sin x)}{\cos^2 x}$
$y'=\frac{-x\sin x-\cos x}{x^2}+\frac{\cos x+x\sin x}{\cos^2 x}$
$y'=\frac{-x\sin x-\cos x}{x^2}+\frac{\cos x}{\cos^2 x}+\frac{x\sin x}{\cos^2 x}$
$y'=\frac{-x\sin x-\cos x}{x^2}+\frac{\cos x}{\cos x}\frac{1}{\cos x}+\frac{1}{\cos x}\frac{\sin x}{\cos x}x$
Applying trigonometric identities:
$y'=\frac{-x\sin x-\cos x}{x^2}+\sec x+x\sec x\tan x$
