Answer
$ \sec x$
Work Step by Step
We have: $z(x)= \ln |\sec x+\tan x|$
We differentiate both sides with respect to $x$.
Use $\dfrac{d}{dx} [\ln |u|]=\dfrac{1}{u}\dfrac{du}{dx}$
$z^{\prime}(x)=\dfrac{d}{dx} [\ln |\sec x+\tan x|] \\= \dfrac{1}{\sec x+\tan x} \dfrac{d}{dx} [\sec x+\tan x]$
Simplify to obtain:
$z^{\prime}(x)=\dfrac{\sec x}{\sec x+\tan x} (\tan x +\sec x)=\sec x$
