Answer
$y'=\tan x $, $\;\; y''=\sec ^2 x$
Work Step by Step
It is given that :- \[y=\ln |\sec x|\;\;\;...(1)\]
Differentiating (1) with respect to $x$ using chain rule
\[y'=\frac{1}{\sec x}\cdot (\sec x)'\]
\[y'=\frac{1}{\sec x}\cdot (\sec x\tan x)\]
\[y'=\tan x\;\;\;...(2)\]
Differentiating (2) with respect to $x$
\[y''=\sec ^2 x\]
Hence $y'=\tan x $ and $ y''=\sec ^2 x$
