Answer
$$y'=5 \sec 5 x$$
Work Step by Step
Given $$y=\ln|\sec 5 x+\tan 5 x|$$
Then
\begin{align*}
y^{\prime}&=\frac{1}{\sec 5 x+\tan 5 x}\left(\sec 5 x \tan 5 x \cdot 5+\sec ^{2} 5 x \cdot 5\right)\\
&=\frac{5 \sec 5 x(\tan 5 x+\sec 5 x)}{\sec 5 x+\tan 5 x}\\
&=5 \sec 5 x
\end{align*}
