Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 30

Determine $y'$ and $y''$ of $y = \ln (\sec x + \tan x)$
Solving for $y'$

$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} 1 (\sec x + \tan x)\\
\\
y' &= \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx} ( \sec x + \tan x )\\
\\
y' &= \frac{1}{\sec x + \tan x} ( \sec x \tan x + \sec^2 x)\\
\\
y' &= \frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \left[ \left( \frac{1}{\cos x} \right) \left( \frac{\sin x}{\cos x} \right) + \frac{1}{\cos^2x} \right]\\
\\
y' &= \frac{1}{\frac{1+ \sin x}{\cos x}} \left[ \frac{\sin x + 1}{\cos^2 x} \right]\\
\\
y' &= \left( \frac{\cos x}{\cancel{1 + \sin x}} \right) \left( \frac{\cancel{1 + \sin x}}{\cos ^2 x} \right)\\
\\
y' &= \frac{1}{\cos x}\\
\\
y' &= \sec x
\end{aligned}
\end{equation}
$


Solving for $y''$

$
\begin{equation}
\begin{aligned}
y'' &= \frac{d}{dx} (\sec x)\\
\\
y'' &= \sec x \tan x
\end{aligned}
\end{equation}
$