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} $

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