To differentiate \(\sec x = \frac{1}{\cos x}\), use the quotient rule: \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\).

Let \(u = 1\) and \(v = \cos x\). Then \(\frac{du}{dx} = 0\) and \(\frac{dv}{dx} = -\sin x\).

Applying the quotient rule:

\(\frac{d}{dx} \left( \frac{1}{\cos x} \right) = \frac{\cos x \cdot 0 - 1 \cdot (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x.\)

Thus, the derivative of \(\sec x\) is \(\sec x \tan x\).

Now, differentiate \(y = \ln(\sec x + \tan x)\) using the chain rule:

Let \(u = \sec x + \tan x\), then \(y = \ln u\).

\(\frac{dy}{du} = \frac{1}{u}\) and \(\frac{du}{dx} = \sec x \tan x + \sec^2 x\).

Thus, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x)\).

Simplifying:

\(\frac{dy}{dx} = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} = \sec x\).