91Ó°ÊÓ

Differentiate each term in the simplified function with respect to \(x\). Keep in mind that the derivative of \(\ln(u)\) is \(\frac{1}{u} \cdot \frac{du}{dx}\), so we will apply the chain rule when differentiating each term. \begin{align*} f'(x) &= \frac{d}{dx} \left( 2\ln(\sec^2 x) + 2\ln(\tan x) \right) \\ &= 2 \cdot \frac{1}{\sec^2 x} \cdot \frac{d}{dx}(\sec^2 x) + 2 \cdot \frac{1}{\tan x} \cdot \frac{d}{dx}(\tan x) \end{align*} Now we just need to compute the derivatives of \(\sec^2 x\) and \(\tan x\). Recall that the derivative of \(\sec u\) is \(\sec u \tan u \cdot \frac{du}{dx}\), and the derivative of \(\tan u\) is \(\sec^2 u \cdot \frac{du}{dx}\). Using these facts, we get: \begin{align*} \frac{d}{dx}(\sec^2 x) &= \frac{d}{dx}\left((\sec x)^2\right) \\ &= 2 \sec x \cdot \frac{d}{dx}(\sec x) \\ &= 2\sec x \cdot (\sec x \tan x) \\ &= 2\sec^2 x \tan x \end{align*} and \begin{align*} \frac{d}{dx}(\tan x) &= \sec^2 x. \end{align*} Now substitute these expressions back into the derivative of \(f(x)\): \begin{align*} f'(x) &= 2 \cdot \frac{1}{\sec^2 x} \cdot (2\sec^2 x \tan x) + 2 \cdot \frac{1}{\tan x} \cdot (\sec^2 x) \\ &= 4\tan x + 2\sec^2 x \cdot \frac{1}{\tan x} \\ &= 4\tan x + 2\left(\frac{\sec^2 x}{\tan x}\right) \end{align*} So, the derivative of the given function is \(f'(x) = 4\tan x + 2\left(\frac{\sec^2 x}{\tan x}\right)\).