Differentiate $$f(x)$$​ = cos(tan $$3x$$​) + sin(tan $$3x$$).

Correct option is D

​$$\begin{aligned}&\textbf{1. Differentiate } \cos(\tan 3x): \\&\text{Let } u = \tan 3x. \\&\text{Then, } \cos(u) \text{ differentiates to } -\sin(u) \cdot \frac{du}{dx}. \\&\text{Compute } \frac{du}{dx}: \\&\frac{d}{dx}(\tan 3x) = \sec^2 3x \cdot 3 = 3 \sec^2 3x \\&\text{So, the derivative is:} \\&\frac{d}{dx}[\cos(\tan 3x)] = -\sin(\tan 3x) \cdot 3 \sec^2 3x = -3 \sin(\tan 3x) \sec^2 3x \\\\&\textbf{2. Differentiate } \sin(\tan 3x): \\&\text{Similarly, let } u = \tan 3x. \\&\text{Then, } \sin(u) \text{ differentiates to } \cos(u) \cdot \frac{du}{dx}. \\&\text{We already have } \frac{du}{dx} = 3 \sec^2 3x. \\&\text{So, the derivative is:} \\&\frac{d}{dx}[\sin(\tan 3x)] = \cos(\tan 3x) \cdot 3 \sec^2 3x = 3 \cos(\tan 3x) \sec^2 3x \\\\&\textbf{Combine the Derivatives} \\&\text{Add the derivatives of the two terms:} \\&f'(x) = -3 \sin(\tan 3x) \sec^2 3x + 3 \cos(\tan 3x) \sec^2 3x \\&\text{Factor out the common term } 3 \sec^2 3x: \\&f'(x) = 3 \sec^2 3x \left( \cos(\tan 3x) - \sin(\tan 3x) \right)\end{aligned}$$​​