1. **State the problem:** We need to find and sketch the derivative of the function $$f(x) = \ln(\tan x)$$. 2. **Recall the formula for the derivative of a composite function:** If $$f(x) = \ln(g(x))$$, then $$f'(x) = \frac{g'(x)}{g(x)}$$. 3. **Identify the inner function:** Here, $$g(x) = \tan x$$. 4. **Find the derivative of the inner function:** $$g'(x) = \sec^2 x$$. 5. **Apply the chain rule:** $$ f'(x) = \frac{\sec^2 x}{\tan x} $$ 6. **Simplify the expression:** Recall that $$\tan x = \frac{\sin x}{\cos x}$$ and $$\sec x = \frac{1}{\cos x}$$, so $$ f'(x) = \frac{\sec^2 x}{\tan x} = \frac{\frac{1}{\cos^2 x}}{\frac{\sin x}{\cos x}} = \frac{1}{\cos^2 x} \times \frac{\cos x}{\sin x} = \frac{1}{\cos x \sin x} $$ 7. **Final simplified derivative:** $$ f'(x) = \frac{1}{\sin x \cos x} $$ This derivative is undefined where $$\sin x = 0$$ or $$\cos x = 0$$, which correspond to the vertical asymptotes of the original function. **Summary:** The derivative of $$f(x) = \ln(\tan x)$$ is $$f'(x) = \frac{1}{\sin x \cos x}$$.