1. **State the problem:** Find the derivative of the function $$f(x) = \frac{\sec x}{\tan x}$$. 2. **Recall the formula:** To differentiate a quotient $$\frac{u}{v}$$, use the quotient rule: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ where $$u = \sec x$$ and $$v = \tan x$$. 3. **Find derivatives of numerator and denominator:** - $$u' = \frac{d}{dx}(\sec x) = \sec x \tan x$$ - $$v' = \frac{d}{dx}(\tan x) = \sec^2 x$$ 4. **Apply the quotient rule:** $$f'(x) = \frac{(\sec x \tan x)(\tan x) - (\sec x)(\sec^2 x)}{(\tan x)^2}$$ 5. **Simplify numerator:** $$\sec x \tan^2 x - \sec x \sec^2 x = \sec x (\tan^2 x - \sec^2 x)$$ 6. **Use the identity:** $$\sec^2 x = 1 + \tan^2 x \implies \tan^2 x - \sec^2 x = \tan^2 x - (1 + \tan^2 x) = -1$$ 7. **Substitute back:** $$\sec x (\tan^2 x - \sec^2 x) = \sec x (-1) = -\sec x$$ 8. **Write the derivative:** $$f'(x) = \frac{-\sec x}{\tan^2 x} = -\frac{\sec x}{\tan^2 x}$$ **Final answer:** $$\boxed{f'(x) = -\frac{\sec x}{\tan^2 x}}$$