1. **Problem:** Differentiate $y = \frac{\sec x}{1 + \tan x}$. 2. **Formula and rules:** Use the quotient rule: if $y = \frac{u}{v}$, then $y' = \frac{u'v - uv'}{v^2}$. 3. Identify $u = \sec x$ and $v = 1 + \tan x$. 4. Compute derivatives: - $u' = \sec x \tan x$ - $v' = \sec^2 x$ 5. Apply quotient rule: $$y' = \frac{(\sec x \tan x)(1 + \tan x) - (\sec x)(\sec^2 x)}{(1 + \tan x)^2}$$ 6. Factor $\sec x$ in numerator: $$y' = \frac{\sec x \left[ \tan x (1 + \tan x) - \sec^2 x \right]}{(1 + \tan x)^2}$$ 7. Simplify inside brackets: $$\tan x + \tan^2 x - \sec^2 x$$ 8. Use identity $\sec^2 x = 1 + \tan^2 x$: $$\tan x + \tan^2 x - (1 + \tan^2 x) = \tan x - 1$$ 9. Final derivative: $$y' = \frac{\sec x (\tan x - 1)}{(1 + \tan x)^2}$$ **Answer:** $y' = \frac{\sec x (\tan x - 1)}{(1 + \tan x)^2}$