1. **Problem:** Find the derivative of $y = \frac{\sin x}{\cos x}$. 2. **Formula:** Use the quotient rule: if $y = \frac{u}{v}$, then $$y' = \frac{u'v - uv'}{v^2}$$ where $u = \sin x$ and $v = \cos x$. 3. **Derivatives:** $$u' = \cos x$$ $$v' = -\sin x$$ 4. **Apply quotient rule:** $$y' = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{(\cos x)^2} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$$ 5. **Simplify using Pythagorean identity:** $$\cos^2 x + \sin^2 x = 1$$ 6. **Final derivative:** $$y' = \frac{1}{\cos^2 x} = \sec^2 x$$ **Answer:** The derivative of $y = \frac{\sin x}{\cos x}$ is $y' = \sec^2 x$.