1. **State the problem:** Given the parametric equations $$x = \tan^2(2t)$$ and $$y = \cos(2t)$$ for $$0 < t < \frac{\pi}{4}$$, show that $$\frac{dy}{dx} = -\frac{1}{2} \cos^3(2t)$$. 2. **Recall the formula for parametric derivatives:** $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ This means we need to find $$\frac{dy}{dt}$$ and $$\frac{dx}{dt}$$ first. 3. **Calculate $$\frac{dy}{dt}$$:** Given $$y = \cos(2t)$$, $$\frac{dy}{dt} = -\sin(2t) \cdot 2 = -2 \sin(2t)$$. 4. **Calculate $$\frac{dx}{dt}$$:** Given $$x = \tan^2(2t)$$, use the chain rule: $$\frac{dx}{dt} = 2 \tan(2t) \cdot \sec^2(2t) \cdot 2 = 4 \tan(2t) \sec^2(2t)$$. 5. **Form the derivative $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = \frac{-2 \sin(2t)}{4 \tan(2t) \sec^2(2t)} = \frac{-2 \sin(2t)}{4 \frac{\sin(2t)}{\cos(2t)} \frac{1}{\cos^2(2t)}}$$. 6. **Simplify the expression:** Cancel $$\sin(2t)$$ in numerator and denominator: $$\frac{dy}{dx} = \frac{-2 \cancel{\sin(2t)}}{4 \frac{\cancel{\sin(2t)}}{\cos(2t)} \frac{1}{\cos^2(2t)}} = \frac{-2}{4 \frac{1}{\cos(2t)} \frac{1}{\cos^2(2t)}} = \frac{-2}{4 \frac{1}{\cos^3(2t)}}$$. 7. **Rewrite and simplify further:** $$\frac{dy}{dx} = -2 \times \frac{\cos^3(2t)}{4} = -\frac{1}{2} \cos^3(2t)$$. **Final answer:** $$\boxed{\frac{dy}{dx} = -\frac{1}{2} \cos^3(2t)}$$ This completes the proof.