1. **State the problem:** Differentiate the function $$y = e^{2x} \times \sqrt{4x^2 - 1}$$ using the chain rule. 2. **Recall the product rule and chain rule:** - Product rule: $\frac{d}{dx}[u \cdot v] = u'v + uv'$. - Chain rule: $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$. 3. **Identify parts:** Let - $u = e^{2x}$ - $v = \sqrt{4x^2 - 1} = (4x^2 - 1)^{1/2}$ 4. **Differentiate $u$:** $$u' = \frac{d}{dx} e^{2x} = e^{2x} \cdot \frac{d}{dx}(2x) = e^{2x} \times 2 = 2e^{2x}$$ 5. **Differentiate $v$ using chain rule:** $$v = (4x^2 - 1)^{1/2}$$ $$v' = \frac{1}{2}(4x^2 - 1)^{-1/2} \times \frac{d}{dx}(4x^2 - 1)$$ $$= \frac{1}{2}(4x^2 - 1)^{-1/2} \times 8x = \frac{8x}{2\sqrt{4x^2 - 1}} = \frac{4x}{\sqrt{4x^2 - 1}}$$ 6. **Apply product rule:** $$\frac{dy}{dx} = u'v + uv' = 2e^{2x} \sqrt{4x^2 - 1} + e^{2x} \times \frac{4x}{\sqrt{4x^2 - 1}}$$ 7. **Final answer:** $$\boxed{\frac{dy}{dx} = 2e^{2x} \sqrt{4x^2 - 1} + \frac{4xe^{2x}}{\sqrt{4x^2 - 1}}}$$