1. **State the problem:** We are given the function $$y = (x^2 + 1)\sqrt{2x - 3}$$. (i) Show that $$\frac{dy}{dx} = \frac{Px^2 + Qx + 1}{\sqrt{2x - 3}}$$ where $P$ and $Q$ are integers. (ii) Find the equation of the normal to the curve at $x=2$ in the form $ax + by + c = 0$. 2. **Find the derivative $\frac{dy}{dx}$:** Use the product rule: $$\frac{d}{dx}[u v] = u' v + u v'$$ where $$u = x^2 + 1$$ and $$v = (2x - 3)^{1/2}$$. Calculate derivatives: $$u' = 2x$$ $$v' = \frac{1}{2}(2x - 3)^{-1/2} \cdot 2 = \frac{1}{(2x - 3)^{1/2}}$$ 3. **Apply product rule:** $$\frac{dy}{dx} = u' v + u v' = 2x (2x - 3)^{1/2} + (x^2 + 1) \frac{1}{(2x - 3)^{1/2}}$$ 4. **Combine terms over common denominator:** Rewrite first term with denominator: $$2x (2x - 3)^{1/2} = \frac{2x (2x - 3)}{(2x - 3)^{1/2}}$$ So, $$\frac{dy}{dx} = \frac{2x (2x - 3)}{(2x - 3)^{1/2}} + \frac{x^2 + 1}{(2x - 3)^{1/2}} = \frac{2x (2x - 3) + x^2 + 1}{(2x - 3)^{1/2}}$$ 5. **Simplify numerator:** $$2x (2x - 3) + x^2 + 1 = 4x^2 - 6x + x^2 + 1 = 5x^2 - 6x + 1$$ 6. **Final derivative form:** $$\frac{dy}{dx} = \frac{5x^2 - 6x + 1}{\sqrt{2x - 3}}$$ Here, $P = 5$ and $Q = -6$. 7. **Find slope of tangent at $x=2$:** Calculate denominator: $$\sqrt{2(2) - 3} = \sqrt{4 - 3} = 1$$ Calculate numerator: $$5(2)^2 - 6(2) + 1 = 5(4) - 12 + 1 = 20 - 12 + 1 = 9$$ So, $$\frac{dy}{dx}\bigg|_{x=2} = \frac{9}{1} = 9$$ 8. **Find $y$ at $x=2$:** $$y = (2^2 + 1) \sqrt{2(2) - 3} = (4 + 1)(1) = 5$$ 9. **Equation of normal line:** Slope of tangent $m = 9$, so slope of normal $m_n = -\frac{1}{9}$. Using point-slope form: $$y - 5 = -\frac{1}{9}(x - 2)$$ Multiply both sides by 9: $$9(y - 5) = -(x - 2)$$ $$9y - 45 = -x + 2$$ Rearranged: $$x + 9y - 47 = 0$$ **Final answers:** (i) $$\frac{dy}{dx} = \frac{5x^2 - 6x + 1}{\sqrt{2x - 3}}$$ (ii) Equation of normal at $x=2$ is $$x + 9y - 47 = 0$$.