UPSC NDA 2 2022 Mathematics Solved Paper

Concept:Differentiate using product rule, chain rule, and derivative of logarithmic function.Explanation:Given $y = \frac{x \sqrt{x^2-16}}{2} - 8 \ln\left|x + \sqrt{x^2-16}\right|$.Differentiate each term with respect to $x$.First term: $\frac{1}{2}$ times derivative of $x\sqrt{x^2-16}$.Using product rule: $\frac{d}{dx}[x\sqrt{x^2-16}] = \sqrt{x^2-16} + x \cdot \frac{1}{2\sqrt{x^2-16}} \cdot (2x) = \sqrt{x^2-16} + \frac{x^2}{\sqrt{x^2-16}}$.So first term's derivative = $\frac{1}{2}\left(\sqrt{x^2-16} + \frac{x^2}{\sqrt{x^2-16}}\right)$.Second term: derivative of $-8\ln\left|x+\sqrt{x^2-16}\right|$ is $-8 \cdot \frac{1}{x+\sqrt{x^2-16}} \cdot \left(1 + \frac{x}{\sqrt{x^2-16}}\right)$.Now combine: $\frac{dy}{dx} = \frac{1}{2}\left(\sqrt{x^2-16} + \frac{x^2}{\sqrt{x^2-16}}\right) - \frac{8}{x+\sqrt{x^2-16}}\left(1 + \frac{x}{\sqrt{x^2-16}}\right)$.Simplify the second term: $1 + \frac{x}{\sqrt{x^2-16}} = \frac{\sqrt{x^2-16}+x}{\sqrt{x^2-16}}$, so it becomes $\frac{8}{x+\sqrt{x^2-16}} \cdot \frac{x+\sqrt{x^2-16}}{\sqrt{x^2-16}} = \frac{8}{\sqrt{x^2-16}}$.Now first term: $\frac{1}{2}\left(\frac{x^2-16 + x^2}{\sqrt{x^2-16}}\right) = \frac{1}{2}\left(\frac{2x^2-16}{\sqrt{x^2-16}}\right) = \frac{x^2-8}{\sqrt{x^2-16}}$.Thus $\frac{dy}{dx} = \frac{x^2-8}{\sqrt{x^2-16}} - \frac{8}{\sqrt{x^2-16}} = \frac{x^2-16}{\sqrt{x^2-16}} = \sqrt{x^2-16}$.Answer:$\sqrt{x^2-16}$ (option C).