UPSC NDA I and II 2020 Math Solved Paper

Concept:Use implicit differentiation: differentiate both sides with respect to $x$, treating $y$ as a function of $x$. Apply the product rule and chain rule.Explanation:Given: $x^{m} y^{n} = a^{m+n}$ where $a$ is a constant.Differentiate both sides with respect to $x$: $\frac{d}{dx}(x^{m} y^{n}) = \frac{d}{dx}(a^{m+n})$.Since $a$ is constant, the right-hand side is $0$.Apply the product rule: $\frac{d}{dx}(x^{m}) \cdot y^{n} + x^{m} \cdot \frac{d}{dx}(y^{n}) = 0$.Now compute the derivatives: $\frac{d}{dx}(x^{m}) = m x^{m-1}$ and $\frac{d}{dx}(y^{n}) = n y^{n-1} \frac{dy}{dx}$ (using chain rule).Substitute: $(m x^{m-1}) y^{n} + x^{m} (n y^{n-1} \frac{dy}{dx}) = 0$.Simplify: $m x^{m-1} y^{n} + n x^{m} y^{n-1} \frac{dy}{dx} = 0$.Isolate the term with $\frac{dy}{dx}$: $n x^{m} y^{n-1} \frac{dy}{dx} = - m x^{m-1} y^{n}$.Divide both sides by $n x^{m} y^{n-1}$ (assuming $x,y \neq 0$):$\frac{dy}{dx} = \frac{- m x^{m-1} y^{n}}{n x^{m} y^{n-1}} = - \frac{m}{n} \cdot \frac{x^{m-1}}{x^{m}} \cdot \frac{y^{n}}{y^{n-1}} = - \frac{m}{n} \cdot \frac{1}{x} \cdot y = - \frac{my}{nx}$.Answer:$\frac{dy}{dx} = -\frac{my}{nx}$, which corresponds to option B.