UPSC NDA II 2015 Mathematics Paper

Concept:Implicit differentiation using logarithms simplifies the derivative of equations with variable exponents.Explanation:Given $x^{a} y^{b} = (x - y)^{a+b}$.Take natural log on both sides:$a \ln x + b \ln y = (a+b) \ln(x - y)$.Differentiate both sides with respect to $x$:$\frac{a}{x} + \frac{b}{y} \cdot \frac{dy}{dx} = \frac{a+b}{x-y} \left(1 - \frac{dy}{dx}\right)$.Bring terms with $\frac{dy}{dx}$ together:$\frac{b}{y} \cdot \frac{dy}{dx} + \frac{a+b}{x-y} \cdot \frac{dy}{dx} = \frac{a+b}{x-y} - \frac{a}{x}$.Factor $\frac{dy}{dx}$: $\frac{dy}{dx} \left( \frac{b}{y} + \frac{a+b}{x-y} \right) = \frac{a+b}{x-y} - \frac{a}{x}$.Combine fractions on both sides using common denominator:Left: $\frac{dy}{dx} \cdot \frac{b(x-y) + y(a+b)}{y(x-y)} = \frac{dy}{dx} \cdot \frac{bx - by + ay + by}{y(x-y)} = \frac{dy}{dx} \cdot \frac{bx + ay}{y(x-y)}$.Right: $\frac{(a+b)x - a(x-y)}{x(x-y)} = \frac{ax + bx - ax + ay}{x(x-y)} = \frac{bx + ay}{x(x-y)}$.Thus, $\frac{dy}{dx} \cdot \frac{bx+ay}{y(x-y)} = \frac{bx+ay}{x(x-y)}$.Cancel the common factor $(bx+ay)$ and $(x-y)$ (assuming they are non‑zero):$\frac{dy}{dx} \cdot \frac{1}{y} = \frac{1}{x}$.Therefore, $\frac{dy}{dx} = \frac{y}{x}$.Now compute the required expression:$\frac{dy}{dx} - \frac{y}{x} = \frac{y}{x} - \frac{y}{x} = 0$.Answer:Option D: $0$.