NDA Mathematics Question (2026) — Solution

Step-by-step solution

Given the equation (e^y)^x - y = 0, we can rewrite it as: e^ xy = y Taking the natural logarithm on both sides: xy = y Differentiating both sides with respect to x using the product rule and chain rule: y + x dy dx = 1 y dy dx Rearranging the terms to solve for dy dx : y = ( 1 y - x ) dy dx y = ( 1 - xy y ) dy dx dy dx = y^2 1 - xy We are given that at x = 1, y = y_0. Substituting these values into the derivative: dy dx = y_0^2 1 - y_0 From the original equation e^ xy = y, substituting x = 1 and y = y_0 gives: e^ y_0 = y_0 To match the given options, we substitute y_0 = e^ y_0 for one y_0 in the numerator and the y_0 in the denominator: dy dx = y_0 e^ y_0 1 - e^ y_0 Answer: y_0 e^ y_0 1-e^ y_0