Q231 / 270 · NDA 2026 (Phase 1) · Mathematics
Passage: Let $\left(e^y\right)^x-y=0$, where $y$ is a function of $x$ whose domain is $(0,10]$.
Question: What is $\dfrac{dy}{dx}$ equal to ?
Question: What is $\dfrac{dy}{dx}$ equal to ?
- A$\dfrac{y}{1-xy}$
- B$\dfrac{y}{1+xy}$
- Correct answerC$\dfrac{y^2}{1-xy}$
- D$\dfrac{y^2}{1+xy}$
Solution
Given equation is $(e^y)^x - y = 0$
$e^{xy} = y$
Taking natural logarithm on both sides:
$xy = \ln(y)$
Differentiating both sides with respect to $x$:
$y + x \dfrac{dy}{dx} = \dfrac{1}{y} \dfrac{dy}{dx}$
$y = \left( \dfrac{1}{y} - x \right) \dfrac{dy}{dx}$
$y = \left( \dfrac{1 - xy}{y} \right) \dfrac{dy}{dx}$
$\dfrac{dy}{dx} = \dfrac{y^2}{1 - xy}$
$e^{xy} = y$
Taking natural logarithm on both sides:
$xy = \ln(y)$
Differentiating both sides with respect to $x$:
$y + x \dfrac{dy}{dx} = \dfrac{1}{y} \dfrac{dy}{dx}$
$y = \left( \dfrac{1}{y} - x \right) \dfrac{dy}{dx}$
$y = \left( \dfrac{1 - xy}{y} \right) \dfrac{dy}{dx}$
$\dfrac{dy}{dx} = \dfrac{y^2}{1 - xy}$
Hence, the correct answer is option: 'C'.