Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Continuity and Differentiability

Question:

If $xy = e^{(x-y)}$, then $\frac{dy}{dx}$ is equal to:

Options:

$\frac{e^{x-y}+y}{x+e^{x-y}}$

$\frac{e^{x-y}-y}{x+e^{x-y}}$

$\frac{e^{x-y}-y}{x-e^{x-y}}$

$\frac{e^{x-y}+y}{x-e^{x-y}}$

Correct Answer:

$\frac{e^{x-y}-y}{x+e^{x-y}}$

Explanation:

The correct answer is Option (2) → $\frac{e^{x-y}-y}{x+e^{x-y}}$

Given equation: $xy = e^{(x - y)}$

Differentiate both sides with respect to $x$:

$y + x\frac{dy}{dx} = e^{(x - y)}(1 - \frac{dy}{dx})$

Rearrange terms:

$y + x\frac{dy}{dx} = e^{(x - y)} - e^{(x - y)}\frac{dy}{dx}$

Combine $\frac{dy}{dx}$ terms:

$(x + e^{(x - y)})\frac{dy}{dx} = e^{(x - y)} - y$

Hence,

$\frac{dy}{dx} = \frac{e^{(x - y)} - y}{x + e^{(x - y)}}$

Final answer: $\frac{dy}{dx} = \frac{e^{(x - y)} - y}{x + e^{(x - y)}}$