CUET UG 2025 — Mathematics Calculus

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

Using implicit differentiation with respect to $x$x, treating $y$y as a function of $x$x.

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The left side, using the product rule:

$$\frac{d}{dx}(xy) = x \cdot \frac{dy}{dx} + y \cdot 1$$dxd​(xy)=x⋅dxdy​+y⋅1

$$\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$$dxd​(xy)=xdxdy​+y

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The right side, using the chain rule:

$$\frac{d}{dx}(e^{(x-y)}) = e^{(x-y)} \cdot \frac{d}{dx}(x-y)$$dxd​(e(x−y))=e(x−y)⋅dxd​(x−y)

$$\frac{d}{dx}(e^{(x-y)}) = e^{(x-y)} \cdot (1 - \frac{dy}{dx})$$dxd​(e(x−y))=e(x−y)⋅(1−dxdy​)

$$\frac{d}{dx}(e^{(x-y)}) = e^{(x-y)} - e^{(x-y)} \cdot \frac{dy}{dx}$$dxd​(e(x−y))=e(x−y)−e(x−y)⋅dxdy​

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Setting both sides equal:

$$x\frac{dy}{dx} + y = e^{(x-y)} - e^{(x-y)} \cdot \frac{dy}{dx}$$xdxdy​+y=e(x−y)−e(x−y)⋅dxdy​

Collecting all terms with $\frac{dy}{dx}$dxdy​ on one side:

$$x\frac{dy}{dx} + e^{(x-y)} \cdot \frac{dy}{dx} = e^{(x-y)} - y$$xdxdy​+e(x−y)⋅dxdy​=e(x−y)−y

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Factoring out $\frac{dy}{dx}$dxdy​:

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

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