CUET UG 2025 — Mathematics Calculus

Given $e^x + e^y = e^{x+y}$ex+ey=ex+y

Differentiating both sides with respect to $x$x:

$$e^x + e^y \cdot \dfrac{dy}{dx} = e^{x+y} \cdot \left(1 + \dfrac{dy}{dx}\right)$$ex+ey⋅dxdy​=ex+y⋅(1+dxdy​)

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$$e^x + e^y \cdot \dfrac{dy}{dx} = e^{x+y} + e^{x+y} \cdot \dfrac{dy}{dx}$$ex+ey⋅dxdy​=ex+y+ex+y⋅dxdy​

$$e^y \cdot \dfrac{dy}{dx} - e^{x+y} \cdot \dfrac{dy}{dx} = e^{x+y} - e^x$$ey⋅dxdy​−ex+y⋅dxdy​=ex+y−ex

$$\dfrac{dy}{dx}\left(e^y - e^{x+y}\right) = e^{x+y} - e^x$$dxdy​(ey−ex+y)=ex+y−ex

$$\dfrac{dy}{dx} = \dfrac{e^{x+y} - e^x}{e^y - e^{x+y}}$$dxdy​=ey−ex+yex+y−ex​

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$$\dfrac{dy}{dx} = \dfrac{e^x(e^y - 1)}{-e^y(e^x - 1)}$$dxdy​=−ey(ex−1)ex(ey−1)​

From the original equation $e^x + e^y = e^x \cdot e^y$ex+ey=ex⋅ey:

$$e^y(e^x - 1) = e^x \quad \cdots (i)$$ey(ex−1)=ex⋯(i)

$$e^x(e^y - 1) = e^y \quad \cdots (ii)$$ex(ey−1)=ey⋯(ii)

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Substituting $(i)$(i) and $(ii)$(ii):

$$\dfrac{dy}{dx} = \dfrac{e^y}{-e^x}$$dxdy​=−exey​

$$= -e^{\,y-x}$$=−ey−x

$$\boxed{\dfrac{dy}{dx} = -e^{\,y-x}}$$dxdy​=−ey−x​