CUET UG 2025 — Mathematics Calculus

Given the implicit equation:

e^x + e^y = e^{x+y}$ 1. **Differentiate both sides with respect to $x$:**\frac{d}{dx}(e^x) + \frac{d}{dx}(e^y) = \frac{d}{dx}(e^{x+y})$$$$e^x + e^y\frac{dy}{dx} = e^{x+y}\left(1 + \frac{dy}{dx}\right)$$2. **Expand the right-hand side:**$$2.∗∗Expandtheright−handside:∗∗e^x + e^y\frac{dy}{dx} = e^{x+y} + e^{x+y}\frac{dy}{dx}$

1. Substitute $e^{x+y} = e^x + e^y$ex+y=ex+ey back into the right side to simplify:

$$e^x + e^y\frac{dy}{dx} = (e^x + e^y) + (e^x + e^y)\frac{dy}{dx}$$ex+eydxdy​=(ex+ey)+(ex+ey)dxdy​

e^x + e^y\frac{dy}{dx} = e^x + e^y + e^x\frac{dy}{dx} + e^y\frac{dy}{dx}$ 4. **Cancel out identical terms on both sides:** Subtract $e^x$ and $e^y\frac{dy}{dx}$ from both sides:0 = e^y + e^x\frac{dy}{dx}$

1. Isolate $\frac{dy}{dx}$dxdy​:

$$e^x\frac{dy}{dx} = -e^y$$exdxdy​=−ey

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