If e^x+e^y=e^{x+y},then frac{dy}{dx}= | vedclass

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(A) Given the equation: $e^x + e^y = e^{x+y}$ Differentiating both sides with respect to $x$: $\frac{d}{dx}(e^x) + \frac{d}{dx}(e^y) = \frac{d}{dx}(e^

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$-e^{y-x}$

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(A) Given the equation: $e^x + e^y = e^{x+y}$ Differentiating both sides with respect to $x$: $\frac{d}{dx}(e^x) + \frac{d}{dx}(e^y) = \frac{d}{dx}(e^{x+y})$ $e^x + e^y \cdot \frac{dy}{dx} = e^{x+y} \cdot (1 + \frac{dy}{dx})$ $e^x + e^y \cdot \frac{dy}{dx} = e^{x+y} + e^{x+y} \cdot \frac{dy}{dx}$ Rearranging the terms to isolate $\frac{dy}{dx}$: $e^y \cdot \frac{dy}{dx} - e^{x+y} \cdot \frac{dy}{dx} = e^{x+y} - e^x$ $\frac{dy}{dx} (e^y - e^{x+y}) = e^{x+y} - e^x$ Since $e^{x+y} = e^x + e^y$,substitute this into the equation: $\frac{dy}{dx} (e^y - (e^x + e^y)) = (e^x + e^y) - e^x$ $\frac{dy}{dx} (e^y - e^x - e^y) = e^y$ $\frac{dy}{dx} (-e^x) = e^y$ $\frac{dy}{dx} = -\frac{e^y}{e^x} = -e^{y-x}$

If $e^x+e^y=e^{x+y}$,then $\frac{dy}{dx}=$

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