If 2^x+2^y=2^{x+y}, then frac{d y}{d x}= | vedclass

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(A) Given equation is $2^x+2^y=2^{x+y}$.Dividing both sides by $2^{x+y}$, we get $\frac{2^x}{2^{x+y}}+\frac{2^y}{2^{x+y}}=1$.This simplifies to $2^{-y

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$1-2^y$

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(A) Given equation is $2^x+2^y=2^{x+y}$.Dividing both sides by $2^{x+y}$, we get $\frac{2^x}{2^{x+y}}+\frac{2^y}{2^{x+y}}=1$.This simplifies to $2^{-y}+2^{-x}=1$.Differentiating both sides with respect to $x$, we get $\frac{d}{dx}(2^{-y})+\frac{d}{dx}(2^{-x})=\frac{d}{dx}(1)$.Using the chain rule, we have $2^{-y} \cdot \log 2 \cdot (-\frac{dy}{dx}) + 2^{-x} \cdot \log 2 \cdot (-1) = 0$.Dividing by $-\log 2$, we get $2^{-y} \frac{dy}{dx} + 2^{-x} = 0$.Thus, $\frac{dy}{dx} = -\frac{2^{-x}}{2^{-y}} = -2^{y-x}$.From the original equation $2^x+2^y=2^{x+y}$, we have $2^x = 2^{x+y}-2^y = 2^y(2^x-1)$, so $2^{-x} = \frac{1}{2^y-1}$ is not direct. Let's use $2^{-x} = 1-2^{-y}$.Then $\frac{dy}{dx} = -\frac{1-2^{-y}}{2^{-y}} = -(\frac{1}{2^{-y}}-1) = -(2^y-1) = 1-2^y$.

If $2^x+2^y=2^{x+y}$, then $\frac{d y}{d x}=$

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