If 2^{x}+2^{y}=2^{x+y},then frac{dy}{dx} is | vedclass

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(B) Given equation is $2^{x}+2^{y}=2^{x+y} \quad ...(i)$ Differentiating both sides with respect to $x$: $\frac{d}{dx}(2^{x}) + \frac{d}{dx}(2^{y}) =

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$-2^{y-x}$

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(B) Given equation is $2^{x}+2^{y}=2^{x+y} \quad ...(i)$ Differentiating both sides with respect to $x$: $\frac{d}{dx}(2^{x}) + \frac{d}{dx}(2^{y}) = \frac{d}{dx}(2^{x+y})$ $2^{x} \ln 2 + 2^{y} \ln 2 \cdot \frac{dy}{dx} = 2^{x+y} \ln 2 \cdot (1 + \frac{dy}{dx})$ Dividing by $\ln 2$: $2^{x} + 2^{y} \frac{dy}{dx} = 2^{x+y} + 2^{x+y} \frac{dy}{dx}$ Rearranging the terms to isolate $\frac{dy}{dx}$: $2^{y} \frac{dy}{dx} - 2^{x+y} \frac{dy}{dx} = 2^{x+y} - 2^{x}$ $\frac{dy}{dx} (2^{y} - 2^{x+y}) = 2^{x+y} - 2^{x}$ From equation $(i)$,we know $2^{x+y} = 2^{x} + 2^{y}$. Substituting this: $\frac{dy}{dx} (2^{y} - (2^{x} + 2^{y})) = (2^{x} + 2^{y}) - 2^{x}$ $\frac{dy}{dx} (-2^{x}) = 2^{y}$ $\frac{dy}{dx} = -\frac{2^{y}}{2^{x}} = -2^{y-x}$

If $2^{x}+2^{y}=2^{x+y}$,then $\frac{dy}{dx}$ is

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