If {2^x} + {2^y} = {2^{x + y}}, then the valu | vedclass

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

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$-1$

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(B) Given equation: ${2^x} + {2^y} = {2^{x + y}}$.Differentiating both sides with respect to $x$:${2^x} \ln(2) + {2^y} \ln(2) \frac{dy}{dx} = {2^{x + y}} \ln(2) \left( 1 + \frac{dy}{dx} \right)$.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 $\frac{dy}{dx}$:$\frac{dy}{dx} ({2^y} - {2^{x + y}}) = {2^{x + y}} - {2^x}$.Thus,$\frac{dy}{dx} = \frac{{2^{x + y}} - {2^x}}{{2^y} - {2^{x + y}}}$.At $x = 1$ and $y = 1$:$\frac{dy}{dx} = \frac{{2^{1 + 1}} - {2^1}}{{2^1} - {2^{1 + 1}}} = \frac{{2^2} - 2}{2 - {2^2}} = \frac{4 - 2}{2 - 4} = \frac{2}{-2} = -1$.

If ${2^x} + {2^y} = {2^{x + y}},$ then the value of $\frac{dy}{dx}$ at $x = y = 1$ is

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