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

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

Vedclass · Accepted Answer

${2^{x - y}} \frac{2^y - 1}{1 - 2^x}$

Vedclass · Answer

(B) Given the equation: ${2^x} + {2^y} = {2^{x + y}}$Differentiating both sides with respect to $x$:$\frac{d}{dx}({2^x}) + \frac{d}{dx}({2^y}) = \frac{d}{dx}({2^{x + y}})$Using the chain rule,we get:${2^x} \ln 2 + {2^y} \ln 2 \cdot \frac{dy}{dx} = {2^{x + y}} \ln 2 \cdot (1 + \frac{dy}{dx})$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}$$\frac{dy}{dx} ({2^y} - {2^{x + y}}) = {2^{x + y}} - {2^x}$$\frac{dy}{dx} = \frac{{2^{x + y}} - {2^x}}{{2^y} - {2^{x + y}}}$Since ${2^{x + y}} = {2^x} + {2^y}$,substitute this into the expression:$\frac{dy}{dx} = \frac{({2^x} + {2^y}) - {2^x}}{{2^y} - ({2^x} + {2^y})} = \frac{{2^y}}{{2^y} - {2^x} - {2^y}} = \frac{{2^y}}{-{2^x}} = -{2^{y - x}}$Wait,let's re-evaluate the simplification:$\frac{dy}{dx} = \frac{{2^x}({2^y} - 1)}{{2^y}(1 - {2^x})} = {2^{x - y}} \frac{{2^y} - 1}{1 - {2^x}}$Thus,the correct option is $B$.

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

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