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

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Question

(A) Given the equation: $2^x + 2^y = 2^{x + y}$.Divide both sides by $2^{x+y}$:$\frac{2^x}{2^{x+y}} + \frac{2^y}{2^{x+y}} = 1$$2^{-y} + 2^{-x} = 1$.Di

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the equation: $2^x + 2^y = 2^{x + y}$.Divide both sides by $2^{x+y}$:$\frac{2^x}{2^{x+y}} + \frac{2^y}{2^{x+y}} = 1$$2^{-y} + 2^{-x} = 1$.Differentiating both sides with respect to $x$:$\frac{d}{dx}(2^{-y}) + \frac{d}{dx}(2^{-x}) = \frac{d}{dx}(1)$Using the chain rule,$\frac{d}{dx}(a^u) = a^u \ln(a) \frac{du}{dx}$:$2^{-y} \ln(2) \cdot (-\frac{dy}{dx}) + 2^{-x} \ln(2) \cdot (-1) = 0$.Divide by $-\ln(2)$:$2^{-y} \frac{dy}{dx} + 2^{-x} = 0$$2^{-y} \frac{dy}{dx} = -2^{-x}$$\frac{dy}{dx} = -\frac{2^{-x}}{2^{-y}} = -\frac{2^y}{2^x}$.Thus,option $A$ is correct.

If $2^x + 2^y = 2^{x + y}$,then $\frac{dy}{dx}$ has the value equal to:

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