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

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2 x + y

- 2^{y - x}

2 x y

2^{x - y}

answer is B.

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Detailed Solution

We have

2^{x} + 2^{y} = 2^{x + y}

divide with $2^{x + y}$ on both sides

$\frac{2^{x}}{2^{x + y}} + \frac{2^{y}}{2^{x + y}} = \frac{2^{x + y}}{2^{x + y}}$

$2^{- x} + 2^{- y} = 1$

derivative with respective of x

-log2. $2^{- x}$ + $2^{- y}$ .log2 $\frac{d y}{d x}$ =0

$\frac{d y}{d x}$ $= - 2^{- x - (- y)}$

$= - 2^{y - x}$

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