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If $2^x+2^y=2^{x+y}$, then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$1-2^y$
It is given that,
$2^x+2^y=2^{x+y} \Rightarrow 2^{-y}+2^{-x}=1$
On differentiating both sides w.r.t. ' $x$ ', we get $\Rightarrow\left(-2^{-y} \frac{d y}{d x}-2^{-x}\right) \log 2=0$
$\Rightarrow \frac{d y}{d x}=-\frac{2^{-x}}{2^{-y}}=-\frac{1-2^{-y}}{2^{-y}}=-2^y+1=1-2^y$
$2^x+2^y=2^{x+y} \Rightarrow 2^{-y}+2^{-x}=1$
On differentiating both sides w.r.t. ' $x$ ', we get $\Rightarrow\left(-2^{-y} \frac{d y}{d x}-2^{-x}\right) \log 2=0$
$\Rightarrow \frac{d y}{d x}=-\frac{2^{-x}}{2^{-y}}=-\frac{1-2^{-y}}{2^{-y}}=-2^y+1=1-2^y$
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