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If $2^{x}+2^{y}=2^{x+y}$, then $\frac{d y}{d x}$ is
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Verified Answer
The correct answer is:
$-2^{y-x}$
We have,
$2^{x}+2^{y}=2^{x+y}...(i)$
On differentiating Eq. (i) w.r.t. $x$, we get
$2^{x} \log 2+2^{y} \log 2 \frac{d y}{d x}$
$=2^{x+y} \log 2\left(1+\frac{d y}{d x}\right)$
$\Rightarrow \quad 2^{x}+2^{y} \frac{d y}{d x}=2^{x+y}\left(1+\frac{d y}{d x}\right)$
$\Rightarrow \quad 2^{x}-2^{x+y}=\frac{d y}{d x}\left(2^{x+y}-2^{y}\right)$
$\Rightarrow \quad-2^{y}=\frac{d y}{d x}\left(2^{x}\right)$
$\Rightarrow \quad \frac{d y}{d x}=-2^{y} / 2^{x}=-2^{y-x}$
$2^{x}+2^{y}=2^{x+y}...(i)$
On differentiating Eq. (i) w.r.t. $x$, we get
$2^{x} \log 2+2^{y} \log 2 \frac{d y}{d x}$
$=2^{x+y} \log 2\left(1+\frac{d y}{d x}\right)$
$\Rightarrow \quad 2^{x}+2^{y} \frac{d y}{d x}=2^{x+y}\left(1+\frac{d y}{d x}\right)$
$\Rightarrow \quad 2^{x}-2^{x+y}=\frac{d y}{d x}\left(2^{x+y}-2^{y}\right)$
$\Rightarrow \quad-2^{y}=\frac{d y}{d x}\left(2^{x}\right)$
$\Rightarrow \quad \frac{d y}{d x}=-2^{y} / 2^{x}=-2^{y-x}$
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