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$\text { If } 2^{x}+2^{y}=2^{x+y} \text {, then } \frac{d y}{d x}=$
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Verified Answer
The correct answer is:
None of these
On differentiating
$\begin{array}{l}
2^{x} \log 2+2^{y} \log 2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \\
=2^{x} \cdot 2^{y} \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \log 2+2^{y} \cdot 2^{x} \log 2 \\
\Rightarrow 2^{x}+2^{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2^{x+y} \frac{\mathrm{dy}}{\mathrm{dx}}+2^{x+y} \\
\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{x+y}-2^{x}}{2^{y}-2^{x+y}} \\
\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{x}+2^{y}-2^{x}}{2^{y}-2^{x}-2^{y}}=-2^{y-x}
\end{array}$
$\begin{array}{l}
2^{x} \log 2+2^{y} \log 2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \\
=2^{x} \cdot 2^{y} \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \log 2+2^{y} \cdot 2^{x} \log 2 \\
\Rightarrow 2^{x}+2^{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2^{x+y} \frac{\mathrm{dy}}{\mathrm{dx}}+2^{x+y} \\
\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{x+y}-2^{x}}{2^{y}-2^{x+y}} \\
\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{x}+2^{y}-2^{x}}{2^{y}-2^{x}-2^{y}}=-2^{y-x}
\end{array}$
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