A) zero
B) \[-1\]
C) 1
D) 2
Correct Answer: B
Solution :
\[{{2}^{x}}+{{2}^{y}}={{2}^{x+y}}\] Differentiating, we get \[{{2}^{x}}\log 2+{{2}^{y}}\log 2\frac{dy}{dx}={{2}^{x+y}}\log 2\left( 1+\frac{dy}{dx} \right)\] \[\Rightarrow \]\[{{2}^{x}}+{{2}^{y}}\frac{dy}{dx}={{2}^{x+y}}+{{2}^{x+y}}\frac{dy}{dx}\] \[\Rightarrow \]\[({{2}^{y}}-{{2}^{x+y}})\frac{dy}{dx}={{2}^{x+y}}-{{2}^{x}}\] \[\Rightarrow \]\[\frac{dy}{dx}=\frac{{{2}^{x+y}}-{{2}^{x}}}{{{2}^{y}}-{{2}^{x+y}}}\] \[\Rightarrow \]\[\frac{dy}{dx}=\frac{{{2}^{2}}-2}{2-{{2}^{2}}}=\frac{2}{-2}=-1\]
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