A) \[\frac{({{2}^{x}}+{{2}^{y}})}{({{2}^{x}}-{{2}^{y}})}\]
B) \[\frac{({{2}^{x}}+{{2}^{y}})}{(1+{{2}^{x+y}})}\]
C) \[{{2}^{x-y}}\left( \frac{{{2}^{y}}-1}{1-{{2}^{x}}} \right)\]
D) \[\frac{{{2}^{x+y}}-{{2}^{x}}}{{{2}^{y}}}\]
Correct Answer: C
Solution :
Given\[{{2}^{x}}+{{2}^{y}}={{2}^{x+y}}\] On differentiating w.r.t.\[x,\]we get \[{{2}^{x}}{{\log }_{e}}2+{{2}^{y}}{{\log }_{e}}\frac{dy}{dx}\] \[={{2}^{x+y}}{{\log }_{e}}2\left( 1+\frac{dy}{dx} \right)\] \[\Rightarrow \] \[\frac{dy}{dx}({{2}^{y}}{{\log }_{e}}2-{{2}^{x+y}}{{\log }_{e}}2)\] \[={{2}^{x+y}}{{\log }_{e}}2(1-{{2}^{x}})\] \[\Rightarrow \] \[\frac{dy}{dx}=[{{2}^{y}}{{\log }_{e}}2(1-{{2}^{x}})]\] \[={{2}^{x}}{{\log }_{e}}2[{{2}^{y}}-1]\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{{{2}^{x}}}{{{2}^{y}}}\frac{{{2}^{y}}-1}{1-{{2}^{x}}}={{2}^{x-y}}\left( \frac{{{2}^{y}}-1}{1-{{2}^{x}}} \right)\]
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