Question

If $ {{2}^{x}}+{{2}^{y}}={{2}^{x+y}}, $ then $ \frac{dy}{dx} $ is equal to

• 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}}} $

Answer 1

Answer: (C)

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) $