A) 0
B) ? 1
C) 1
D) 2
Correct Answer: B
Solution :
\[{{2}^{x}}+{{2}^{y}}={{2}^{x+y}}\]; Differentiating w.r.t. x, we get \[{{2}^{x}}(\log 2)+{{2}^{y}}(\log 2)\frac{dy}{dx}\] =\[{{2}^{(x+y)}}.(\log 2)\,\left( 1+\frac{dy}{dx} \right)\] Þ \[{{2}^{x}}+{{2}^{y}}\frac{dy}{dx}={{2}^{x+y}}+{{2}^{x+y}}\left( \frac{dy}{dx} \right)\] Þ \[\frac{dy}{dx}({{2}^{y}}-{{2}^{x+y}})={{2}^{x+y}}-{{2}^{x}}\]Þ\[\frac{dy}{dx}=\frac{{{2}^{x+y}}-{{2}^{x}}}{{{2}^{y}}-{{2}^{x+y}}}\]. \[\therefore \] \[{{\left( \frac{dy}{dx} \right)}_{x=y=1}}=\frac{{{2}^{2}}-2}{2-{{2}^{2}}}=\frac{2}{-2}=-1.\]
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