A) 0
B) \[\frac{1}{2}\]
C) 1
D) 2
Correct Answer: A
Solution :
Given, \[{{x}^{y}}={{e}^{x-y}}\] Taking log on both sides, we get \[y\text{ }log\text{ }x=x-y\] \[\Rightarrow\] \[\frac{y}{x}+\log x\frac{dy}{dx}=1-\frac{dy}{dx}\] \[\Rightarrow\] \[\frac{dy}{dx}(1+\log x)=1-\frac{y}{x}\] \[\Rightarrow\] \[{{\left( \frac{dy}{dx} \right)}_{(1,1)}}(1+\log 1)=1-\frac{1}{1}\] \[\Rightarrow\] \[{{\left( \frac{dy}{dx} \right)}_{(1,1)}}=0\]
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