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