A) \[\frac{y(x+y\log x)}{x(y+x\log y)}\]
B) \[\frac{y(y+x\log y)}{x(y\log x+y)}\]
C) \[-\frac{y(y+x\log y)}{x(y\log x+x)}\]
D) \[-\frac{y(x+y\log x)}{x(y+x\log y)}\]
Correct Answer: C
Solution :
\[{{x}^{y}}.{{y}^{x}}=1\] Taking log on both sides, \[y\text{ }log\text{ }x+x\text{ }log\text{ }y=log\text{ }1\] \[\Rightarrow \] \[y\text{ }log\text{ }x+x\text{ }log\text{ }y=0\] On differentiating w.r.t.\[x,\]we get \[y.\frac{1}{x}+\log x\frac{dy}{dx}+x.\frac{1}{y}\frac{dy}{dx}+\log y=0\] \[\Rightarrow \] \[\frac{y}{x}+\frac{dy}{dx}\left( \log x+\frac{x}{y} \right)+\log y=0\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{-(\log y+y/x)}{(\log x+x/y)}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{-y(x\log y+y)}{x(y\log x+x)}\]
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