A) \[\frac{y(y+x\log y)}{x(y\log x+x)}\]
B) \[\frac{y(x+y\log x)}{x(y+x\log y)}\]
C) \[\frac{-y(y+x\log y)}{x(x+y\log x)}\]
D) None of these
Correct Answer: C
Solution :
We have \[{{x}^{y}}\cdot {{y}^{x}}=1\] On taking log on both sides, we get \[y\log x+x\log y=0\] On differentiating w.r.t. x, we get \[\frac{y}{x}+\log x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}+\log y=0\] \[\Rightarrow \] \[\frac{dy}{dx}\left( \log x+\frac{x}{y} \right)=-\left( \log y+\frac{y}{x} \right)\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{-y(x\log y+y)}{x(y\log x+x)}\]
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