A) \[y(y-x\log \,y)\]
B) \[y(y+x\,\log \,y)\]
C) \[x(x+y\log x)\]
D) \[x(y-x\log y)\]
Correct Answer: A
Solution :
Since, \[{{x}^{y}}={{y}^{x}}\] Taking log on both sides, we get \[y\log x=x\log y\] On differentiating w.r.t. \[x,\]we get \[y.\frac{1}{x}+\log x\frac{dy}{dx}=x.\frac{1}{y}\frac{dy}{dx}+\log y\] \[\Rightarrow \] \[\frac{(x-y\log x)dy}{y}\frac{dy}{dx}=\frac{-x\log y+y}{x}\] \[\Rightarrow \]\[x(x-y\log x)\frac{dy}{dx}=y(-x\log y+y)\]You need to login to perform this action.
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