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BCECE Engineering BCECE Engineering Solved Paper-2008

  • question_answer
    If \[{{x}^{y}}={{y}^{x}},\]then \[x(x-y\log x)\frac{dy}{dx}\]is equal to

    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 \[y.\frac{1}{x}+\log x\frac{dy}{dx}=x.\frac{1}{y}\frac{dy}{dx}+\log y\] \[\Rightarrow \]               \[\frac{(x-y\log x)dy}{y}\frac{dy}{dx}=\frac{-x\log y+y}{x}\]                 \[\Rightarrow \]\[x(x-y\log x)\frac{dy}{dx}=y(-x\log y+y)\]


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