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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2005

  • question_answer
        If\[{{x}^{y}}={{e}^{x-y}}\]then\[\frac{dy}{dx}\]is:

    A)  \[\frac{1+x}{1+\log x}\]                               

    B)  \[\frac{1-\log x}{1+\log x}\]

    C)  not defined      

    D)  \[\frac{logx}{(1+\log x)}\]

    Correct Answer: D

    Solution :

                    \[{{x}^{y}}={{e}^{x-y}}\] Taking the log of both sides \[y\text{ }log\text{ }x=x-y\] \[\Rightarrow \]               \[y=\frac{x}{1+\log x}\]                 \[\frac{dy}{dx}=\frac{(1+\log x)-x\times \frac{1}{x}}{{{(1+\log x)}^{2}}}\]                 \[\frac{dy}{dx}=\frac{\log x}{{{(1+\log x)}^{2}}}\]


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