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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2011

  • question_answer
        If\[x={{e}^{y+{{e}^{y+......\infty }}}},x>0,\]then\[\frac{dy}{dx}\]is equal to

    A)  \[\frac{1-x}{x}\]                              

    B)  \[\frac{1}{x}\]

    C)  \[\frac{x}{1+x}\]                             

    D)  \[\frac{1+x}{x}\]

    Correct Answer: A

    Solution :

                    Given, \[x={{e}^{y+x}}\] \[\Rightarrow \]          \[log\text{ }x=x+y\] On differentiating w.r.t.\[x,\]we get                 \[\frac{1}{x}=1+\frac{dy}{dx}\] \[\Rightarrow \]               \[\frac{dy}{dx}=\frac{1-x}{x}\]


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