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JEE Main & Advanced Sample Paper JEE Main Sample Paper-25

  • question_answer
    If \[y={{x}^{2}}+\frac{1}{{{x}^{2}}+\frac{1}{{{x}^{2}}+\frac{1}{{{x}^{2}}+\frac{1}{{{x}^{2}}+......\infty }}}}\] then \[\frac{dy}{dx}\] is equal to

    A)  \[\frac{2xy}{2y-{{x}^{2}}}\]              

    B)  \[\frac{xy}{2y-{{x}^{2}}}\]

    C)  \[\frac{xy}{y-{{x}^{2}}}\]                             

    D)  \[\frac{2xy}{2+\frac{{{x}^{2}}}{y}}\]

    Correct Answer: A

    Solution :

    \[y={{x}^{2}}+\frac{1}{y}\Rightarrow \,\frac{dy}{dx}=\frac{2xy}{2y-{{x}^{2}}}\]


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