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JEE Main & Advanced Mathematics Differentiation Question Bank Derivative at a point Standard Differentiation

  • question_answer
    If \[f(x)={{\cos }^{-1}}\left[ \frac{1-{{(\log x)}^{2}}}{1+{{(\log x)}^{2}}} \right]\,,\]then the value of \[f'(e)=\]            [Karnataka CET 1999; Pb. CET 2000]

    A)            1

    B)            1/e

    C)            2/e

    D)            \[\frac{2}{{{e}^{2}}}\]

    Correct Answer: B

    Solution :

               \[f(x)={{\cos }^{-1}}\left[ \frac{1-{{(\log x)}^{2}}}{1+{{(\log x)}^{2}}} \right]\]\[=2{{\tan }^{-1}}(\log x)\]            Þ  \[{f}'(x)=2.\frac{1}{1+{{(\log x)}^{2}}}.\frac{1}{x}.\text{Therefore }{f}'(e)=\frac{1}{e}\].


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