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JEE Main & Advanced Mathematics Differentiation Question Bank Differentiation by Substitution

  • question_answer
    The derivative of \[{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\,\]w.r.t. \[{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\] is    [Karnataka CET 2000; Pb. CET 2004]

    A)            ?1

    B)            1

    C)            2

    D)            4

    Correct Answer: B

    Solution :

               Let \[p={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}=2{{\tan }^{-1}}x\]                     and \[q={{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}}=2{{\tan }^{-1}}x\]; \ \[\frac{dp}{dq}=\frac{dp/dx}{dq/dx}=1\].


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