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

  • question_answer
    If \[y=\sec ({{\tan }^{-1}}x),\]then \[\frac{dy}{dx}\] is [DCE 2002; Kurukshetra CEE 2001]

    A)            \[\frac{x}{\sqrt{1+{{x}^{2}}}}\]

    B)            \[\frac{-x}{\sqrt{1+{{x}^{2}}}}\]

    C)            \[\frac{x}{\sqrt{1-{{x}^{2}}}}\]

    D)            None of these

    Correct Answer: A

    Solution :

               \[y=\sec ({{\tan }^{-1}}x)\]\[\therefore \theta =\frac{1}{2}{{\cos }^{-1}}x\]            \[=\frac{x}{1+{{x}^{2}}}\,.\,\sqrt{1+{{x}^{2}}}=\frac{x}{\sqrt{1+{{x}^{2}}}}\], \[({{\tan }^{-1}}x={{\sec }^{-1}}\sqrt{1+{{x}^{2}}})\].


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