2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Differentiation

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

  • question_answer
    If \[y=(1+{{x}^{2}}){{\tan }^{-1}}x-x,\]then \[\frac{dy}{dx}=\]             [Karnataka CET 2001]

    A)            \[{{\tan }^{-1}}x\]

    B)            \[2x{{\tan }^{-1}}x\]

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

    D)            \[\frac{2x}{{{\tan }^{-1}}x}\]

    Correct Answer: B

    Solution :

               \[y=(1+{{x}^{2}}){{\tan }^{-1}}x-x\]                                 Þ \[\frac{dy}{dx}=(1+{{x}^{2}}).\frac{1}{(1+{{x}^{2}})}+{{\tan }^{-1}}x(2x)-1=2x{{\tan }^{-1}}x.\]


You need to login to perform this action.
You will be redirected in 3 sec spinner