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

  • question_answer
    If \[y={{\cot }^{-1}}({{x}^{2}})\], then \[\frac{dy}{dx}\] is equal to        [Pb. CET 2002]

    A)            \[\frac{2x}{1+{{x}^{4}}}\]

    B)            \[\frac{2x}{\sqrt{1+4x}}\]

    C)            \[\frac{-2x}{1+{{x}^{4}}}\]

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

    Correct Answer: C

    Solution :

               \[y={{\cot }^{-1}}\left( {{x}^{2}} \right)\]                    \[\frac{dy}{dx}=\frac{-1}{1+{{({{x}^{2}})}^{2}}}\frac{d}{dx}({{x}^{2}})=\frac{-1}{1+{{x}^{4}}}(2x)=\frac{-2x}{1+{{x}^{4}}}\].


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