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

  • question_answer
    If \[y={{\cot }^{-1}}\left( \frac{1+x}{1-x} \right)\], then \[\frac{dy}{dx}=\]                       [DSSE 1984]

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

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

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

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

    Correct Answer: B

    Solution :

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


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