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

  • question_answer
    \[\frac{d}{dx}\left( \frac{{{\cot }^{2}}x-1}{{{\cot }^{2}}x+1} \right)=\]

    A)            \[-\sin 2x\]

    B)            \[2\sin 2x\]

    C)            \[2\cos 2x\]

    D)            \[-2\sin 2x\]

    Correct Answer: D

    Solution :

               \[\frac{d}{dx}\left[ \frac{{{\cot }^{2}}x-1}{{{\cot }^{2}}x+1} \right]=\frac{d}{dx}\left[ \frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x+{{\sin }^{2}}x} \right]\]                                     \[=\frac{d}{dx}[\cos 2x]=-2\sin 2x\].


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