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

  • question_answer
    \[\frac{d}{dx}\sqrt{{{\sec }^{2}}x+\text{cose}{{\text{c}}^{2}}x}=\]                                [DSSE 1981]

    A)            \[4\cos \text{ec 2}x.\cot 2x\]

    B)            \[-4\cos \text{ec 2}x.\cot 2x\]

    C)            \[-4\cos \text{ec }x.\cot 2x\]

    D)            None of these

    Correct Answer: B

    Solution :

               \[\frac{d}{dx}[\sqrt{{{\sec }^{2}}x+\text{cose}{{\text{c}}^{2}}x}]=\frac{d}{dx}\left[ \sqrt{\left( \frac{1}{{{\cos }^{2}}x}+\frac{1}{{{\sin }^{2}}x} \right)} \right]\] \[=\frac{d}{dx}[2\,\text{cosec}2x]=-4\,\text{cosec}2x\cot 2x\].


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