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JEE Main & Advanced Mathematics Differentiation Question Bank Self Evaluation Test - Limits and Derivatives

  • question_answer
    If \[f(x)={{\left( \frac{{{\sin }^{m}}x}{{{\sin }^{n}}x} \right)}^{m+n}}.{{\left( \frac{{{\sin }^{n}}x}{{{\sin }^{p}}x} \right)}^{n+p}}.{{\left( \frac{{{\sin }^{p}}\,x}{{{\sin }^{m}}x} \right)}^{p+m}}\] Then \[f'(x)\]is equal to

    A) 0

    B) 1

    C) \[{{\cos }^{m+n+px}}\]

    D) None of these

    Correct Answer: A

    Solution :

    [a] we have, \[f(x)={{(si{{n}^{m-n}}x)}^{m+n}}.{{(si{{n}^{n-p}}x)}^{n+p}}.{{(si{{n}^{p-m}}x)}^{p+m}}\] \[={{\sin }^{{{m}^{2}}-{{n}^{2}}}}x.{{\sin }^{{{n}^{2}}-{{p}^{2}}}}x.{{\sin }^{{{p}^{2}}-{{m}^{2}}}}x\] \[={{(sinx)}^{{{m}^{2}}-{{n}^{2}}+{{n}^{2}}-{{p}^{2}}+{{p}^{2}}-{{m}^{2}}}}={{(sinx)}^{0}}=1.\] \[\therefore f'(x)=0.\]


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