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

  • question_answer
    \[\frac{d}{dx}{{\sin }^{-1}}(3x-4{{x}^{3}})=\]                                                     [RPET 2003]

    A)            \[\frac{3}{\sqrt{1-{{x}^{2}}}}\]

    B)            \[\frac{-3}{\sqrt{1-{{x}^{2}}}}\]

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

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

    Correct Answer: A

    Solution :

               Put \[x=\sin \theta ,\] we get \[\frac{d}{dx}{{\sin }^{-1}}(3x-4{{x}^{3}})\]                            \[=\frac{d}{dx}{{\sin }^{-1}}(\sin 3\theta )=\frac{3}{\sqrt{1-{{x}^{2}}}}\].


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