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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Self Evaluation Test - Application of Derivatives

  • question_answer
    What is the slope of the tangent to the curve\[y={{\sin }^{-1}}({{\sin }^{2}}x)at\,\,x=0\]?

    A) 0

    B) 1

    C) 2

    D) None of these

    Correct Answer: A

    Solution :

    [a] \[y={{\sin }^{-1}}({{\sin }^{2}}x)\] \[\frac{dy}{dx}=\frac{2\sin x\cos x}{\sqrt{1-{{\sin }^{4}}x}}\Rightarrow \frac{dy}{dx}=\frac{\sin 2x}{\sqrt{1-{{\sin }^{4}}x}}\] \[at\,\,x=0,\frac{dy}{dx}=0\]


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