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BCECE Engineering BCECE Engineering Solved Paper-2002

  • question_answer
    If\[y={{\sin }^{-1}}(\cos x),\] then \[\frac{dy}{dx}\]is equal to:

    A) \[\frac{1}{\sin x}\]          

    B)         \[{{\cos }^{-1}}x\]          

    C)         \[-1\]                                   

    D)  \[\frac{1}{2}\]

    Correct Answer: C

    Solution :

    \[y={{\sin }^{-1}}(\cos x)\] \[={{\sin }^{-1}}\left( \sin \left( \frac{\pi }{2}-x \right) \right)\] \[\Rightarrow \]               \[y=\frac{\pi }{2}-x\] On differentiating both sides, w.r.t. \[x,\] we get \[\frac{dy}{dx}=-1\]


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