2. Solved Papers
  3. BCECE Engineering

BCECE Engineering BCECE Engineering Solved Paper-2015

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

    A) \[\frac{1}{\sqrt{2}}\]                                     

    B) \[\frac{1}{2}\]

    C) \[1\]                                     

    D) \[\sqrt{2}\]

    Correct Answer: A

    Solution :

    Given \[Y=\sec ({{\tan }^{-1}}x)\] \[=\sec ({{\sec }^{-1}}\sqrt{1+{{x}^{2}}})\] \[\Rightarrow \]               \[y=\sqrt{1+{{x}^{2}}}\] \[\therefore \]  \[\frac{dy}{dx}=\frac{x}{\sqrt{1+{{x}^{2}}}}\] \[\Rightarrow \]  \[{{\left( \frac{dy}{dx} \right)}_{x=1}}=\frac{1}{\sqrt{2}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner