2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Applications of Derivatives

JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    If \[y=\frac{x}{2}\sqrt{{{a}^{2}}+{{x}^{2}}}+\frac{{{a}^{2}}}{2}\log (x+\sqrt{{{x}^{2}}+{{a}^{2}}})\],then \[\frac{dy}{dx}=\]  [AISSE 1983]

    A) \[\sqrt{{{x}^{2}}+{{a}^{2}}}\]

    B) \[\frac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}\]

    C) \[2\sqrt{{{x}^{2}}+{{a}^{2}}}\]

    D) \[\frac{2}{\sqrt{{{x}^{2}}+{{a}^{2}}}}\]

    Correct Answer: A

    Solution :

    • \[y=\frac{x}{2}\sqrt{{{a}^{2}}+{{x}^{2}}}+\frac{{{a}^{2}}}{2}\log (x+\sqrt{{{x}^{2}}+{{a}^{2}}})\]                   
    • Þ \[\frac{dy}{dx}=\frac{1}{2}\left[ \sqrt{{{a}^{2}}+{{x}^{2}}}+x\frac{1}{2}{{({{a}^{2}}+{{x}^{2}})}^{-1/2}}2x \right]\]           
    • \[+\frac{{{a}^{2}}}{2}\frac{1}{(x+\sqrt{({{x}^{2}}+{{a}^{2}})}}\left[ 1+\frac{1}{2}{{({{x}^{2}}+{{a}^{2}})}^{-1/2}}2x \right]\]                     
    • \[=\frac{{{n}^{2}}[{{({{\sec }^{n}}\theta -{{\cos }^{n}}\theta )}^{2}}+4{{\sec }^{n}}\theta {{\cos }^{n}}\theta ]}{{{(\sec \theta -\cos \theta )}^{2}}+4\sec \theta .\cos \theta }=\frac{{{n}^{2}}({{y}^{2}}+4)}{{{x}^{2}}+4}\].


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