JEE Main & Advanced Mathematics Differentiation Question Bank Mock Test - Limits and Derivatives

  • question_answer
    If \[y={{\left( x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right)}^{n}}\], then \[\frac{dy}{dx}\]is

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

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

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

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

    Correct Answer: A

    Solution :

    [a] \[\frac{dy}{dx}=\frac{d}{dx}\left[ {{(x+\sqrt{{{x}^{2}}+{{a}^{2}}})}^{n}} \right]\] \[=n{{\left( x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right)}^{n-1}}.\frac{d}{dx}\left( x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right)\] \[=n{{\left( x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right)}^{n-1}}\left( \frac{\sqrt{{{x}^{2}}+{{a}^{2}}}+x}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right)\] \[=\frac{n{{(x+\sqrt{{{x}^{2}}+{{a}^{2}}})}^{n}}}{\sqrt{{{x}^{2}}+{{a}^{2}}}}\] \[=\frac{ny}{\sqrt{{{n}^{2}}+{{a}^{2}}}}\]


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