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JEE Main & Advanced Mathematics Differentiation Question Bank Differentiation by Substitution

  • question_answer
    If \[y=a{{x}^{n+1}}+b{{x}^{-n}}\], then \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=\]             [Karnataka CET 1993]

    A)            \[n\,(n-1)y\]

    B)            \[n\,(n+1)y\]

    C)            ny

    D)            \[{{n}^{2}}y\]

    Correct Answer: B

    Solution :

               \[y=a{{x}^{n+1}}+b{{x}^{-n}}\Rightarrow \frac{dy}{dx}=(n+1)a{{x}^{n}}-nb{{x}^{-n-1}}\]            Þ \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=n(n+1)a{{x}^{n-1}}+n(n+1)b{{x}^{-n-2}}\]            Þ \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=n(n+1)y\].


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