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JEE Main & Advanced Mathematics Differentiation Question Bank Derivative at a point Standard Differentiation

  • question_answer
    If \[y\sqrt{{{x}^{2}}+1}=\log \left\{ \sqrt{{{x}^{2}}+1}-x \right\}\], then\[({{x}^{2}}+1)\frac{dy}{dx}+xy+1=\]  [Roorkee 1978; Kurukshetra CEE 1998]

    A)            0

    B)            1

    C)            2

    D)            None of these

    Correct Answer: A

    Solution :

               \[y\sqrt{{{x}^{2}}+1}=\log \left\{ \sqrt{{{x}^{2}}+1}-x \right\}\]            Differentiating both sides w.r.t. x, we get \[\frac{dy}{dx}\sqrt{{{x}^{2}}+1}+y.\frac{1}{2\sqrt{{{x}^{2}}+1}}.2x=\frac{1}{\sqrt{{{x}^{2}}+1}-x}\times \left\{ \frac{1}{2}\frac{2x}{\sqrt{{{x}^{2}}+1}}-1 \right\}\]            Þ  \[({{x}^{2}}+1)\frac{dy}{dx}+xy=\sqrt{{{x}^{2}}+1}.\frac{-1}{\sqrt{{{x}^{2}}+1}}\]            Þ  \[({{x}^{2}}+1)\frac{dy}{dx}+xy+1=0\].


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