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

  • question_answer
    If \[x=\exp \left\{ {{\tan }^{-1}}\left( \frac{y-{{x}^{2}}}{{{x}^{2}}} \right) \right\}\,\,\], then \[\frac{dy}{dx}\]equals [MP PET 2002]

    A)            \[2x\,[1+\tan \,(\log x)]+x{{\sec }^{2}}(\log x)\]

    B)            \[x\,[1+\tan \,(\log x)]+{{\sec }^{2}}(\log x)\]

    C)            \[2x\,[1+\tan \,(\log x)]+{{x}^{2}}\,\,{{\sec }^{2}}(\log x)\]

    D)            \[2x\,[1+\tan \,(\log x)]+{{\sec }^{2}}(\log x)\]

    Correct Answer: A

    Solution :

               \[x=\exp \left\{ {{\tan }^{-1}}\left( \frac{y-{{x}^{2}}}{{{x}^{2}}} \right) \right\}\]                    Þ  \[\log x={{\tan }^{-1}}\left( \frac{y-{{x}^{2}}}{{{x}^{2}}} \right)\]            Þ  \[\frac{y-{{x}^{2}}}{{{x}^{2}}}\,=\,\tan (\log x)\]  Þ  \[y={{x}^{2}}\tan (\log x)+{{x}^{2}}\]            Þ  \[\frac{dy}{dx}=2x\,.\,\tan (\log x)+{{x}^{2}}\,.\,\frac{{{\sec }^{2}}(\log x)}{x}+2x\]            Þ  \[\frac{dy}{dx}=2x\tan (\log x)+x{{\sec }^{2}}(\log x)+2x\]            Þ  \[\frac{dy}{dx}=2x[1+\tan (\log x)]+x{{\sec }^{2}}(\log x)\].


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