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

  • question_answer
    If \[y=\frac{{{e}^{2x}}+{{e}^{-2x}}}{{{e}^{2x}}-{{e}^{-2x}}}\], then \[\frac{dy}{dx}=\]                   [AI CBSE 1988]

    A)            \[\frac{-8}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]

    B)            \[\frac{8}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]

    C)            \[\frac{-4}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]

    D)            \[\frac{4}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]

    Correct Answer: A

    Solution :

               \[y=\frac{{{e}^{2x}}+{{e}^{-2x}}}{{{e}^{2x}}-{{e}^{-2x}}}\]            \[\therefore \frac{dy}{dx}=\frac{({{e}^{2x}}-{{e}^{-2x}})2({{e}^{2x}}-{{e}^{-2x}})-({{e}^{2x}}+{{e}^{-2x}})2({{e}^{2x}}+{{e}^{-2x}})}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]                      \[=\frac{-8}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\].


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