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

  • question_answer
    If \[y=\sqrt{\frac{1+{{e}^{x}}}{1-{{e}^{x}}}}\], then \[\frac{dy}{dx}=\]                                         [AI CBSE 1986]

    A)            \[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1-{{e}^{2x}}}}\]

    B)            \[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1-{{e}^{x}}}}\]

    C)            \[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1+{{e}^{2x}}}}\]

    D)            \[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1-{{e}^{2x}}}}\]

    Correct Answer: A

    Solution :

               \[y=\sqrt{\frac{1+{{e}^{x}}}{1-{{e}^{x}}}}\]or \[{{y}^{2}}=\frac{1+{{e}^{x}}}{1-{{e}^{x}}}\]                    \[2y\frac{dy}{dx}=\frac{(1-{{e}^{x}}){{e}^{x}}+(1+{{e}^{x}}){{e}^{x}}}{{{(1-{{e}^{x}})}^{2}}}=\frac{2{{e}^{x}}}{{{(1-{{e}^{x}})}^{2}}}\]                 \[\therefore \frac{dy}{dx}=\frac{{{e}^{x}}}{{{(1-{{e}^{x}})}^{2}}}\sqrt{\left[ \frac{1-{{e}^{x}}}{1+{{e}^{x}}} \right]\left[ \frac{1-{{e}^{x}}}{1-{{e}^{x}}} \right]}\]\[=\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1-{{e}^{2x}}}}\].


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