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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2005

  • question_answer
    If \[y=\frac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}}\], then \[\frac{dy}{dx}\] is equal to :

    A)  \[\sec \,{{h}^{2}}x\]                      

    B)  \[\cos ec\,{{h}^{2}}x\]

    C)  \[\sec \,\,{{h}^{2}}x\]                                  

    D)  \[-\cos ec\,{{h}^{2}}x\]

    Correct Answer: D

    Solution :

    \[y=\frac{({{e}^{x}}+{{e}^{-x}})}{({{e}^{x}}-{{e}^{-x}})/2}=\frac{\cosh x}{\sinh x}\]\[y=\cot \,h\,x\] On differentiating with respect to \[x\], we get                 \[\frac{dy}{dx}=-\cos ec{{h}^{2}}x\]


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