CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If\[y={{\log }_{2}}{{\log }_{2}}(x),\]then\[\frac{dy}{dx}\]is equal to

    A)  \[\frac{{{\log }_{2}}e}{{{\log }_{e}}x}\]                 

    B)  \[\frac{{{\log }_{2}}e}{x{{\log }_{x}}2}\]

    C)  \[\frac{{{\log }_{2}}x}{{{\log }_{e}}2}\] 

    D)         \[\frac{{{\log }_{2}}e}{{{\log }_{2}}x}\]

    E)  \[\frac{{{\log }_{2}}e}{x{{\log }_{e}}x}\]

    Correct Answer: E

    Solution :

    Given,   \[y={{\log }_{2}}{{\log }_{2}}(x)\] \[=\frac{{{\log }_{e}}{{\log }_{2}}(x)}{{{\log }_{e}}2}\] \[=\frac{{{\log }_{e}}\left[ \frac{{{\log }_{e}}x}{{{\log }_{e}}2} \right]}{{{\log }_{e}}2}\] \[\Rightarrow \]               \[y=\frac{{{\log }_{e}}{{\log }_{e}}x-{{\log }_{e}}{{\log }_{e}}2}{{{\log }_{e}}2}\] On differentiating w.r.t.\[x,\]we get                 \[\frac{dy}{dx}=\frac{1}{{{\log }_{e}}2}\left[ \frac{1}{x{{\log }_{e}}x}-0 \right]\]                 \[=\frac{{{\log }_{2}}e}{x{{\log }_{e}}x}\]


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