JEE Main & Advanced Mathematics Limits, Continuity and Differentiability Question Bank Self Evaluation Test - Continuity and Differentiability

  • question_answer
    If \[y=lo{{g}_{10}}x+lo{{g}_{x}}10+lo{{g}_{x}}x+lo{{g}_{10}}10\] then what is \[{{\left( \frac{dy}{dx} \right)}_{x=10}}\] equal to?

    A) 10

    B) 2

    C) 1

    D) 0

    Correct Answer: D

    Solution :

    [d] \[y={{\log }_{10}}x+{{\log }_{x}}10+{{\log }_{x}}x+{{\log }_{10}}10\] \[y={{\log }_{10}}x+{{\log }_{x}}10+1+1\] Differentiating equation w.r.t.x \[\frac{dy}{dx}=\frac{1}{x{{\log }_{e}}10}-\frac{1}{{{({{\log }_{10}}x)}^{2}}}.\frac{1}{(x\log 10)}\] \[=\frac{1}{x{{\log }_{e}}10}\left[ 1-\frac{1}{{{({{\log }_{10}}x)}^{2}}} \right]\] \[{{\left( \frac{dy}{dx} \right)}_{x=10}}=\frac{1}{10\,{{\log }_{e}}10}[1-1]=0\] \[\left[ \begin{align}   & Note:{{\log }_{x}}10=\frac{{{\log }_{10}}10}{{{\log }_{10}}x}=\frac{1}{{{\log }_{10}}x} \\  & \frac{d}{dx}\left[ \frac{1}{{{\log }_{10}}x} \right]=-{{({{\log }_{10}}x)}^{-2}}\times \frac{1}{x{{\log }_{e}}10} \\  & =-\frac{1}{{{({{\log }_{10}}x)}^{2}}x{{\log }_{e}}10} \\ \end{align} \right]\]


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