Blog: find dy/dx when y=(logx)logx``

You need to differentiate the function with respect to x, using product rule such that:

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{{d}{\left({\log{{x}}}\right)}}}{{{\left.{d}{x}\right.}}}\cdot{\log{{x}}}+{\log{{x}}}\cdot\frac{{{d}{\left({\log{{x}}}\right)}}}{{{\left.{d}{x}\right.}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={\left(\frac{{1}}{{{x}\cdot{\ln{{10}}}}}\right)}\cdot{\log{{x}}}+{\log{{x}}}\cdot{\left(\frac{{1}}{{{x}\cdot{\ln{{10}}}}}\right)}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{{2}{\log{{x}}}}}{{{x}\cdot{\ln{{10}}}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={\left(\frac{{2}}{{{\ln{{10}}}}}\right)}\cdot{\left(\frac{{{\log{{x}}}}}{{x}}\right)}$

Hence, differentiating the given function yields $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={\left(\frac{{2}}{{{\ln{{10}}}}}\right)}\cdot{\left(\frac{{{\log{{x}}}}}{{x}}\right)}.$

***This is assuming you are using the base 10 log***

If you are using the natural log, ln(x), then your answer...

You need to differentiate the function with respect to x, using product rule such that:

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{{2}{\log{{x}}}}}{{{x}\cdot{\ln{{10}}}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={\left(\frac{{2}}{{{\ln{{10}}}}}\right)}\cdot{\left(\frac{{{\log{{x}}}}}{{x}}\right)}$

***This is assuming you are using the base 10 log***

If you are using the natural log, ln(x), then your answer would be

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={\left(\frac{{2}}{{x}}\right)}\cdot{\left({\ln{{\left({x}\right)}}}\right)}$

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find dy/dx when y=(logx)logx

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find dy/dx when y=(logx)logx

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