Thursday, November 11, 2010

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

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


`(dy)/(dx) = (d(logx))/(dx)*log x + log x*(d(log x))/(dx)`



`(dy)/(dx) = (1/(x*ln 10))*log x + log x*(1/(x*ln 10))`


`(dy)/(dx) = (2log x)/(x*ln 10)`


`(dy)/(dx) = (2/(ln 10))*((log x)/x)`


Hence, differentiating the given function yields `(dy)/(dx) = (2/(ln 10))*((log x)/x).`


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


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


`(dy)/(dx) = (2/x)*(ln(x))`

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