find dy/dx when y=(logx)logx``      

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sciencesolve eNotes educator| Certified Educator

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))`


vkinard eNotes educator| Certified Educator

If you simplify the original equation to `y=(logx)^2` then you can find `(dy)/(dx)` using the chain rule:

`d/(dx)((logx)^2)=2(logx)^1 * (1/(xln10))`

Then you can clean up your answer into one simplified fraction:


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