How to Play with Log Derivatives

Actually the reason why we learn the number e  is becase we need to use it for calculus. So how to play with the log derivatives, and some derivatives related to e (natural log)?

Here is a list of log derivatives you can see in exams very often:

(elnx)'= 1

(lnx)' = 1/x

(ax)=ax * lna

logax = (1/ lna ) * (1/x)

(xx)'= (xx) * (lnx +1)

The first three are easy to memorize, but that last two are kind of tricky to memorize, so let's proove it.

For logax = (1/ lna ) * (1/x) , we first have to know that a ^ (logax) = loga a=x  as a common sense.

So  take derivative on a ^ (logax  =x we get

[a ^ (logax) ]'  =x'

From chain rule we get:

[a ^ (logax)  * ln a] * (logax)' =1

(logax)'=1 / [a ^ (logax)  * ln a]

Again, since a ^ (logax)  =x , finally we get:

(logax)'=1 / [x  * ln a]


For (xx)'= (xx) * (lnx +1) , we need to know xx = (elnx)x=exlnx

Take derivative on both sides of xx =exlnx   , so we get:

(xx)' =(exlnx )'

(xx)' =(exlnx ) * (xlnx)'           (This is chain rule!!!!!!!!!!)

(xx)' =(xx) * (xlnx)'                 (Since I said xx =exlnx)

(xx)' =(xx) * (lnx + x/x)          (We use product rule here)

Finally through simplification we get:

  (xx)'= (xx) * (lnx +1)


If you understand all of the above and do some more practices, you will be able to enjoy playing with log derivatives!




  • Subject : Math
  • Topic : Calculus
  • Posted By : Jason

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