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:

(e^lnx)'= 1

(lnx)' = 1/x

(a^x)=a^x * lna

log_a^x = (1/ lna ) * (1/x)

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

So  take derivative on a ^ (loga^x)   =x we get

[a ^ (loga^x) ]'  =x'

From chain rule we get:

[a ^ (loga^x)  * ln a] * (loga^x)' =1

(loga^x)'=1 / [a ^ (loga^x)  * ln a]

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

(loga^x)'=1 / [x  * ln a]

For (x^x)'= (x^x) * (lnx +1) , we need to know x^x = (e^lnx)^x=e^xlnx

Take derivative on both sides of x^x =e^xlnx   , so we get:

(x^x)' =(e^xlnx )'

(x^x)' =(e^xlnx ) * (xlnx)'           (This is chain rule!!!!!!!!!!)

(x^x)' =(x^x) * (xlnx)'                 (Since I said x^x =exlnx)

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

Finally through simplification we get:

  (x^x)'= (x^x) * (lnx +1)

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