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!

**Subject :**Math**Topic :**Calculus-
**Posted By :**Jason **Created on :**03-04-2011

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