Question

How do you find the derivative of `ln x`?

Answer 1

`1/x`

Consider f(x) = lnx and using the first principles work out `f '(x)= lim h->0 {f(x+h)-f(x)}/h`

= `lim h->0 {ln(x+h) -lnx}/h`

=`limh->0{ln((x+h)/x)}/h`

= `lim h->0{ln(1+h/x)}/h`

= `limh->0 1/x {ln(1+h/x)}/ (h/x)`

= `1/x` `limh->0 {ln(1+h/x)}/(h/x)`

= `1/x`, because `[limh->0{ln(1+h/x)}/(h/x) =1]`

Answer 2

The "how" depends on the definition you are using for `lnx`.

I like: `lnx = int_1^x 1/t dt` and the result is immediate useing the Fundamental Theorem of Calculus.

If you start with a definition of `e^x` and find its derivative, then you probably defined `lnx` by:

`y = lnx` `iff` `x = e^y`

Differentiate implicitely:

`d/(dx)(x) = d/(dx)(e^y)`

`1 = e^y dy/dx`

So
`dy/dx = 1/ e^y = 1/x`