Prove that dy/dx=1/x (y=lnx)

I would use the limit definition of a derivative.

where h is a very small change in x

y' = [ln (x+h) - ln (x) ]/h

Means the same as the:

Limit: (ln (x+h) - ln (x)) / h

as h - - > 0

Using the properties of logarithms, combine them into the logarithm function.

Limit: ln ((x+h) / x) / h

ln (1+h/x) * (1/h)

change of variable

h/x = n; h = nx

as h--> 0 so does n-->0

So now we have ...

Limit: (1/nx) * ln (1+n)

as n--> 0

now the definition of Euler's number "e"

Limit: (1+n) ^ (1/n) as n-->0

So we change the limit to

Limit: (1/x) * ln ((1+n) ^ (1/n))

as n-->0

"limit of a log is the log of a limit"

(1/x) * ln (Limit: (1+n) ^ (1/n) as n-->0)

= (1/x) * ln (e)

= (1/x) * (1)

= 1/x