# Proof of the derivative of ln(x)

Anonymous

### Question Description

I'm trying to prove that ddxlnx=1x.

Here's what I've got so far:

ddxlnx=limh→0ln(x+h)−ln(x)h=limh→0ln(x+hx)h=limh→0ln(1+hx)h

To simplify the logarithm:

limh→0(1+hx)1h=e1x

^This is the line I have trouble with. I can see that it is true by putting numbers in, but I can't prove it. I know that e1x=limh→0(1+h)1xh, but I can't work out how to get from the above line to that.

limh→0((1+hx)1h)h=ehx

Going back to the derivative:

ddxlnx=limh→0ln(ehx)h=limh→0hxln(e)h=limh→0hx÷h=1x

This proof seems fine, apart from the middle step to get e1x. How could I prove that part?

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