The derivative of $\ln(x)$

Question

How can one prove the following by elementary means? $$\ln(x)'=\frac{1}{x}$$

Say we know that $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}.$$

Answer 1

Let $f(x)=e^x$, so $f^{-1}(x)=\ln x$. We know that $(f \circ f^{-1})(x)=x$. Differentiate by the chain rule to get $$ f'(f^{-1}(x)) \frac{d}{dx}f^{-1}(x)=1. $$ As $f'=f$ this boils down to $$ x \frac{d}{dx}f^{-1}(x)=1 $$ giving what you want.

Answer 2

If we know that, then\begin{align}\exp'(x)&=\left(\sum_{n=0}^\infty\frac{x^n}{n!}\right)'\\&=\sum_{n=0}^\infty\frac{x^n}{n!}\\&=\exp(x).\end{align}Therefore,\begin{align}\ln'(x)&=\frac1{\exp'(\ln(x))}\\&=\frac1{\exp(\ln(x))}\\&=\frac1x.\end{align}

Answer 3

Let $f(x)=e^x$. Then:

$${\partial \log (x) \over \partial x} = {\partial f^{-1}(x) \over \partial x} = {1 \over e^{\log(x)}} = 1/x$$.