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Proof of the Derivative of $\ln(x)$

The derivative of $f\left(x\right)=\ln x$ is $\displaystyle f'(x)=\frac{1}{x}$.
To prove this, we will use the following properties:
  1. $\displaystyle f'(x)=\lim \limits_{h \to 0} \frac{f\left(x+h\right)-f\left(x\right)}{h}$
  2. $\ln{x}-\ln{y}=\ln$
  3. $m \ln x= \ln$
  4. $\displaystyle \lim \limits_{n \to \infty} \left(1+\frac{1}{n}\right)^n=$
Find the derivative of $f\left(x\right)=\ln x$ from first principles (1):
$f'(x)=\lim \limits_{h \to 0}$

Write the denominator as $\displaystyle \frac{1}{h}$:
$\displaystyle f'(x)=\lim \limits_{h \to 0} \frac{1}{h} \Bigl($ $\Bigl)$

Use property 2:
$\displaystyle f'(x)=\lim \limits_{h \to 0} \frac{1}{h} \; \ln\Bigl($ $\Bigl)$

Divide both terms by $x$ and use property 3:
$\displaystyle f'(x)=\lim \limits_{h \to 0} \ln \left(1+\frac{h}{x}\right)$

Consider the substitution $\displaystyle n=\frac{x}{h}$. Then:
$h=$

Also, if $h \rightarrow 0$, then:
$n \rightarrow$

Substitute $h$ and write in terms of $n$ and $x$:
$\displaystyle f'(x)=\lim \limits_{n \to \infty} \ln$

Write $\displaystyle \frac{n}{x}$ as $n$ raised to an exponent:
$\displaystyle f'(x)=\lim \limits_{n \to \infty} \ln \left\{\left(1+\frac{1}{n}\right)^n\right\}$

Use property 3:
$\displaystyle f'(x)=\lim \limits_{n \to \infty}$ $\displaystyle \ln \left(1+\frac{1}{n}\right)^n$

Use property 4:
$\displaystyle f'(x)=\frac{1}{x} \ln \bigl($ $\bigr)$

Therefore, the derivative of $f\left(x\right)=\ln x$ is:
$f'(x)=$