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Proof of the Derivative of $e^x$

The derivative of $f\left(x\right)=e^x$ is $\displaystyle f'(x)=e^x$.
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. $e^{x}e^{y}=e$
  3. $m \ln x= \ln$
  4. $\displaystyle \lim \limits_{n \to \infty} \left(1+\frac{1}{n}\right)^n=$
Find the derivative of $f\left(x\right)=e^x$ from first principles (1):
$f'(x)=\lim \limits_{h \to 0}$

Use property 2:
$f'(x)=\lim \limits_{h \to 0}$

Factorize $e^x$ and bring outside the limit:
$\displaystyle f'(x)=e^x \lim \limits_{h \to 0}$

Consider the substitution $\displaystyle e^h -1=n$. Then:
$h=$

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

Substitute $e^h-1$ and $h$ and write in terms of $n$:
$\displaystyle f'(x)=e^x \lim \limits_{n \to 0}$

Multiply the numerator and denominator by $\displaystyle \frac{1}{n}$:
$\displaystyle f'(x)=e^x \lim \limits_{n \to 0}$

Use property 3 in the denominator:
$\displaystyle f'(x)=e^x \lim \limits_{n \to 0}$

Consider the substitution $\displaystyle \frac{1}{n}=m$. Then:
$n=$

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

Substitute $n$ and $\displaystyle \frac{1}{n}$ and write in terms of $m$:
$\displaystyle f'(x)=e^x \lim \limits_{m \to \infty}$

Use property 4:
$\displaystyle f'(x)=e^x $
$1$ / $\ln \bigl($ $\bigl)$

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