Derivative of exp x, e^x

Derivative of exponential x

Derivative $f’$ of the function $f(x)=\exp x= e^{x}$ is:

$$ \forall x \in ]-\infty, +\infty[ , f’(x) = \exp x = e^{x}$$

Proof/Demonstration

$$ \begin{aligned} f^\prime(x)=(e^x)^\prime &=\lim _{h \rightarrow 0} \frac{e^{x+h}-e^{x}}{h} \\ &=\lim _{h \rightarrow 0} \frac{e^{x} \cdot e^{h}-e^{x}}{h} \\ &=\lim _{h \rightarrow 0} \frac{e^{x}(e^{h}-1)}{h} \\ &=e^{x}\cdot \lim _{h \rightarrow 0} \frac{e^{h}-1}{h} \\ &=e^{x}\cdot f^\prime(0) \end{aligned} $$

We need to find the derivative $f^\prime(0)$.
We fix: $n=e^{h}-1$, ie $n+1=e^{h}$ ie $h=\ln(1+n)$

$$ \begin{aligned} f^\prime(0)&= \lim _{h \rightarrow 0} \frac{e^{h}-1}{h} \\ &=\lim _{n \rightarrow 0} \frac{n}{\ln (1+n)} \\ &=\lim _{n \rightarrow 0} \frac{1}{\displaystyle\frac{1}{n}\ln (1+n)} \\ &=\lim _{n \rightarrow 0} \frac{1}{\displaystyle\ln \left((1+n)^{\frac{1}{n}}\right)} \\ &=\frac{1}{\ln e} = 1 \end{aligned} $$

because

$$ \lim _{n \rightarrow 0}(1+n)^{\frac{1}{n}}=e $$

see Proof.

We conclude since $f^\prime(x)=e^{x}\cdot f^\prime(0)$

$$ f^\prime(x)=e^{x} $$