# From First Principles

## Key Questions

• #### Answer:

$\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

#### Explanation:

We seek:

$\frac{d}{\mathrm{dx}} {e}^{x}$

Method 1 - Using the limit definition:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

We have:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{{e}^{x + h} - {e}^{x}}{h}$
$\text{ } = {\lim}_{h \to 0} \frac{{e}^{x} {e}^{h} - {e}^{x}}{h}$
$\text{ } = {\lim}_{h \to 0} {e}^{x} \frac{\left({e}^{h} - 1\right)}{h}$
$\text{ } = {e}^{x} {\lim}_{h \to 0} \frac{\left({e}^{h} - 1\right)}{h}$

Think about this limit for a moment and we can rewrite it as:

${\lim}_{h \to 0} \frac{\left({e}^{h} - 1\right)}{h} = {\lim}_{h \to 0} \frac{\left({e}^{h} - {e}^{0}\right)}{h}$
$\text{ } = {\lim}_{h \to 0} \frac{\left({e}^{0 + h} - {e}^{0}\right)}{h}$
$\text{ } = f ' \left(0\right)$ (by the derivative definition)

Hence,

$f ' \left(x\right) = {e}^{x} f ' \left(0\right)$

Now, It can be shown that this limit:

$f ' \left(0\right) = {\lim}_{h \to 0} \frac{\left({e}^{h} - 1\right)}{h}$

both exists and is equal to unity. Additionly, the number $2.718281 \ldots$, which we call Euler's number) denoted by $e$ is extremely important in mathematics, and is in fact an irrational number (like $\pi$ and $\sqrt{2}$,

And so:

$\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

This special exponential function with Euler's number, $e$, is the only function that remains unchanged when differentiated.

Method 2 - Power Series

We can use the power series:

 e^x = 1 +x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ...

Then we can differentiate term by term using the power rule:

 d/dx e^x = d/dx{1 +x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... }

 \ \ \ \ \ \ \ \ \ = 0 +1 + (2x)/(2!) + (3x^2)/(3!) + (4x^3)/(4!) + (5x^4)/(5!) + ...

 \ \ \ \ \ \ \ \ \ = 1 + (x)/(1!) + (3x^2)/(2! * 2) + (4x^3)/(3! * 4) + (5x^4)/(4! * 5) + ...

 \ \ \ \ \ \ \ \ \ = 1 +x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ...

• By rewriting a bit,

$y = {c}^{x} = {e}^{\left(\ln c\right) x}$.

By Chain Rule,

$y ' = {e}^{\left(\ln c\right) x} \cdot \left(\ln c\right) = \left(\ln c\right) {c}^{x}$

I hope that this was helpful.