# The derivative of e to the x

## Statement

The derivative of $e^x$ is $e^x$:

$$\tfrac{d}{dx}(e^x) = e^x$$

## The definition of e

Take the definition of $e$.

$$e = \lim_{n \to \infin}\bigg(\left(1 + \tfrac{1}{n}\right)^n\bigg)$$

When substituting $n = \frac{1}{h}$, $h$ goes to $0$ as $n$ goes to $\infin$.

$$e = \lim_{h \to 0}\bigg((1 + h)^\frac{1}{h}\bigg)$$

Later in the proof this definition of $e$ is used.

## Proof

Define function $f$.

$$f(x) = e^x$$

Take the definition of the derivative and substitute.

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

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

By the rule of adding exponents, write $e^{x + h} = e^x * e^h$ and factor out $e^x$.

$$f'(x) = \lim_{h \to 0}\left(\frac{e^x * e^h - e^x}{h}\right)$$

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

Since $e^x$ does not include the limit variable, $h$, bring it out of the limit.

$$f'(x) = \lim_{h \to 0}\left(e^x * \frac{e^h - 1}{h}\right)$$

$$f'(x) = e^x * \lim_{h \to 0}\left(\frac{e^h - 1}{h}\right)$$

Now substitute $e$ in the limit with what is defined above. Since that definition uses $h \to 0$ and this limit does too, it is not required to write the limit again.

$$f'(x) = e^x * \lim_{h \to 0}\left(\frac{\big((1 + h)^\frac{1}{h}\big)^h - 1}{h}\right)$$

Multiply the exponents and cancel like terms.

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

$$f'(x) = e^x * \lim_{h \to 0}\left(\frac{h}{h}\right)$$

Finally, since $\frac{h}{h}$ is $1$, the limit disappears.

$$f'(x) = e^x * \lim_{h \to 0}\big(1\big)$$

$$f'(x) = e^x$$

## Proofs building upon this proof

### Exponent rule in calculus

This proofs show the derivative of a^x is a^x * ln(a).

### Power rule in calculus

This proofs shows that the derivative of x^r is r * x^(r - 1).

### The derivative of ln(x)

This proof shows that the derivative of ln(x) is 1/x.