12A

The exponential function base e

Why the gradient of y = a^x at the origin defines e, approximating e, the derivative of e^x and e^(ax + b), and the graph of y = e^x.

Defining e

Every graph $y = a^x$ passes through $(0, 1)$. For $a = 2$ the tangent there has gradient about $0.69$; for $a = 3$ it is about $1.10$. Somewhere between, there is a base whose gradient at $(0, 1)$ is exactly $1$. That base is Euler's number:

Euler's number

$$e \approx 2.718\,281\,828$$

$e$ is irrational. One way to approximate it: $\left(1 + \dfrac1n\right)^n$ gets closer to $e$ as $n$ grows; for $n = 1000$ it is $2.7169$.

The derivative of e^x

Gradient equals height

Blue dashed: the tangent at x = a. Its gradient always equals the height of the curve. Each tangent meets the x-axis exactly 1 unit to the left of the point.

Try this

Where does the tangent at $x = 1$ meet the $x$-axis?

Example 1●

Differentiate $y = e^{3x}$, $y = 5e^{-x}$ and $y = e^{2x + 1}$.

$\dfrac{d}{dx}e^{3x} = 3e^{3x}$

$\dfrac{d}{dx}5e^{-x} = -5e^{-x}$

$\dfrac{d}{dx}e^{2x + 1} = 2e^{2x + 1}$

Example 2: chain rule●●

Differentiate $y = e^{x^2}$.

$\dfrac{dy}{dx} = e^{x^2} \times 2x = 2xe^{x^2}$

The graph of y = e^x

Passes through $(0, 1)$ with gradient $1$ there.
Horizontal asymptote $y = 0$ as $x \to -\infty$.
As $x \to \infty$, both $y$ and $\dfrac{dy}{dx}$ grow without bound.
Always positive, always increasing, concave up.

Flashcards

$e$

$\approx 2.718$

$rac{d}{dx}e^x$

$e^x$

$rac{d}{dx}e^{ax + b}$

$ae^{ax + b}$

Gradient of $y = e^x$ at $(0, 1)$

$1$

Practice

$e$ is approximately:
$\dfrac{d}{dx}e^x = $
For $y = e^{3x}$, find $\dfrac{dy}{dx}$ at $x = 0$.
$\dfrac{d}{dx}e^{2x + 1} = $
Find the gradient of $y = e^x$ at $x = 0$.
The tangent to $y = e^x$ at $x = 1$ passes through:
For $y = 5e^{-x}$, find $\dfrac{dy}{dx}$ at $x = 0$.
$\dfrac{d}{dx}e^{x^2} = $
Use your calculator to find $e^2$, correct to 3 decimal places.