Review of the exponential function base e

The number e, the derivative of e^x and e^(ax+b), and tangents and normals to exponential curves.

The number $e$

$e \approx 2.718$ is the base for which the gradient of $y = e^x$ at every point equals its height. In particular, the gradient at the $y$-intercept $(0, 1)$ is exactly $1$.

Gradient equals height

The dashed line is the tangent at x = P. Its gradient always equals the height of the curve.

Try this

Where does the tangent cut the $x$-axis? Compare it with $P$ as you move the slider.

Derivatives

$$\frac{d}{dx}e^x = e^x \qquad \frac{d}{dx}e^{ax + b} = ae^{ax + b}$$

Tangents and normals

Example 1●●

Find the tangent and the normal to $y = e^x$ at $x = 1$.

The point is $(1, e)$ and the gradient is $e^1 = e$.

Point and gradient

$y - e = e(x - 1)$, so $y = ex$. It passes through the origin.

Tangent

The gradient is $-\dfrac1e$: $y - e = -\dfrac1e(x - 1)$, so $y = -\dfrac{x}{e} + \dfrac1e + e$.

Normal

Example 2●

Find the tangent to $y = e^{2x}$ at its $y$-intercept.

At $x = 0$: $y = 1$ and $\dfrac{dy}{dx} = 2e^0 = 2$.

Tangent: $y = 2x + 1$.

Flashcards

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

$e^x$

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

$ae^{ax + b}$

$e$ to 3 decimal places

$2.718$

Practice

$\dfrac{d}{dx}e^{3x} = $
$\dfrac{d}{dx}5e^{-2x} = $
Find the gradient of $y = e^x$ at $x = 0$.
The tangent to $y = e^x$ at $x = 1$ is:
The gradient of the normal to $y = e^x$ at $x = 1$ is:
Find the gradient of the tangent to $y = e^{2x}$ at $x = 0$.
The tangent to $y = e^x$ at $x = 2$ cuts the $x$-axis at $x = ?$
Write $e$ correct to 2 decimal places.
Which statement about $y = e^x$ is true?