The derivative at a point: the limit definition

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is $y = x^2$ . The blue line goes through its points at $x = a$ and $x = a + h$ . Move $a$ , then shrink $h$ .

The parabola y = x² and a secant line through its points at x = a and x = a + h. As h shrinks, the secant turns into the tangent line at x = a, of slope 2a.

With $a = 1$ , the slope of the blue line is $2 + h$ . What does it approach as $h \to 0$ ?

2
1

Key idea

The derivative at a point is the limit of the secant slopes.

Step 1h = 2: the secant through P and Q has slope 2.
Step 2h = 1: slope 1.5.
Step 3h = 0.5: slope 1.25.
Step 4h = 0.1: slope 1.05 — the slopes approach 1, the slope of the tangent at P.

f'(a) = \displaystyle\lim_{h \to 0} \dfrac{f(a + h) - f(a)}{h}

, when this limit exists. It is the instantaneous rate of change of

f

at

a

and the slope of the tangent line at

(a, f(a))

.

Worked example

$f(x) = x^2 + 3x$ . Find $f'(2)$ .

f(2 + h) - f(2) = 7h + h^2

f(2 + h) = 4 + 4h + h² + 6 + 3h; f(2) = 10.

\dfrac{f(2 + h) - f(2)}{h} = 7 + h

Divide by h ≠ 0.

f'(2) = 7

Let h → 0.

Your turn, guided

Your turn: $f(x) = x^2 + 3x$ at $a = 1$ .

f(1 + h) - f(1) = 5h + h^2

f(1 + h) = 1 + 2h + h² + 3 + 3h; f(1) = 4.

f'(1) = 5

Divide by h, then let h → 0.

Practice ★

$f(x) = x^2 + 4x$ . Use $f'(4) = \displaystyle\lim_{h \to 0} \dfrac{f(4 + h) - f(4)}{h}$ to find $f'(4)$ .
$f(x) = x^2 + 4x$ . Use $f'(2) = \displaystyle\lim_{h \to 0} \dfrac{f(2 + h) - f(2)}{h}$ to find $f'(2)$ .
$f(x) = x^2 + 5x$ . Use $f'(5) = \displaystyle\lim_{h \to 0} \dfrac{f(5 + h) - f(5)}{h}$ to find $f'(5)$ .

Watch out!

Practice ★★

Which expression is $f'(2)$ ?
1. $\displaystyle\lim_{h \to 0} \dfrac{f(2 + h) - f(2)}{h}$
2. $\displaystyle\lim_{h \to 0} \big(f(2 + h) - f(2)\big)$
3. $\dfrac{f(2 + 1) - f(2)}{1}$
4. $f(2)$

$f(x) = \dfrac{ 4 }{x}$ . Use the limit definition to find $f'(3)$ .
$f(x) = \dfrac{ 6 }{x}$ . Use the limit definition to find $f'(4)$ .

Exit ticket

Three questions to check you have got it.

$f(x) = x^2 + 4x$ . Use $f'(4) = \displaystyle\lim_{h \to 0} \dfrac{f(4 + h) - f(4)}{h}$ to find $f'(4)$ .
$f(x) = x^2 + 4x$ . Use $f'(2) = \displaystyle\lim_{h \to 0} \dfrac{f(2 + h) - f(2)}{h}$ to find $f'(2)$ .
$f(x) = x^2 + 5x$ . Use $f'(5) = \displaystyle\lim_{h \to 0} \dfrac{f(5 + h) - f(5)}{h}$ to find $f'(5)$ .

Summary card

The derivative at a point

Write $f(a + h) - f(a)$ and simplify.
Divide by $h$ .
Let $h \to 0$ : the result is $f'(a)$ .

Watch out: $f'(a)$ is a slope, not the value $f(a)$ .

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