The Derivative at a Point

The secant line through (a, f(a)) and (a + h, f(a + h)) rises f(a + h) − f(a) over a run of h.

Example 1

The table gives the amount of water

W(t)

, in liters, in a tank

t

hours after noon.

$t$ (hours)	$W(t)$ (liters)
$0$	$150$
$2$	$182$
$5$	$236$
$9$	$266$
$12$	$248$

(a) Find the average rate of change of

W

on each of the intervals

[0, 2]

[2, 5]

[5, 9]

, and

[9, 12]

, with units.

(b) Find the interval on which the average rate of change of

W

is greatest, and interpret the rate on

[9, 12]

(a) Each rate is the change in

W

over the change in

t

. The four rates are

\frac{182 - 150}{2 - 0} = 16

\frac{236 - 182}{5 - 2} = 18

\frac{266 - 236}{9 - 5} = 7.5

, and

\frac{248 - 266}{12 - 9} = -6

, all in liters per hour. The units are the output's units over the input's.

(b) The greatest rate,

18

liters per hour, is on

[2, 5]

. On

[9, 12]

the rate is

-6

liters per hour. From 9 p.m. to midnight, the tank lost water at an average of

6

liters per hour.

From average to instantaneous

Keep

a

fixed and let

h

shrink. The second point slides along the curve toward

(a, f(a))

, and the secant line turns. When the secant slopes approach a limit as

h \to 0

, that limit is the instantaneous rate of change of

f

a

. The line through

(a, f(a))

with that slope is the tangent line.

h = 2.4

h = 1.2

h = 0.4: the secant (blue) is close to the tangent (dashed).

Interactive: From Secant Line to Tangent Line

The three pictures stop

h

at three values. In the graph below,

f(x) = x^2

with

A

fixed at

x = 1

, and

B

slides along the curve toward

A

. As

B

closes in, the secant slopes approach

2

, the slope of the tangent line at

A

Example 2

For

f(x) = x^3

, find the slope of the secant line from

x = -1

x = -1 + h

for

h = \pm 0.1

\pm 0.01

, and

\pm 0.001

. Then find the limit of these slopes as

h \to 0

The slopes are

\frac{(-1 + h)^3 - (-1)}{h}

$h$	slope
$-0.1$	$3.31$
$-0.01$	$3.0301$
$-0.001$	$3.003001$
$0.001$	$2.997001$
$0.01$	$2.9701$
$0.1$	$2.71$

They close in on

3

from both sides. Algebra confirms it. Since

(-1 + h)^3 = -1 + 3h - 3h^2 + h^3

\begin{aligned} &\frac{(-1 + h)^3 + 1}{h} \\ &\qquad = \frac{3h - 3h^2 + h^3}{h} \\ &\qquad = 3 - 3h + h^2 \quad (h \ne 0), \end{aligned}

which approaches

3

h \to 0

. The instantaneous rate of change of

f

x = -1

3

The derivative at a point

The derivative of f at a

The derivative of

f

x = a

f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h},

provided the limit exists. Then

f

is differentiable at

a

. Setting

x = a + h

gives the same number in a second form:

f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}.

The number

f'(a)

is the instantaneous rate of change of

f

a

and the slope of the graph of

f

(a, f(a))

Read backward, a limit in this form is a derivative. For example,

\lim_{x \to -2} \frac{g(x) - g(-2)}{x + 2}

g'(-2)

, and it exists exactly when

g

is differentiable at

-2

Example 3

Let

f(x) = x^2 - 3x

. Find

f'(4)

with each form of the definition.

First,

f(4) = 16 - 12 = 4

. With

h

\begin{aligned} f(4 + h) &= 16 + 8h + h^2 - 12 - 3h \\ &= 4 + 5h + h^2, \end{aligned}

\begin{aligned} f'(4) &= \lim_{h \to 0} \frac{f(4 + h) - f(4)}{h} \\ &= \lim_{h \to 0} \frac{5h + h^2}{h} \\ &= \lim_{h \to 0} (5 + h) = 5. \end{aligned}

With

x

, the numerator

x^2 - 3x - 4

factors:

\begin{aligned} f'(4) &= \lim_{x \to 4} \frac{f(x) - f(4)}{x - 4} \\ &= \lim_{x \to 4} \frac{(x - 4)(x + 1)}{x - 4} \\ &= \lim_{x \to 4} (x + 1) = 5. \end{aligned}

Either way,

f'(4) = 5

. The second form is quicker when

f(x) - f(a)

factors. When

f

is a polynomial,

x - a

is always one of the factors.

Example 4

Find an equation of the tangent line to

y = \frac{1}{x + 1}

x = 1

Let

f(x) = \frac{1}{x + 1}

, so

f(1) = \frac{1}{2}

. Combine the fractions in the numerator over the common denominator

2(2 + h)

\begin{aligned} &\frac{f(1 + h) - f(1)}{h} \\ &\qquad = \frac{1}{h}\left(\frac{1}{2 + h} - \frac{1}{2}\right) \\ &\qquad = \frac{1}{h} \cdot \frac{2 - (2 + h)}{2(2 + h)} \\ &\qquad = \frac{-1}{2(2 + h)} \quad (h \ne 0). \end{aligned}

h \to 0

this approaches

-\frac{1}{4}

, so

f'(1) = -\frac{1}{4}

. The tangent line is

y = \frac{1}{2} - \frac{1}{4}(x - 1).

The tangent line at (1, ½) has slope −¼ and crosses the x-axis at x = 3.

Example 5

Let

f(x) = x^3 - 2x

. Find the tangent line at

x = 1

, and find where else it meets the graph of

f

Here

f(1) = -1

, and expanding

(1 + h)^3

gives

f(1 + h) = -1 + h + 3h^2 + h^3

. So

\begin{aligned} &\frac{f(1 + h) - f(1)}{h} \\ &\qquad = \frac{h + 3h^2 + h^3}{h} \\ &\qquad = 1 + 3h + h^2 \to 1. \end{aligned}

The tangent line is

y = -1 + 1(x - 1)

, or

y = x - 2

. It meets the graph where

x^3 - 2x = x - 2

, that is,

x^3 - 3x + 2 = 0

. Since

x = 1

is a root,

x - 1

is a factor:

\begin{aligned} x^3 - 3x + 2 &= (x - 1)(x^2 + x - 2) \\ &= (x - 1)^2(x + 2). \end{aligned}

So the line meets the graph again at

x = -2

, at the point

(-2, -4)

. A tangent line touches the graph at the point of tangency. Away from that point, it's free to cross the graph.

The tangent line y = x − 2 touches the graph at (1, −1) and crosses it at (−2, −4).