The Derivative as a Function

A derivative can be given by a formula, a graph, a table, or in words. Let

V(t)

be the volume of water in a tank, in liters,

t

minutes after it starts filling. In words,

V'(t)

is the rate at which the volume is changing at time

t

, in liters per minute.

Example 1

Let

f(x) = x^3 - 3x

. Find

f'(x)

from the definition. Then find

f'(-1)

f'(0)

, and

f'(2)

Since

(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3

, the

x^3

and

3x

terms cancel in the numerator:

\begin{aligned} &f(x + h) - f(x) \\ &\qquad = 3x^2h + 3xh^2 + h^3 - 3h. \end{aligned}

Divide by

h

and let

h \to 0

\begin{aligned} f'(x) &= \lim_{h \to 0}\left(3x^2 + 3xh + h^2 - 3\right) \\ &= 3x^2 - 3. \end{aligned}

f'(-1) = 0

f'(0) = -3

, and

f'(2) = 9

. The graph of

f

has a horizontal tangent at

x = -1

, falls at

x = 0

, and rises steeply at

x = 2

Reading the graph of f′ from the graph of f

The height of the graph of

f'

x

is the slope of the graph of

f

x

. Where

f

has a horizontal tangent,

f'

0

. Where the tangent line slopes up,

f'

is positive, and where it slopes down,

f'

is negative. Read the other way, the instantaneous rate of change of

f

equals

c

wherever the graph of

f'

meets the line

y = c

The graph of f(x) = x³ − 3x with its tangent lines at x = −1.5, −1, 0, 1, 1.5.

The graph of f′(x) = 3x² − 3. Each dot's height is the slope of the tangent above it: 3.75, 0, −3, 0, 3.75.

Interactive: The Tangent Line and Its Slope

The two graphs above give the slope of

x^3 - 3x

at five points. Below,

f(x) = x^2

, and the point can travel along the curve. Compare each slope with the point's

x

-coordinate.

Example 2

Let

f(x) = \sqrt{2x + 1}

. Find

f'(x)

, and give the domains of

f

and

f'

Multiply the difference quotient by the conjugate of its numerator:

\begin{aligned} &\frac{\sqrt{2x + 2h + 1} - \sqrt{2x + 1}}{h} \\ &\quad = \frac{(2x + 2h + 1) - (2x + 1)}{h\left(\sqrt{2x + 2h + 1} + \sqrt{2x + 1}\right)} \\ &\quad = \frac{2}{\sqrt{2x + 2h + 1} + \sqrt{2x + 1}}. \end{aligned}

For

x > -\frac{1}{2}

, let

h \to 0

\begin{aligned} &f'(x) \\ &\quad = \lim_{h \to 0} \frac{2}{\sqrt{2x + 2h + 1} + \sqrt{2x + 1}} \\ &\quad = \frac{2}{2\sqrt{2x + 1}} = \frac{1}{\sqrt{2x + 1}}. \end{aligned}

The domain of

f

x \ge -\frac{1}{2}

. At

x = -\frac{1}{2}

only

h > 0

is allowed, and the quotient is

\frac{\sqrt{2h}}{h} = \frac{\sqrt{2}}{\sqrt{h}}

, which grows without bound as

h \to 0^+

. So

f'\left(-\frac{1}{2}\right)

doesn't exist, and the domain of

f'

x > -\frac{1}{2}

Tangents to y = √(2x + 1) at x = 0, 3/2, 4 have slopes 1, ½, ⅓. They flatten as x grows and steepen without bound as x → −½⁺.

Example 3

The graph of

f

is made of four line segments, with corners at

(-2, 3)

(1, 3)

, and

(3, -1)

The graph of f on [−4, 5].

(a) Sketch the graph of

f'

(-4, 5)

(b) Find

f'(0)

and

f'(2)

f'(1)

doesn't exist.

(a) On each segment the slope is constant, so

f'

is constant there:

1

(-4, -2)

0

(-2, 1)

-2

(1, 3)

, and

\frac{1}{2}

(3, 5)

. The graph of

f'

is four horizontal pieces, with open circles at the corners, where

f'

is undefined, and at the ends of the interval.

The graph of f′: a step function with no value at the corners x = −2, 1, 3 or at the ends.

(b)

f'(0) = 0

, since

f

is constant near

0

, and

f'(2) = -2

, the slope of the segment from

(1, 3)

(3, -1)

(c) For

h < 0

the difference quotient at

1

is the slope to the left,

0

, and for

h > 0

it's the slope to the right,

-2

. The one-sided limits disagree, so the limit defining

f'(1)

doesn't exist.

Tangent lines from values of f and f′

Example 5

The function

f

is differentiable, and selected values of

f

and

f'

are given in the table.

$x$	$-1$	$0$	$2$	$3$
$f(x)$	$4$	$1$	$-2$	$5$
$f'(x)$	$-3$	$0$	$-1$	$4$

(a) Find an equation of the tangent line at

x = 3

(b) Find the listed value of

x

at which the tangent line is horizontal.

x

at which the tangent line passes through the origin.

(a) The point is

(3, 5)

and the slope is

f'(3) = 4

y = 5 + 4(x - 3)

(b) At

x = 0

, where

f'(0) = 0

. The tangent line there is

y = 1

(c) The tangent at

x = a

y = f(a) + f'(a)(x - a)

. At

x = 0

it gives

y = f(a) - a f'(a)

, so it passes through the origin exactly when

f(a) = a f'(a)

. Checking each column,

4 \ne 3

1 \ne 0

-2 = -2

, and

5 \ne 12

. Only

x = 2

works, and its tangent line is

y = -x