Defining the Derivative

Now that we have both a conceptual understanding of a limit and the practical ability to compute limits, we have established the foundation for our study of calculus, the branch of mathematics in which we compute derivatives and integrals. Most mathematicians and historians agree that calculus was developed independently by the Englishman Isaac Newton

(1643–1727)

whose images appear in [link]. When we credit Newton and Leibniz with developing calculus, we are really referring to the fact that Newton and Leibniz were the first to understand the relationship between the derivative and the integral. Both mathematicians benefited from the work of predecessors, such as Barrow, Fermat, and Cavalieri. The initial relationship between the two mathematicians appears to have been amicable; however, in later years a bitter controversy erupted over whose work took precedence. Although it seems likely that Newton did, indeed, arrive at the ideas behind calculus first, we are indebted to Leibniz for the notation that we commonly use today.

Tangent Lines

We begin our study of calculus by revisiting the notion of secant lines and tangent lines. Recall that we used the slope of a secant line to a function at a point

(a, f (a))

to estimate the rate of change, or the rate at which one variable changes in relation to another variable. We can obtain the slope of the secant by choosing a value of

x

as shown in [link]. The slope of this line is given by an equation in the form of a difference quotient:

We can also calculate the slope of a secant line to a function at a value a by using this equation and replacing

x

is a value close to 0. We can then calculate the slope of the line through the points

(a, f (a))

In this case, we find the secant line has a slope given by the following difference quotient with increment

h :

These two expressions for calculating the slope of a secant line are illustrated in [link]. We will see that each of these two methods for finding the slope of a secant line is of value. Depending on the setting, we can choose one or the other. The primary consideration in our choice usually depends on ease of calculation.

the slopes of the secant lines provide better estimates of the rate of change of the function at

a .

Furthermore, the secant lines themselves approach the tangent line to the function at

a,

which represents the limit of the secant lines. Similarly, [link](b) shows that as the values of

h

the secant lines also approach the tangent line. The slope of the tangent line at

a

As the intervals become narrower, the graph of the function and its tangent line appear to coincide, making the values on the tangent line a good approximation to the values of the function for choices of

x

Just as we have used two different expressions to define the slope of a secant line, we use two different forms to define the slope of the tangent line. In this text we use both forms of the definition. As before, the choice of definition will depend on the setting. Now that we have formally defined a tangent line to a function at a point, we can use this definition to find equations of tangent lines.

The Derivative of a Function at a Point

The type of limit we compute in order to find the slope of the line tangent to a function at a point occurs in many applications across many disciplines. These applications include velocity and acceleration in physics, marginal profit functions in business, and growth rates in biology. This limit occurs so frequently that we give this value a special name: the derivative. The process of finding a derivative is called differentiation.

Velocities and Rates of Change

Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if

s (t)

is the position of an object moving along a coordinate axis, the average velocity of the object over a time interval

[a, t]

To better understand the relationship between average velocity and instantaneous velocity, see [link]. In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time

t = a

The slope of the secant line (shown in green) is the average velocity of the object over the time interval

[a, t] .

We can use [link] to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using [link].

As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or the rate of change of a function at any point along the function.

Key Concepts

Key Equations

Average velocities using values of t approaching 0
$t$	$\frac{sin t - sin 0}{t - 0} = \frac{sin t}{t}$
$−0.1$	$0.998334166$
$−0.01$	$0.9999833333$
$−0.001$	$0.999999833$
$0.001$	$0.999999833$
$0.01$	$0.9999833333$
$0.1$	$0.998334166$

$v (t)$ at different values of t
$t$	$v (t)$
$0$	$0$
$3.05$	$88$
$5.88$	$147.67$
$14.51$	$293.33$
$19.96$	$337.19$

Average acceleration
$t$	$\frac{v (t) - v (19.96)}{t - 19.96} = \frac{v (t) - 337.19}{t - 19.96}$
$0.0$	$16.89$
$3.05$	$14.74$
$5.88$	$13.46$
$14.51$	$8.05$

For the following exercises, use [link] to find the slope of the secant line between the values $x_{1}$

and $x_{2}$

for each function $y = f (x) .$

f (x) = 4 x + 7; x_{1} = 2, x_{2} = 5

4

f (x) = 8 x - 3; x_{1} = −1, x_{2} = 3

f (x) = x^{2} + 2 x + 1; x_{1} = 3, x_{2} = 3.5

8.5

f (x) = − x^{2} + x + 2; x_{1} = 0.5, x_{2} = 1.5

f (x) = \frac{4}{3 x - 1}; x_{1} = 1, x_{2} = 3

- \frac{3}{4}

f (x) = \frac{x - 7}{2 x + 1}; x_{1} = −2, x_{2} = 0

f (x) = \sqrt{x}; x_{1} = 1, x_{2} = 16

0.2

f (x) = \sqrt{x - 9}; x_{1} = 10, x_{2} = 13

f (x) = x^{1 / 3} + 1; x_{1} = 0, x_{2} = 8

0.25

f (x) = 6 x^{2 / 3} + 2 x^{1 / 3}; x_{1} = 1, x_{2} = 27

For the following functions,

use [link] to find the slope of the tangent line $m_{tan} = f^{'} (a),$
and
find the equation of the tangent line to $f$
at
$x = a .$

f (x) = 3 - 4 x, a = 2

a. $−4$

b. $y = 3 - 4 x$

f (x) = \frac{x}{5} + 6, a = −1

f (x) = x^{2} + x, a = 1

a. $3$

b. $y = 3 x - 1$

f (x) = 1 - x - x^{2}, a = 0

f (x) = \frac{7}{x}, a = 3

a. $\frac{−7}{9}$

b. $y = \frac{−7}{9} x + \frac{14}{3}$

f (x) = \sqrt{x + 8}, a = 1

f (x) = 2 - 3 x^{2}, a = −2

a. $12$

b. $y = 12 x + 14$

f (x) = \frac{−3}{x - 1}, a = 4

f (x) = \frac{2}{x + 3}, a = −4

a. $−2$

b. $y = −2 x - 10$

f (x) = \frac{3}{x^{2}}, a = 3

For the following functions $y = f (x),$

find $f^{'} (a)$

using [link].

f (x) = 5 x + 4, a = −1

5

f (x) = −7 x + 1, a = 3

f (x) = x^{2} + 9 x, a = 2

13

f (x) = 3 x^{2} - x + 2, a = 1

f (x) = \sqrt{x}, a = 4

\frac{1}{4}

f (x) = \sqrt{x - 2}, a = 6

f (x) = \frac{1}{x}, a = 2

- \frac{1}{4}

f (x) = \frac{1}{x - 3}, a = −1

f (x) = \frac{1}{x^{3}}, a = 1

−3

f (x) = \frac{1}{\sqrt{x}}, a = 4

For the following exercises, given the function $y = f (x),$

find the slope of the secant line $P Q$
for each point
$Q (x, f (x))$
with
$x$
value given in the table.
Use the answers from a. to estimate the value of the slope of the tangent line at $P .$
Use the answer from b. to find the equation of the tangent line to $f$
at point
$P .$

[T] $f (x) = x^{2} + 3 x + 4, P (1, 8)$

(Round to $6$

decimal places.)

x	Slope $m_{P Q}$

x	Slope $m_{P Q}$


{: valign=”top”}	———-
1.1	(i)	0.9	(vii)
1.01	(ii)	0.99	(viii)
1.001	(iii)	0.999	(ix)
1.0001	(iv)	0.9999	(x)
1.00001	(v)	0.99999	(xi)
1.000001	(vi)	0.999999	(xii)

a. $(i) 5.100000,$

(ii) 5.010000,

(iii) 5.001000,

(iv) 5.000100,

(v) 5.000010,

(vi) 5.000001,

(vii) 4.900000,

(viii) 4.990000,

(ix) 4.999000,

(x) 4.999900,

(xi) 4.999990,

(x) 4.999999

b. $m_{tan} = 5$

c. $y = 5 x + 3$

[T] $f (x) = \frac{x + 1}{x^{2} - 1}, P (0, −1)$

x	Slope $m_{P Q}$

x	Slope $m_{P Q}$


{: valign=”top”}	———-
0.1	(i)	$−0.1$

(vii)
0.01	(ii)	$−0.01$

(viii)
0.001	(iii)	$−0.001$

(ix)
0.0001	(iv)	$−0.0001$

(x)
0.00001	(v)	$−0.00001$

(xi)
0.000001	(vi)	$−0.000001$

(xii)

[T] $f (x) = 10 e^{0.5 x}, P (0, 10)$

(Round to $4$

decimal places.)

x	Slope $m_{P Q}$


{: valign=”top”}	———-
$−0.1$

(i)

−0.01

(ii)

−0.001

(iii)

−0.0001

(iv)

−0.00001

(v)
−0.000001	(vi)

a. $(i) 4.8771,$

(ii) 4.9875 (iii) 4.9988,

(iv) 4.9999,

(v) 4.9999,

(vi) 4.9999

b. $m_{tan} = 5$

c. $y = 5 x + 10$

[T] $f (x) = tan (x), P (π, 0)$

x	Slope $m_{P Q}$


{: valign=”top”}	———-
3.1	(i)
3.14	(ii)
3.141	(iii)
3.1415	(iv)
3.14159	(v)
3.141592	(vi)

[T] For the following position functions $y = s (t),$

an object is moving along a straight line, where $t$

is in seconds and $s$

is in meters. Find

the simplified expression for the average velocity from $t = 2$
to
$t = 2 + h;$
the average velocity between $t = 2$
and
$t = 2 + h,$
where
$(i) h = 0.1,$ $(ii) h = 0.01,$ $(iii) h = 0.001,$
and
$(iv) h = 0.0001;$
and
use the answer from a. to estimate the instantaneous velocity at $t = 2$
second.

s (t) = \frac{1}{3} t + 5

a. $\frac{1}{3};$

b. $(i) 0. \bar{3}$

m/s, $(ii) 0. \bar{3}$

m/s, $(iii) 0. \bar{3}$

m/s, $(iv) 0. \bar{3}$

m/s; c. $0. \bar{3} = \frac{1}{3}$

m/s

s (t) = t^{2} - 2 t

s (t) = 2 t^{3} + 3

a. $2 (h^{2} + 6 h + 12);$

b. $(i) 25.22$

m/s, $(ii) 24.12$

m/s, $(iii) 24.01$

m/s, $(iv) 24$

m/s; c. $24$

m/s

s (t) = \frac{16}{t^{2}} - \frac{4}{t}

Use the following graph to evaluate a. $f^{'} (1)$

and b. $f^{'} (6) .$

This graph shows two connected line segments: one going from (1, 0) to (4, 6) and the other going from (4, 6) to (8, 8).

a. $1.25;$

b. $0.5$

Use the following graph to evaluate a. $f^{'} (−3)$

and b. $f^{'} (1.5) .$

This graph shows two connected line segments: one going from (−4, 3) to (1, 3) and the other going from (1, 3) to (1.5, 4).

For the following exercises, use the limit definition of derivative to show that the derivative does not exist at $x = a$

for each of the given functions.

f (x) = x^{1 / 3}, x = 0

lim_{x \to 0^{-}} \frac{x^{1 / 3} - 0}{x - 0} = lim_{x \to 0^{-}} \frac{1}{x^{2 / 3}} = \infty

f (x) = x^{2 / 3}, x = 0

f (x) = {\begin{matrix} 1, x < 1 \\ x, x \geq 1 \end{matrix}, x = 1

lim_{x \to 1^{-}} \frac{1 - 1}{x - 1} = 0 \neq 1 = lim_{x \to 1^{+}} \frac{x - 1}{x - 1}

f (x) = \frac{\| x \|}{x}, x = 0

[T] The position in feet of a race car along a straight track after $t$

seconds is modeled by the function $s (t) = 8 t^{2} - \frac{1}{16} t^{3} .$

Find the average velocity of the vehicle over the following time intervals to four decimal places:
1. [4, 4.1]
2. [4, 4.01]
3. [4, 4.001]
4. [4, 4.0001]
Use a. to draw a conclusion about the instantaneous velocity of the vehicle at $t = 4$
seconds.

a. $(i) 61.7244$

ft/s, $(ii) 61.0725$

ft/s $(iii) 61.0072$

ft/s $(iv) 61.0007$

ft/s b. At $4$

seconds the race car is traveling at a rate/velocity of $61$

ft/s.

[T] The distance in feet that a ball rolls down an incline is modeled by the function $s (t) = 14 t^{2},$

where t is seconds after the ball begins rolling.

Find the average velocity of the ball over the following time intervals:
1. [5, 5.1]
2. [5, 5.01]
3. [5, 5.001]
4. [5, 5.0001]
Use the answers from a. to draw a conclusion about the instantaneous velocity of the ball at $t = 5$
seconds.

Two vehicles start out traveling side by side along a straight road. Their position functions, shown in the following graph, are given by $s = f (t)$

and $s = g (t),$

where $s$

is measured in feet and $t$

is measured in seconds.

Two functions s = g(t) and s = f(t) are graphed. The first function s = g(t) starts at (0, 0) and arcs upward through roughly (2, 1) to (4, 4). The second function s = f(t) is a straight line passing through (0, 0) and (4, 4).

Which vehicle has traveled farther at $t = 2$
seconds?
What is the approximate velocity of each vehicle at $t = 3$
seconds?
Which vehicle is traveling faster at $t = 4$
seconds?
What is true about the positions of the vehicles at $t = 4$
seconds?

a. The vehicle represented by $f (t),$

because it has traveled $2$

feet, whereas $g (t)$

has traveled $1$

foot. b. The velocity of $f (t)$

is constant at $1$

ft/s, while the velocity of $g (t)$

is approximately $2$

ft/s. c. The vehicle represented by $g (t),$

with a velocity of approximately $4$

ft/s. d. Both have traveled $4$

feet in $4$

seconds.

[T] The total cost $C (x),$

in hundreds of dollars, to produce $x$

jars of mayonnaise is given by $C (x) = 0.000003 x^{3} + 4 x + 300 .$

Calculate the average cost per jar over the following intervals:
1. [100, 100.1]
2. [100, 100.01]
3. [100, 100.001]
4. [100, 100.0001]
Use the answers from a. to estimate the average cost to produce $100$
jars of mayonnaise.

[T] For the function $f (x) = x^{3} - 2 x^{2} - 11 x + 12,$

do the following.

Use a graphing calculator to graph f in an appropriate viewing window.
Use the ZOOM feature on the calculator to approximate the two values of $x = a$
for which
$m_{tan} = f^{'} (a) = 0 .$

a.* * *

The function starts in the third quadrant, passes through the x axis at x = −3, increases to a maximum around y = 20, decreases and passes through the x axis at x = 1, continues decreasing to a minimum around y = −13, and then increases through the x axis at x = 4, after which it continues increasing.

b. $a \approx - 1.361, 2.694$

[T] For the function $f (x) = \frac{x}{1 + x^{2}},$

do the following.

Use a graphing calculator to graph $f$
in an appropriate viewing window.
Use the ZOOM feature on the calculator to approximate the values of $x = a$
for which
$m_{tan} = f^{'} (a) = 0 .$

Suppose that $N (x)$

computes the number of gallons of gas used by a vehicle traveling $x$

miles. Suppose the vehicle gets $30$

mpg.

Find a mathematical expression for $N (x) .$
What is $N (100)?$
Explain the physical meaning.
What is $N^{'} (100) ?$
Explain the physical meaning.

a. $N (x) = \frac{x}{30}$

b. $\sim 3.3$

gallons. When the vehicle travels $100$

miles, it has used $3.3$

gallons of gas. c. $\frac{1}{30} .$

The rate of gas consumption in gallons per mile that the vehicle is achieving after having traveled $100$

miles.

[T] For the function $f (x) = x^{4} - 5 x^{2} + 4,$

do the following.

Use a graphing calculator to graph $f$
in an appropriate viewing window.
Use the $nDeriv$
function, which numerically finds the derivative, on a graphing calculator to estimate
$f^{'} (−2), f^{'} (−0.5), f^{'} (1.7),$
and
$f^{'} (2.718) .$

[T] For the function $f (x) = \frac{x^{2}}{x^{2} + 1},$

do the following.

Use a graphing calculator to graph $f$
in an appropriate viewing window.
Use the $nDeriv$
function on a graphing calculator to find
$f^{'} (−4), f^{'} (−2), f^{'} (2),$
and
$f^{'} (4) .$

a.* * *

The function starts in the second quadrant and gently decreases, touches the origin, and then it increases gently.

b. $−0.028, −0.16, 0.16, 0.028$

Defining the Derivative

Tangent Lines

The Derivative of a Function at a Point

Velocities and Rates of Change

Key Concepts

Key Equations

Glossary