facts together with the quotient rule from above:

cos x(d/dx) sin x ’ sin x(d/dx) cos x

d sin x

d

3. tan x = =

(cos x)2

dx cos x

dx

(cos x)2 + (sin x)2 1

= = = (sec x)2 .

(cos x)2 (cos x)2

Similarly we have

d

cot x = ’(csc x)2 .

4.

dx

d

sec x = sec x tan x.

5.

dx

d

csc x = ’ csc x cot x.

6.

dx

C Derivatives of ln x and ex : We conclude our library of derivatives of basic

functions with

dx

e = ex

dx

and

1

d

ln x = .

dx x

We may apply the Chain Rule to obtain the following particularly useful general-

ization of this logarithmic derivative:

d • (x)

ln •(x) = .

dx •(x)

Now it is time to learn to differentiate the functions that we will commonly

encounter in our work. We do so by applying the rules for sums, products, quotients,

and compositions to the formulas for the derivatives of the elementary functions.

73

CHAPTER 2 Foundations of Calculus

Practice is the essential tool in mastery of these ideas. Be sure to do all the You Try

It problems in this section.

EXAMPLE 2.14

Calculate the derivative

d

[(sin x + x) · (x 3 ’ ln x)].

dx

SOLUTION

We know that (d/dx) sin x = cos x, (d/dx)x = 1, (d/dx)x 3 = 3x 2 , and

(d/dx) ln x = (1/x). Therefore, by the addition rule,

d d d

(sin x + x) = sin x + x = cos x + 1

dx dx dx

and

1

d3 d3 d

(x ’ ln x) = x’ ln x = 3x 2 ’ .

dx dx dx x

Now we may conclude the calculation by applying the product rule:

d

(sin x + x) · (x 3 ’ ln x)

dx

d d3

= (sin x + x) · (x 3 ’ ln x) + (sin x + x) · (x ’ ln x)

dx dx

1

= (cos x + 1) · (x 3 ’ ln x) + (sin x + x) · 3x 2 ’

x

1

= (4x 3 ’ 1) + x 3 cos x + 3x 2 sin x ’ sin x ’ (ln x cos x + ln x).

x

EXAMPLE 2.15

Calculate the derivative

ex + x sin x

d

.

tan x

dx

SOLUTION

We know that (d/dx)ex = ex , (d/dx)x = 1, (d/dx) sin x = cos x, and

(d/dx) tan x = sec2 x. By the product rule,

d d d

[x · sin x] = x · sin x + x · sin x = 1 · sin x + x · cos x.

dx dx dx

74 CHAPTER 2 Foundations of Calculus

Therefore, by the quotient rule,

ex + x sin x tan x · (d/dx)(ex + x sin x) ’ (ex + x sin x)(d/dx) tan x

d

=

(tan x)2

tan x

dx

tan x · (ex + sin x + x cos x) ’ (ex + x sin x) · (sec x)2

=

(tan x)2

ex tan x + tan x sin x + x sin x ’ ex sec2 x ’ x sin x sec2 x

= .

tan2 x

d x

sin x · cos x ’

You Try It: Calculate the derivative .

ex + ln x

dx

EXAMPLE 2.16

Calculate the derivative

d

(sin(x 3 ’ x 2 )).

dx

SOLUTION

This is the composition of functions, so we must apply the Chain Rule. It is

essential to recognize what function will play the role of f and what function

will play the role of g.

Notice that, if x is the variable, then x 3 ’ x 2 is applied ¬rst and sin

applied next. So it must be that g(x) = x 3 ’ x 2 and f (s) = sin s. Notice that

(d/ds)f (s) = cos s and (d/dx)g(x) = 3x 2 ’ 2x. Then

sin(x 3 ’ x 2 ) = f —¦ g(x)

and

d d

(sin(x 3 ’ x 2 )) = (f —¦ g(x))

dx dx

df d

= (g(x)) · g(x)

ds dx

= cos(g(x)) · (3x 2 ’ 2x)

= [cos(x 3 ’ x 2 )] · (3x 2 ’ 2x).

EXAMPLE 2.17

Calculate the derivative

x2

d

ln .

x’2

dx

75

CHAPTER 2 Foundations of Calculus

SOLUTION

Let

x2

h(x) = ln .

x’2

Then

h = f —¦ g,

where f (s) = ln s and g(x) = x 2 /(x ’ 2). So (d/ds)f (s) = 1/s and

(d/dx)g(x) = (x ’ 2) · 2x ’ x 2 · 1/(x ’ 2)2 = (x 2 ’ 4x)/(x ’ 2)2 . As a

result,

d d

h(x) = (f —¦ g)(x)

dx dx

df d

= (g(x)) · g(x)

ds dx

x 2 ’ 4x

1

= ·

g(x) (x ’ 2)2

x 2 ’ 4x

1

=2 ·

x /(x ’ 2) (x ’ 2)2

x’4

= .

x(x ’ 2)

You Try It: Perform the differentiation in the last example by ¬rst applying a rule

of logarithms to simplify the function to be differentiated.

You Try It: Calculate the derivative of tan(ex ’ x).

EXAMPLE 2.18

2

Calculate the tangent line to the graph of f (x) = x · ex at the point (1, e).

SOLUTION

The slope of the tangent line will be the derivative. Now

2 2 2 2

f (x) = [x] · ex + x · ex = ex + x · 2x · ex .

In the last derivative we have of course used the Chain Rule. Thus f (1) =

e + 2e = 3e. Therefore the equation of the tangent line is

(y ’ e) = 3e(x ’ 1).

You Try It: Calculate the equation of the tangent line to the graph of g(x) =

cos((x 2 ’ 2)/ln x) at the point (2, cos[2/ ln 2]).

76 CHAPTER 2 Foundations of Calculus

Math Note: Calculate (d/dx)(x 2 /x) using the quotient rule. Of course x 2 /x = x,

and you may calculate the derivative directly. Observe that the two answers are

the same. The calculation con¬rms the validity of the quotient rule by way of an

example. Use a similar example to con¬rm the validity of the product rule.

2.5.1 THE DERIVATIVE OF AN INVERSE

An important formula in the calculus relates the derivative of the inverse of a

function to the derivative of the function itself. The formula is

1

[f ’1 ] (t) = ()

.

f (f ’1 (t))

We encourage you to apply the Chain Rule to the formula f (f ’1 (x)) = x to obtain

a formal derivation of the formula ( ).

EXAMPLE 2.19

Calculate the derivative of g(t) = t 1/3 .

SOLUTION

Set f (s) = s 3 and apply formula ( ). Then f (s) = 3s 2 and f ’1 (t) = t 1/3 .

With s = f ’1 (t) we then have

1 1 1 1

[f ’1 ] (t) = = · t ’2/3 .

= =

f (f ’1 (t)) 3 · [t 1/3 ]2

3s 2 3

Formula ( ) may be applied to obtain some interesting new derivatives to add

to our library. We record some of them here:

1

d

arcsin x = √

I.

1 ’ x2

dx

1

d

arccos x = ’ √

II.

1 ’ x2

dx

1

d

arctan x =

III.

1 + x2

dx

√

You Try√ Calculate the derivative of f (x) =

It: x. Calculate the derivative of

g(x) = x for any k ∈ {2, 3, 4, . . . }.

k

2.6 The Derivative as a Rate of Change

If f (t) represents the position of a moving body, or the amount of a changing quan-

tity, at time t, then the derivative f (t) (equivalently, (d/dt)f (t)) denotes the rate

of change of position (also called velocity) or the rate of change of the quantity.

77

CHAPTER 2 Foundations of Calculus

When f (t) represents velocity, then sometimes we calculate another derivative”

(f ) (t)”and this quantity denotes the rate of change of velocity, or acceleration.

In specialized applications, even more derivatives are sometimes used. For

example, sometimes the derivative of the acceleration is called jerk and sometimes

the derivative of jerk is called surge.

EXAMPLE 2.20

The position of a body moving along a linear track is given by p(t) = 3t 2 ’

5t + 7 feet. Calculate the velocity and the acceleration at time t = 3 seconds.

SOLUTION

The velocity is given by

p (t) = 6t ’ 5.

At time t = 3 we therefore ¬nd that the velocity is p (3) = 18 ’ 5 = 13 ft/sec.

The acceleration is given by the second derivative:

p (t) = (p ) (t) = (6t ’ 5) = 6.

The acceleration at time t = 3 is therefore 6 ft/sec2 .

Math Note: As previously noted, velocity is measured in feet per second (or

ft/sec). Acceleration is the rate of change of velocity with respect to time; therefore

acceleration is measured in “feet per second per second” (or ft/sec2 ).

EXAMPLE 2.21

A massive ball is dropped from a tower. It is known that a falling body

descends (near the surface of the earth) with an acceleration of about 32

ft /sec. From this information one can determine that the equation for the

position of the ball at time t is

p(t) = ’16t 2 + v0 t + h0 ft.

Here v0 is the initial velocity and h0 is the initial height of the ball in feet.1

Also t is time measured in seconds. If the ball hits the earth after 5 seconds,

then determine the height from which the ball is dropped.

SOLUTION

Observe that the velocity is

v(t) = p (t) = ’32t + v0 .

Obviously the initial velocity of a falling body is 0. Thus

0 = v(0) = ’32 · 0 + v0 .

1 We shall say more about this equation, and this technique, in Section 3.4.

78 CHAPTER 2 Foundations of Calculus

It follows that v0 = 0, thus con¬rming our intuition that the initial velocity

is 0. Thus

p(t) = ’16t 2 + h0 .

Now we also know that p(5) = 0; that is, at time t = 5 the ball is at height 0.

Thus

0 = p(5) = ’16 · 52 + h0 .

We may solve this equation for h0 to determine that h0 = 400.

We conclude that

p(t) = ’16t 2 + 400.

Furthermore, p(0) = 400, so the initial height of the ball is 400 feet.

You Try It: Suppose that a massive ball falls from a height of 600 feet. After how

many seconds will it strike the ground?

Y

FL

Exercises

AM

1. Calculate, if possible, each of these limits. Give reasons for each step of

your solution.

lim x · ex

(a)

TE

x’0

x2 ’ 1

(b) lim

x’1 x ’ 1

lim (x ’ 2) · cot(x ’ 2)

(c)

x’2

lim x · ln x

(d)

x’0

t 2 ’ 7t + 12

(e) lim

t ’3

t’3

s 2 ’ 3s ’ 4

(f) lim

s’4

s’4

ln x

(g) lim

x’1 x ’ 1

x2 ’ 9

(h) lim

x’’3 x + 3

79

CHAPTER 2 Foundations of Calculus

2. Determine whether the given function f is continuous at the given point c.

Give careful justi¬cations for your answers.

x’1

f (x) = c = ’1

(a)

x+1

x’1

f (x) = c=3

(b)

x+1

f (x) = x · sin(1/x) c=0

(c)

f (x) = x · ln x c=0

(d)

if x ¤ 1

x2

f (x) = c=1

(e)

if 1 < x

x

if x ¤ 1

x2

f (x) = c=1

(f)

2x if 1 < x

if x ¤ π

sin x

f (x) = c=π

(g)

(x ’ π ) if π < x

f (x) = eln x+x c=2

(h)

3. Use the de¬nition of derivative to calculate each of these derivatives.

f (2) when f (x) = x 2 + 4x

(a)

f (1) when f (x) = ’1/x 2

(b)

4. Calculate each of these derivatives. Justify each step of your calculation.

x

(a)

x2 + 1

d

sin(x 2 )

(b)

dx

d

t · tan(t 3 ’ t 2 )

(c)

dt