3.1The Derivative

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Кузбасский государственный технический университет

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Chapter 3

Di erentiation

In Deﬁnition 2.2.2, we deﬁned the slope function of a function f at c by

slope(f(x), c) = lim	f(x) − f(c)
slope(f(x), c) = lim	x − c
x→c	x − c
= lim	f(c + h) − f(c)	.
h→0	h

The slope (f(x), c) is called the derivative of f at c and is denoted f0(c). Thus,

f0(c) = lim f(c + h) − f(c).

h→0 h

Link to another ﬁle.

(kf)0(x) = lim

kf(x + h) − kf(x)

f(x + h) − f(x)

Deﬁnition 3.1.1 Let f be deﬁned on a closed interval [a, b] and a < x < b. Then the derivative of f at x, denoted f0(x), is deﬁned by

f0(x) = lim f(x + h) − f(x)

h→0 h

whenever the limit exists. When f0(x) exists, we say that f is di erentiable at x. At the end points a and b, we deﬁne one-sided derivatives as follows:

(i) f0(a+) = lim	f(x) − f(a)	=	lim	f(a + h) − f(a)	.
x→a+	x − a		h→0+	h

100				CHAPTER 3. DIFFERENTIATION
We call f0	(a+) the right-hand derivative of f at a.
(ii) f0(b−) = lim		f(x) − f(b)	=	lim	f(b + h) − f(b)	.
We call f0	x→b−	x − b		h→0−	h
	(b) the left-hand derivative of f at b.

Example 3.1.1 In Example 28 of Section 2.2, we proved that if f(x) = sin x, then f0(c) = slope (sin x, c) = cos c. Thus, f0(x) = cos x if f(x) = sin x.

Example 3.1.2 In Example 29 of Section 2.2, we proved that if f(x) = cos x, then f0(c) = − sin c. Thus, f0(x) = − sin x if f(x) = cos x.

Example 3.1.3 In Example 30 of Section 2.2, we proved that if f(x) = xn for a natural number n, then f0(c) = ncn−1. Thus f0(x) = nxn−1, when f(x) = xn, for any natural number n.

In order to ﬁnd derivatives of functions obtained from the basic elementary functions using the operations of addition, subtraction, multiplication and division, we state and prove the following theorem.

Theorem 3.1.1 If f is di erentiable at c, then f is continuous at c. The converse is false.

Proof. Suppose that f is di erentiable at c. Then

lim

f(x) − f(c)

= f0(c)

x→c

x − c

and f0(c) is a real number. So,

− c

x→c

−

lim f(x) = lim

f(x)

f(c)

c) + f(c)

−

= lim

f(x) − f(c)

lim(x

c) + f(c)

→

−

· x

→

−

=f0(c) · 0 + f(c)

=f(c).

3.1. THE DERIVATIVE

101

Therefore, if f is di erentiable at c, then f is continuous at c.

To prove that the converse is false we consider the function f(x) = |x|.

This function is continuous at x = 0. But

( ) = h→0

h| − | |

f0 x

lim

x + h

(|x + h| − |x|)(|x + h| + |x|)

= lim

h→0

| |

)

h( x + h + x

= lim

x2 + 2xh + h2 − x2

h→0

h( x + h +

)

= lim

2x + h

x + h

h→0

|x|

for x > 0

−1

for x < 0

undeﬁned for x = 0.

Thus, |x| is continuous at 0 but not di erentiable at 0. This completes the proof of Theorem 3.1.1.

Theorem 3.1.2 Suppose that functions f and g are deﬁned on some open interval (a, b) and f0(x) and g0(x) exist at each point x in (a, b). Then

(i)	(f + g)0(x) = f0(x) + g0(x)						(The Sum Rule)
(ii)	(f − g)0(x) = f0(x) − g0(x)						(The Di erence Rule)
(iii)	(kf)0(x) = kf0(x), for each constant k.						(The Multiple Rule)
(iv)	(f · g)0(x) = f0(x) · g(x) + f(x) · g0(x)						(The Product Rule)
(v)		f	0	(x) =	g(x)f0(x) − f(x)g0(x)	, if g(x) = 0.	(The Quotient Rule)
	g				(g(x))2
						6

Proof.

102

Part (i)

Part (ii)

Part (iii)

Part (iv)

	CHAPTER 3. DIFFERENTIATION
(f + g)0(x) = lim	[f(x + h) + g(x + h)] − [f(x) + g(x)]
h→0		h
= lim	f(x + h) − f(x)	+ lim	g(x + h) − g(x)
= lim	h	+ lim	h
h→0	h	h→0	h

= f0(x) + g0(x).

(f	−	g)0(x) = lim			[f(x + h) − g(x + h)] − [f(x) − g(x)]
	−	h	→	0		h
			→
		= lim			f(x + h) − f(x)	lim	g(x + h) − g(x)
		h→0			h	− h→0	h
		= f0(x) − g0(x).

= kf0(x).

g)0(x) = lim

f(x + h)g(x + h) − f(x)g(x)

→

lim

f x

g x

f x

g x

= h

→

[( ( +

) −

(

))

( +

) +

(

)( (

) −

( ))]

f(x + h) − f(x)

g(x + h) − g(x)

= lim

lim

g(x + h) + f(x) lim

→

= f0(x)g(x) + f(x)g0(x).

3.1. THE DERIVATIVE

103

Part (v)

) = h→0

1 f(x + h)

f(x)

(

g(x + h)

− g(x)

lim

g·(x + h)g(x)

h→0

= lim

f(x + h) g(x) − g(x + h)f(x)

= (g(x))2

h→0

h −

( ) −

( )

h −

lim

(f(x + h)

f(x))

g x

f x

(g(x + h)

g(x))

· [f0(x)g(x) − f(x)g0(x)]

(g(x))2

= g(x)f0(x) − g(x)g0(x) , if g(x) 6= 0. (g(x))2

To emphasize the fact that the derivatives are taken with respect to the independent variable x, we use the following notation, as is customary:

f0(x) = dxd (f(x)).

Based on Theorem 3.1.2 and the deﬁnition of the derivative, we get the following theorem.

Theorem 3.1.3

(i)ddx(k) = 0, where k is a real constant.

(ii)dxd (xn) = nxn−1, for each real number x and natural number n.

(iii)dxd (sin x) = cos x, for all real numbers (radian measure) x.

(iv)dxd (cos x) = − sin x, for all real numbers (radian measure) x.

(v)dxd (tan x) = sec2 x, for all real numbers x 6= (2n + 1) π2 , n = integer.

104	CHAPTER 3. DIFFERENTIATION

(vi)dxd (cot x) = − csc2 x, for all real numbers x 6= nπ, n = integer.

(vii)dxd (sec x) = sec x tan x, for all real numbers x 6= (2n+1)π2 , n = integer.

(viii)dxd (csc x) = − csc x cot x, for all real numbers x 6= nπ, n = integer.

Proof.

Part(i)

d(k)

(k) = lim

k − k

h→0

= lim

h→0

= 0.

Part (ii) For each natural n, we get

(xn) = lim

(x + h)n − xn

(Binomial Expansion)

h→0

n(n − 1)

= lim

xn + nxn−1h +

xn−2h2 +

· · ·

+ hn

−

h→0

= lim

nxn−1 +

n(n − 1)

xn−2h +

· · ·

+ hn−1

h→0

= nxn−1.

Part (iii) By deﬁnition, we get

(sin x) = lim

sin(x + h) − sin x

h→0

= lim

sin x cos h + cos x sin h − sin x

h→0

sin h

cos h

lim

sin x

cos

−

−h

= h→0

=cos x · 1 − sin x · 0

=cos x

3.1. THE DERIVATIVE					105
since
lim	sin h	= 1,	lim	1 − cos h	= 0. (Why?)
	h			h
h→0			h→0

Part (iv) By deﬁnition, we get

(cos x) = lim

cos(x + h) − cos x

h→0

lim

− sin

− cos

= h

→

[cos

cos

sin

]

− sin

−

−h

= h→0

lim

sin h

cos x

cos h

= − sin x · 1 − cos x · 0

(Why?)

= − sin x.

Part (v) Using the quotient rule and parts (iii) and (iv), we get


d	(tan x) =	d		sin x
dx	(tan x) =	dx cos x
dx		dx cos x
	=	cos x(sin x)0 − sin x(cos x)0
				(cos x)2
	=	cos2 x + sin2 x
	=
			cos2 x
	1
	=			(Why?)
	=	cos2 x		(Why?)	π
	= sec2 x,			x 6= (2n + 1)	π	, n = integer.

					2

106						CHAPTER 3.		DIFFERENTIATION
Part (vi) Using the quotient rule and Parts (iii) and (iv), we get
	d	(cot x) =	d		cos x

	dx		dx		sin x
		=	(sin x)(cos x)0 − (cos x)(sin x)0
							(sin x)2
		=	− sin2 x − cos2 x					(Why?)
				(sin x)2
		=	−1				(why?)
		=	(sin x)2				(why?)
			(sin x)2
		= − csc2 x,					x 6= nπ, n =	integer.

Part (vii) Using the quotient rule and Parts (iii) and (iv), we get

(sec x) =

cos x

(cos x) · 0 − 1 · (cos x)0

(cos x)2

sin x

(Why?)

cos x

= sec x tan x,

x 6= (2n + 1)

, n = integer.

Part (viii) Using the quotient rule and Parts (iii) and (iv), we get

d
dx	(csc x) =	dx	sin x
dx		dx	sin x
	=	sin x · 0 − 1 · (sin x)0
			(sin x)2
	=	1	− cos x	(Why?)
	=	sin x	· sin x	(Why?)
		sin x	· sin x

= − csc x cot x, x 6= nπ, n = integer.

This concludes the proof of Theorem 3.1.3.

3.1. THE DERIVATIVE

107

Example 3.1.4 Compute the following derivatives:

(i)

(4x3 − 3x2 + 2x + 10)

(ii)

(4 sin x − 3 cos x)

+ 1

(iii)

(x sin x + x2 cos x)

(iv)

+ 4

Part (i)

Using the sum, di erence and constant multiple rules, we get

(4x3 − 3x2 + 2x + 10) = 4

(x3) − 3

(x2) + 2

+ 0

= 12x2 − 6x + 2.

Part (ii)	d	(4 sin x − 3 cos x) = 4	d	(sin x) − 3	d	(cos x)

	dx		dx		dx

=4 cos x − 3(− sin x)

=4 cos x + 3 sin x.

Part (iii) Using the sum and product rules, we get

d dx

(x sin x + x2 cos x) =

(x sin x) +

(x2 cos x)

(Sum Rule)

sin x + x

(sin x)

(cos x)

(x2) cos x + x2

= 1 · sin x + x cos x + 2x cos x + x2(− sin x) = sin x + 3x cos x − x2 sin x.

108

CHAPTER 3.

DIFFERENTIATION

Part (iv). Using the sum and quotient rules, we get

x3 + 1

(x2 + 4)

(x3 + 1) − (x3 + 1)

(x2 + 4)

(Why?)

x2 + 4

(x2 + 4)2

(x2 + 4)(3x2) − (x3 + 1) = x

(Why?)

(x2 + 4)2

3x4 + 12x2 − 2x3 − 2x

(Why?)

(x2 + 4)2

3x4 − 2x3 + 12x2 − 2x

(x2 + 4)2

Exercises 3.1

From the deﬁnition, prove that

(x3) = 3x2.

From the deﬁnition, prove that

−1

Compute the following derivatives:

(x5 − 4x2 + 7x − 2)

(4 sin x + 2 cos x − 3 tan x)

2x + 1

x4 + 2

x2 + 1

3x + 1

(3x sin x + 4x2 cos x)

(4 tan x − 3 sec x)

(3 cot x + 5 csc x)

10.

(x2 tan x + x cot x)

Recall that the equation of the line tangent to the graph of f at (c, f(c)) has slope f0(c) and equations.

Tangent Line: y − f(c) = f0(c)(x − c)

The normal line has slope −1/f0(c), if f0(c) 6= 0 and has the equation:

3.1. THE DERIVATIVE							109
Normal Line:	y	−	f(c) =	−1	(x	−	c).
				f0(c)

In each of the following, ﬁnd the equation of the tangent line and the equation of the normal line for the graph of f at the given c.

11.	f(x) = x3 + 4x − 12, c = 1		12.	f(x) = sin x, c = π/6
13.	f(x) = cos x,	c = π/3	14.	f(x) = tan x, c = π/4
15.	f(x) = cot x,	c = π/4	16.	f(x) = sec x, c = π/3
17.	f(x) = csc x, c = π/6		18.	f(x) = 3 sin x + 4 cos x, c = 0.

Recall that Newton’s Method solves f(x) = 0 for x by using the ﬁxed point iteration algorithm:

f(xn)

xn+1 = g(xn) = xn − f0(xn) , x0 = given, with the stopping rule, for a given natural number n,

|xn+1 − xn| < 10−n.

In each of the following, set up Newton’s Iteration and perform 3 calculations for a given x0.

19.f(x) = 2x − cos x , x0 = 0.5

20.f(x) = x3 + 2x + 1 , x0 = −0.5

21.f(x) = x3 + 3x2 − 1 = 0, x0 = 0.5

22.Suppose that f0(c) exists. Compute each of the following limits in terms of f0(c)

(a)	lim	f(x) − f(c)
		x − c
	x→c
(c)	lim	f(c − h) − f(c)
		h
	h→0
(e)	lim	f(c + h) − f(c − h)
		2h
	h→0

(b) lim

h→0

(d) lim

t→c

(f) lim

h→0

f(c + h) − f(c) h

f(c) − f(t) t − c

f(c + 2h) − f(c − 2h) h

110		CHAPTER 3. DIFFERENTIATION
23. Suppose that g is di erentiable at c and
(g0		(c−)	if t = c.
	g(t)−g(c)		if t 6= c
f(t) =		t c

Show that f is continuous at c.

Suppose that a business produces and markets x units of a commercial item. Let

C(x) = The total cost of producing x-units.

p(x) = The sale price per item when x-units are on the market.

R(x) = xp(x) = The revenue for selling x-units.

P (x) = R(x) − C(x) = The gross proﬁt for selling x-items.

C0(x) = The marginal cost.

R0(x) = The marginal revenue.

P 0(x) = The marginal proﬁt.

In each of the problems 24–26, use the given functions C(x) and p(x) and compute the revenue, proﬁt, marginal cost, marginal revenue and marginal proﬁt.

24.C(x) = 100x − (0.2)x2, 0 ≤ x ≤ 5000, p(x) = 10 − x

25.C(x) = 5000 + x2 , 1 ≤ x ≤ 5000, p(x) = 20 + x1

26.C(x) = 1000 + 4x − 0.1x2, 1 ≤ x ≤ 2000, p(x) = 10 − x1

In exercises 27–60, compute the derivative of the given function.

27.	f(x) = 4x3	− 2x2 + 3x − 10	28.	f(x) = 2 sin x − 3 cos x + 4
29.	f(x) = 3 tan x − 4 sec x		30.	f(x) = 2 cot x + 3 csc x
31.	f(x) = 2x2	+ 4x + 5	32.	f(x) = x2/3 − 4x1/3 + 5
33.	f(x) = 3x−4/3 + 3x−2/3 + 10		34.	f(x) = 2√		+ 4
					x

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