講義請點此

2.1 The Tangent and Velocity Problem goo.gl/gD8vfy 1

Chapter 2

Limits and Derivatives

2.1

The Tangent and Velocity Problem, page 78

cRZjFuNIsKs

Question 1. What will we learn in the Calculus course?

The tangent problem, page 78

Example 2. Plot the parabola f (x) = x2_{. Observe all secant lines (割線) passing through}

the point P (1, f (1)) and Q∆x(1 + ∆x, f (1 + ∆x)), where ∆x 6= 0 is a number close to 0.

x y

1

Figure 1: The parabola f (x) = x2 _{and secant lines passing through P (1, 1).}

Solution. We can compute the slope of secant line LP Q∆x to get

mP Q∆x =

=

So the equation of secant line LP Q∆x is . When ∆x is close to 0,

the slope mP Q∆x is close to 2. That means the family of secant lines LP Q∆x is close to the

line y − 1 = 2(x − 1), which passes through P (1, f (1)) and the slope is 2. We call y − 1 = 2(x − 1) the tangent line (切線) of f (x) = x2 at x = 1.

2 2.1 The Tangent and Velocity Problem goo.gl/gD8vfy

The velocity problem, page 80

WeT8xTC5zS8





Example 3. Suppose that a ball is dropped from the upper observation deck of Taipei 101. Find the velocity of the ball after 5 seconds.

Solution. If the distance fallen after t seconds is denoted by s(t) and measured in meters, then Galileo’s law is expressed by the equation

s(t) = 1

2 · 9.8 · t

2_{= 4.9t}2_.

We can approximate the velocity at instant time t = 5 by computing the average velocity over the brief time interval

average velocity = change in position time elapsed = s(5 + 10−n_{) − s(5)} (5 + 10−n_{) − 5} = 4.9 · ((5 + 10−n) 2_{− 5}2₎ 10−n = 4.9 · (5 + 10−n_{+ 5)(5 + 10}−n_{− 5)} 10−n = 4.9 · (10 + 10−n) = 49 + 4.9 · 10−n. That is,

Time interval Average velocity (m/s) 5 ≤ t ≤ 5.1 49.49

5 ≤ t ≤ 5.01 49.049 5 ≤ t ≤ 5.001 49.0049 5 ≤ t ≤ 5.0001 49.00049 5 ≤ t ≤ 5.00001 49.000049

It appears that as we shorten the time period, the average velocity is becoming closer to 49 m/s. The instantaneous velocity (瞬時速度) when t = 5 is defined to be the limiting value of these average velocities over shorter an shorter time periods that start at t = 5. Thus the instantaneous velocity after 5 second is v = 49 m/s.

Remark 4. Time periods 10−n we choose in Example 3 are just some samples. In general, we can use ∆t to represent any time interval and do the same calculation to get the average velocity form 5 to 5 + ∆t is 4.9 · (10 + ∆t). The average velocity is becoming closer to 49 m/s as well when we shorten the time period.

2.2 The Limit of a Function goo.gl/EqF4nq 3

2.2

The Limit of a Function, page 83

(One sided) Limit

qI6ZLkjJq1U

Definition 1 (page 88). We write

lim

x_→a−f (x) = L

and say the left-hand limit of f (x) as x approaches a (or the limit of f (x) as x approaches a from the left) (左極限) is equal to L if we can make the values of f (x) arbitrarily close to L by taking x to be sufficiently close to a and x less than a.

Similarly, if we require that x be greater than a, we get “the right-hand limit of f (x) as x approaches a (右極限) is equal to L” and we write lim

x→a+f (x) = L.



Figure 1: Left-hand limit and right-hand limit.

Definition 2 (The limit of a function, page 83). Suppose f (x) is defined when x is near the number a. Then we write lim

x_→af (x) = L if we can make the value of f (x) arbitrarily close to

L by taking x to be sufficiently close to a but not equal to a.







Figure 2: Limit of a function.



x→af (x) = L 若且唯若(if and only if) limx_→a−f (x) = L

且 lim

x_→a+f (x) = L

4 2.2 The Limit of a Function goo.gl/EqF4nq

Example 3. Find the limit of the Heaviside function at x = 0.

NaXDbWd6dOk

1

x y

Figure 3: The Heaviside function H(x).

Solution. lim

x_→0−

H(x) = ; lim

x_→0+H(x) = ; lim_x→0H(x) .

Example 4. The graph of a function f (x) is shown in Figure 4. Use it to state the values (if they exist) of the following:

x y 1 1 2 2 3 3

Figure 4: The graph of f (x).

(a1) lim x→1−f (x) (b1) lim_x →1+f (x) (c1) lim_x→1f (x) (d1) f (1) (a2) lim x_→2−f (x) (b2) lim_x →2+f (x) (c2) lim_x→2f (x) (d2) f (2) (a3) lim x→3−f (x) (b3) lim_x →3+f (x) (c3) lim_x→3f (x) (d3) f (3)

Example 5. Observe the function f (x) = sin x_x and guess the value of lim

x→0 sin x x . 1 x y π 2π f(x) = sin x x

2.2 The Limit of a Function goo.gl/EqF4nq 5

Example 6. Guess the limit lim

x_→0sin 1

x
and lim_x→0x sin 1 x
. ESfzxFDYkMw 1 π 1 π _x x y y f(x) = sin _x1
g(x) = x sin 1_x

Figure 6: The graph of f (x) = sin 1_x
and g(x) = x sin 1 x
.



Infinite Limits

Definition 7 (page 89). Let f be a function defined on both sides of a, except possibly at a itself. Then

lim

x→af (x) = ∞

means that the values of f (x) can be made arbitrarily large (as large as we please) by taking x sufficiently close to a, but not equal to a.

Definition 8 (page 94). Let f be a function defined on both sides of a, except possibly at a itself. Then lim

x→af (x) = −∞ means that the values of f (x) can be made arbitrarily negative

by taking x sufficiently close to a, but not equal to a.

a a x x y y

Figure 7: Infinite limit lim

x→af(x) = ∞ and limx→af(x) = −∞.



6 2.2 The Limit of a Function goo.gl/EqF4nq

Similar definition can be given for the one-sided infinite limits: lim

x→a−f (x) = ∞ _xlim

→a+f (x) = ∞ _xlim_→a−f (x) = −∞ _xlim

→a+f (x) = −∞.

Definition 9 (page 90). The line x = a is called a vertical asymptote (垂直漸近線) of the curve y = f (x) if at least one of the following statement is true:

lim

x_→af (x) = ∞ xlim_→a−f (x) = ∞ _xlim

→a+f (x) = ∞

lim

x→af (x) = −∞ xlim→a−f (x) = −∞ _xlim

→a+f (x) = −∞.

Example 10.

(a) f (x) = tan x has vertical asymptotes . (b) f (x) = sec x has vertical asymptotes . (c) f (x) = 1_x has a vertical asymptote .

2.3 The Limit of a Function goo.gl/hNb5fo 7

2.3

Calculating Limits Using the Limit Laws, page

95

Theorem 1. If lim

x_→af (x) exists, then it is unique.

MBNlhc0SmVY

Theorem 2 (Limit laws, page 95). Suppose that c is a constant and the limits lim

x→af (x) and

lim

x_→ag(x) exist. Then

(1) lim

x→a(f (x) + g(x)) = limx→af (x) + limx→ag(x). (Sum Law)

(2) lim

x_→a(f (x) − g(x)) = limx_→af (x) − limx_→ag(x). (Difference Law)

(3) lim

x→a(cf (x)) = c limx→af (x). (Constant Multiple Law)

(4) lim

x→a(f (x)g(x)) = limx→af (x) · limx→ag(x). (Product Law)

(5) lim x_→a f(x) g(x) = lim x→a f(x) lim x→ag(x) if lim

x_→ag(x) 6= 0. (Quotient Law)

The followings are some special limits: (6) lim

x_→a(f (x))

n₌_lim x_→af (x)

n

where _{n ∈ N.} (Power Law) (7) lim x_→ac = c. (8) lim x→ax = a. (9) lim x_→ax n_{= a}n _{where n ∈ N.} (10) lim x_→a n √ x = √n

a where n ∈ N. (If n is even, we assume a > 0.) (11) lim x→a n pf (x) = n q lim

x→af (x) where n ∈ N. (If n is even, we assume limx→af (x) > 0.)







Example 3. Find the limit lim

x→1 x2₋₁

x−1. (比較 Section 2.1, Example 1.)

jbykJ6niFnc

Solution. _{Let x = 1 + ∆x, then x → 1 is equivalent to ∆x → 0, and}

Solution 2.

8 2.3 The Limit of a Function goo.gl/hNb5fo

Example 4. Find the limit lim

t_→0 √ t2₊₉₋₃ t2 . Solution.



Example 5. Prove that lim

x→0 |x|

x does not exist. IgYe0g0pVgw Solution.



Uvarp0VMF0Y

Theorem 6_{(page 101). If f (x) ≤ g(x) when x is near a (except possibly at a) and the limits} of f and g both exist as x approaches a, then

lim

x_→af (x) ≤ limx_→ag(x).



x→af (x) ≤ limx→ag(x)。



Theorem 7 (The Squeeze Theorem (夾擠定理;三明治定理), page 101). If (1) f (x) ≤ g(x) ≤ h(x) when x is near a (except possibly at a), and (2) lim

x_→af (x) = limx_→ah(x) = L,

then lim

x→ag(x) = L.

x y

Figure 1: The Squeeze Theorem.

2.3 The Limit of a Function goo.gl/hNb5fo 9

Example 8. Show that lim

x_→0x sin 1 x = 0.

Solution.

wXafFIZLbTA

Example 9. Show that lim

x_→0 sin x

x = 1.

Solution (page 192). Assume first that x lies between 0 and π₂. Figure 2 shows a sector of a circle with center O, central angle x, and radius 1.

O

Figure 2: A sector of a circle with center O, central angle x, and radius 1.

Since “area of △OAB” < “area of sector OAB” < “area of △OAC”, we have

If −π2 < x < 0, since sin x, x, tan x are odd functions, we get tan x < x < sin x < 0, then





Example 10. Find the limit lim

x_→0 1−cos x

x2 .

10 2.3 The Limit of a Function goo.gl/hNb5fo

Example 11. Find the following limits: lim

x→0 x3_sin(1 x) sin(x2₎ . Solution. Solution 2. VnFcngyxWXg

Example 12. Is there a number a such that lim

x→−2

3x2_{+ ax + a + 3}

x2_{+ x − 2}

exists? If so, find the value of a and the value of the limit. Solution.

Example 13(夾擠定理的應用).

zx4Mey6vKEg

(a) Show that: If |f (x)| ≤ |g(x)| and lim_x

→a|g(x)| = 0, then limx→af (x) = 0.

(b) Show that lim

x_→0sin x = 0.

(c) Show that lim

x→acos x = cos a and limx→asin x = sin a for a ∈ R.

Solution.

Question 14. How do we show that the limit lim

2.4 The Limit of a Function goo.gl/Ac2SyV 11

2.4

The Precise Definition of a Limit, page 104

1galLkYJM-s

Definition 1 (ε-δ language, page 106). Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then we say that the limit of f (x) as x approaches a is L, and we write

lim

x_→af (x) = L

if for every number ε > 0 there is a number δ > 0 such that

if _{0 < |x − a| < δ then |f (x) − L| < ε.}

x y

Figure 1: Limit of f (x) as x approaches a is L.









5TfkXX1np8k

Example 2. Prove that lim

x→1(2x + 3) = 5.

Solution.

• Observation: We calculate |(2x + 3) − 5| = |2x − 2| = 2|x − 1|. We want to find δ > 0 such that

if _{0 < |x − 1| < δ,} then _{2|x − 1| < ε.} That is, if _{0 < |x − 1| < δ,} then _{|x − 1| <} ε

2. This suggests that we can choose δ = ε₂. (or smaller)

12 2.4 The Limit of a Function goo.gl/Ac2SyV

Example 3. Prove that lim

x_→3x 2 _{= 9.}

OS-3Kd_kV78 Solution.

• Observation: We calculate |x2− 9| = |x + 3||x − 3| < ε. We want to find δ > 0 such that

if _{0 < |x − 3| < δ,} then _{|x + 3||x − 3| < ε.}

Notice that if we can find a positive constant M such that |x + 3| < M, then |x + 3||x − 3| < M|x − 3|, and then we can make M|x − 3| < ε by taking |x − 3| < ε

M = δ.

Since we are interested only in values of x that close to 3, it is reasonable to assume |x − 3| < 1, then |x + 3| < 7, so M = 7 is a choice.

• Proof:



5ZNKwxGoyyg

Example 4. Prove the Limit Sum Law: Suppose that the limits lim

x→af (x) and limx→ag(x) exist.

Then lim

x_→a(f (x) + g(x)) = limx_→af (x) + limx_→ag(x).

Proof. For all ε > 0, since

Question 5. How do we show that the limit lim

x→af (x) does not exist?

Solution. 證明極限不存在的其中一招是反證法: 假設極限存在, 記為 lim

x_→af (x) = L, 然後要證明

「任何的」_{L ∈ R} 都會產生矛盾。 那麼怎樣才有矛盾呢? 按極限定義, 必須證明“there exists ε > 0, for all δ > 0, there exists 0 < |x′_{− a| < δ s.t. |f (x}′_{) − L| ≥ ε”}。

Solution 2. 因為極限存在必唯一, 所以利用反證法: 若能設法找到 「兩種數列」, 其極限值不同, 那 麼就得到極限不存在。

2.4 The Limit of a Function goo.gl/Ac2SyV 13

Example 6. The Dirichlet function is defined by

PjVS0OsSDkE

f (x) = (

0 if x is rational 1 if x is irrational. Prove that lim

x→af (x) does not exist for every a ∈ R.

Solution. Suppose lim

x→af (x) = L. Notice that both rational numbers and irrational numbers

are dense in real numbers. If L ≥ 12,

If L < 1₂,



Example 7 (同學若有興趣可想一想這個例子). The Riemann function is defined by f (x) = ( ₁ q if x = p q, (p, q) = 1 is rational 0 if x is irrational.

Then the limit of f (x) exists as x approaches to any irrational number. Definition 8 (Definition of left-hand limit, page 109).

lim

x_→a−f (x) = L

if for every number ε > 0 there is a number δ > 0 such that

if a − δ < x < a then |f (x) − L| < ε. Definition 9 (Definition of right-hand limit, page 109).

lim

x_→a+f (x) = L

if for every number ε > 0 there is a number δ > 0 such that

14 2.4 The Limit of a Function goo.gl/Ac2SyV

Definition 10 (page 112). Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then

8IpPDC-7-QY

lim

x→af (x) = ∞

means that for every number M there is a number δ > 0 such that if _{0 < |x − a| < δ then f (x) > M.}



Example 11. Prove that lim

x→0 1 x2 = ∞.

Solution.

• Observation: Let M be a given positive number. We want to find a number δ > 0 such that if 0 < |x| < δ, then x12 > M . Notice that

1 x2 > M ⇔ x 2_< 1 M ⇔ |x| < 1 √ M. This suggests us to choose δ = √1

M (or smaller).

• Proof:

Definition 12 (page 112). Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then

lim

x_→af (x) = −∞

means that for every number N there is a number δ > 0 such that if _{0 < |x − a| < δ then f (x) < N.}

2.5 Continuity goo.gl/FPvEAN 15

2.5

Continuity, page 114

TDC_ydWmW2U

Definition 1 (continuous at a point, page 114). A function f (x) is continuous at x = a (在

x = a 處連續) if

lim

x→af (x) = f (a).

We say that f (x) is discontinuous at x = a (or f (x) has a discontinuity at x = a) (在 x = a

處不連續) if f (x) is not continuous at a. x y Figure 1: f (x) is continuous at x = a.



x→af (x)存在,即左極限 xlim_→a−f (x)

與 lim x_→a+f (x) 存在且相等。 (3) lim x→af (x) = f (a);極限值等於函數值。



x_→af (x) = f ( limx_→ax) = f (a)。

There are three types of discontinuity:

(1) removable discontinuity (可移不連續點): We can “redefine” the value of the function f (x) at x = a such that f (x) is continuous at x = a.

(2) infinite discontinuity (無限不連續點). (3) jump discontinuity (跳躍不連續點).

x x x

y y y

16 2.5 Continuity goo.gl/FPvEAN

Definition 2 (continuous from the right (or left) (右連續與左連續), page 116).

e-nCP9MlVeI (a) A function f (x) is continuous from the right at x = a if lim

x_→a+f (x) = f (a).

(b) A function f (x) is continuous from the left at x = a if lim

x→a−f (x) = f (a).

x x x

y y y

Figure 3: f (x) is continuous (a) from the right; (b) from the left; (c) at endpoints.

Example 3. Discuss the continuity of the following functions:

f (x) = x 2_{− x − 2} x − 2 , g(x) = ( x2_−x−2 x−2 if x 6= 2 1 if x = 2, , h(x) = ( x2_−x−2 x−2 if x 6= 2 3 if x = 2 . Solution.

Example 4 (同學若有興趣可想一想這個例子). The Riemann function is defined by

f (x) = ( 1 q if x = p q, (p, q) = 1 is rational 0 if x is irrational . Then the Riemann function is continuous at irrational numbers.

7QUHBUSjD1I

Definition 5 (continuous on an interval, page 117). A function f (x) is continuous on an interval(在區間上連續) if it is continuous at every point in the interval. If f (x) is defined only on one side of an endpoint of the interval, we understand continuous at the endpoint to mean continuous from the right or continuous from the left.

2.5 Continuity goo.gl/FPvEAN 17

Theorem 6 (properties of continuous functions, page 117). If f (x) and g(x) are continuous at x = a, and c is a constant, then the following functions are also continuous x = a:

(1) (f ± g)(x) = f (x) ± g(x) (2) cf (x), cg(x)

(3) f (x)g(x)

(4) f_g(x)(x) if _{g(a) 6= 0.}

Proof of (1). Since f (x) and g(x) are continuous at x = a, we have

Therefore



Theorem 7 (page 120). The following type of functions are continuous at every number in their domains:

polynomials rational functions root functions

trigonometric functions inverse trigonometric functions exponential functions logarithmic functions

wOGAe70Rjig

Theorem 8 (page 120). If f is continuous at b and lim

x→ag(x) = b, then limx→af (g(x)) = f (b).

In other words,

lim

x_→af (g(x)) = f ( limx_→ag(x)) = f (b).

Example 9. Evaluate lim

x→1sin

−11−√x 1−x

 Solution.

Theorem 10 (page 121). If g is continuous at a and f is continuous at g(a), then the composition function _{f ◦ g given by (f ◦ g)(x) = f (g(x)) is continuous at a.}

Proof. The function g(x) is continuous at x = a implies

18 2.5 Continuity goo.gl/FPvEAN

Theorem 11 (The Intermediate Value Theorem, 中間值定理, page 122). Suppose that f (x) is continuous on the closed interval [a, b] and let N be any number between f (a) and f (b), where _{f (a) 6= f (b). Then there exists a number c in (a, b) such that f (c) = N.}

U_tO4EB72dU

x y

Figure 4: The Intermediate Value Theorem.







Applications of Intermediate Value Theorem









7nfPoNfjTeY

Example 12. _{Suppose f is a continuous function on [a, b] and a ≤ f (x) ≤ b for all x ∈ [a, b].} Show that there exists c ∈ [a, b] such that f (c) = c.

2.6 Limits at Infinity: Horizontal Asymptotes goo.gl/NHp2Wm 19

2.6

Limits at Infinity: Horizontal Asymptotes, page

126

Lxl5r6-I4GY

Definition 1 (page 127–128). Let f be a function defined on real numbers. (a) lim

x→∞f (x) = L means that the values of f (x) can be made arbitrarily close to L by

taking x sufficiently large (無窮遠處之極限). (b) lim

x→−∞f (x) = L means that the value of f (x) can be made arbitrarily close to L by

taking x sufficiently large negative. (負無窮遠處之極限)



x→∞f (x) = L 的另一種表達法為“f (x) → L as x → ∞”。



x→−∞f (x) = L的另一種表達法為 “f (x) → L as x → −∞”。

Definition 2 (page 128). The line y = L is called a horizontal asymptote (水平漸近線) of the curve y = f (x) if either lim x→∞f (x) = L or x→−∞lim f (x) = L. x x y y

Figure 1: Limit lim

x→∞f(x) = L, limx→−∞f(x) = L, and horizontal asymptotes.



Theorem 3 (page 129).

(a) If r > 0 is a rational number, then lim

x→∞ 1 xr = 0.

(b) If r > 0 is a rational number such that xr _{is defined for all x, then lim} x→−∞

1 xr = 0.

Example 4. Evaluate lim

x_→∞ 3x2

−x−2 5x2_+4x+1.

20 2.6 Limits at Infinity: Horizontal Asymptotes goo.gl/NHp2Wm



Example 5. Evaluate lim

x→∞(2 x_{+ 3}x_{+ 5}x₎1 x. Solution.



Example 6. Find the following limit:

(a) lim x_→0 sin x x (b) limx_→∞ sin x x (c) limx_→0x sin 1 x (d) limx_→∞x sin 1 x. Solution. (a) lim x→0 sin x

x = . (See section 2.3 Example 9.)

(b) (c) (d)



gvcC5ZBEBH8

Example 7. Evaluate the limit lim

x_→∞ 3x2₊₅ 5x+3 sin 2 x. Solution.

2.6 Limits at Infinity: Horizontal Asymptotes goo.gl/NHp2Wm 21

Example 8. Let lim

x→∞ 1 + 1 x
x = e. Find lim x→−∞ 1 + 1 x
x . Solution.

Example 9. Find the limit lim

x→0+(cos √ x)1x. Solution.



Example 10. Let f (x) =√x2_{+ x. Compute the following limits:}

(a) A = lim x_→∞ f(x) x . (b) lim x_→∞(f (x) − Ax). Solution.



22 2.6 Limits at Infinity: Horizontal Asymptotes goo.gl/NHp2Wm

Example 11. Suppose α, β are two constants and lim

x→−∞( √ x2_{+ 3x + 2 − αx − β) = 0. Find} α and β. fy4_4G-WAbw Solution.





Infinite Limits at Infinity, Precise Definition

Definition 12 (page 134–137).

(a) Let f be a function defined on some interval (a, ∞). Then lim_x

→∞f (x) = L means that

for every ε > 0 there is a corresponding number N such that if x > N _{then |f (x) − L| < ε.}

(b) Let f be a function defined on some interval (−∞, a). Then lim_x

→−∞f (x) = L means

that for every ε > 0 there is a corresponding number N such that if x < N _{then |f (x) − L| < ε.}

(c) Let f be a function defined on some interval (a, ∞). Then lim_x

→∞f (x) = ∞ means that

for every positive number M there is a corresponding positive number N such that if x > N , then f (x) > M .

2.7 Derivatives and Rates of Change goo.gl/zRHkv6 23

2.7

Derivatives and Rates of Change, page 140

-evUkuaTh74

Definition 1(page 141). The tangent line (切線) to the curve y = f (x) at the point P (a, f (a)) is the line through P with slope (斜率)

m = lim x→a f (x) − f (a) x − a = limh→0 f (a + h) − f (a) h

provided that this limit exists.

x y

a

... f(x)

Figure 1: Tangent line is the limiting position of the secant line (

割線



Example 2. Find an equation of the tangent line to the hyperbola y = 1_x at the point (1, 1). Solution. Let f (x) = 1_x. Then the slope of the tangent at (1, 1) is

Therefore, an equation of the tangent at the point (1, 1) is .

YVo7X6gl44I

Definition 3 (page 143). If f (x) is the position function, then the average velocity (平均速 度) is

average velocity = displacement time =

f (a + h) − f (a)

h ,

and the velocity (or instantaneous velocity, 瞬時速度) v(a) at time t = a be the limit of these average velocities:

v(a) = lim

h_→0

f (a + h) − f (a)

h .

24 2.7 Derivatives and Rates of Change goo.gl/zRHkv6

Example 4. Suppose that a ball is dropped from the upper observation deck of Taipei 101, 508m above the ground.

(a) What is the velocity of the ball after 5 seconds? (b) How fast is the ball traveling when it hits the ground? Solution. Using the equation of motion s = f (t) = 4.9t2, we have

(a) The velocity after 5 is .

(b) First we solve 4.9t2₁ = 508. This gives t1 =

q

508

4.9. The velocity of the ball as it hits the

ground is

KfYjrUE5RtA

Definition 5 (page 144). The derivative of a function f (x) at a number x = a (函數 f (x)在

x = a的導數), denoted by f′(a), is f′(a) = lim x→a f (x) − f (a) x − a = limh→0 f (a + h) − f (a) h

if this limit exists.

If we use the point-slope form (點斜式) of the equation of a line, we can write an equation of the tangent line to the curve y = f (x) at the point P (a, f (a)):

y − f (a) = f′(a)(x − a).



2.7 Derivatives and Rates of Change goo.gl/zRHkv6 25

Rates of Change

5-aJdX5hQhA

Suppose y is a quantity that depends on another quantity x. Thus y is a function of x and we write y = f (x). If x changes from x1 to x2, then the change in x (also called the increment

(增加量) of x) is ∆x = x2− x1, and the corresponding change in y is ∆y = f (x2) − f (x1).

The difference quotient

∆y ∆x =

f (x2) − f (x1)

x2− x1

is called the average of the change of y with respect to x (平均變化率) over the interval [x1, x2].

We say

instantaneous rate of change = lim

∆x→0 ∆y ∆x = limx2→x1 f (x2) − f (x1) x2− x1 .

The derivative f′(a) is the instantaneous rate of change of y = f (x) with respect to x when x = a (瞬間變化率).

Examples of rates of change:

(1) Velocity of an object: the rate of change of displacement with respect to time.

(2) Marginal cost (邊際成本): the rate of change of production cost with respect to the number of items produced.

(3) Interest (in economics): the rate of change of the debt with respect to time. (4) Power (in physics,功率): the rate of change of work with respect to time.

(5) Rate of reaction (in chemistry): the rate of change in the concentration (濃度) of a reactant with respect to time.

26 2.8 The Derivative as a Function goo.gl/DHdCPZ

2.8

The Derivative as a Function, page 152

UtCufTsRzv8

Definition 1 (page 152). The derivative of f (x) (f (x) 的導函數) is f′(x) = lim h_→0 f (x + h) − f (x) h .







Example 2. Let f (x) = x3. Find f′(x). Solution. By definition,

Other Notations

If we use the traditional notation y = f (x) to indicate that the independent variable is x and the dependent variable is y, then some common alternative notations for the derivative are as follows: f′(x) = y′ = dy dx = df dx = d dxf (x) = Df (x) = Dxf (x).

The symbols D and _dxd are called differentiation operators (微分算子) because thy indicate the operation of differentiation.

We use the notation

dy dx     x=a or dy dx  x=a

to indicate the value of a variable _dxdy at a specific number a, which is a synonym for f′(a).

HF9eAGoF_Bg

Definition 3 (page 155). A function f is differentiable at a (在 x = a 處可微分) if f′(a) exists. It is differentiable on an open interval (a, b) (or (a, ∞) or (−∞, a) or (−∞, ∞)) if it is differentiable at every number in the interval.



2.8 The Derivative as a Function goo.gl/DHdCPZ 27

Theorem 4 (page 157). If f is differentiable at a, then f is continuous at a. Proof of Theorem 4 is in the Appendix.





How Can a Function Fail to Be Differentiable?

(1) corner or kink: the graph of f has no tangent at this point and f is not differentiable there.

(2) discontinuity: f is not continuous at a, then f is not differentiable at a. (3) vertical tangent line: f is continuous at a and lim

x→a|f

′_{(x)| = ∞.}

x x x

y y y

Figure 1: Three ways for f (x) not to be differentiable at x = a.

Higher Derivatives

b261rRqbKmA

If f is differentiable function, then its derivative f′ is also a function, so f′ may have a derivative of its own, denoted by (f′)′ = f′′. This new function f′′ is called the second derivative of f (二次導數). We write the second derivative of y = f (x) as

d dx  dy dx  = d 2_y dx2.



The third derivative f′′′ ₍三次導數) is the derivative of the second derivative: f′′′_{= (f}′′₎′_.

If y = f (x), then alternative notations for the third derivative are

y′′′= f′′′(x) = d dx  d2_y dx2  = d 3_y dx3.

In general, the n-th derivative (n ≥ 4) of f is denoted by f(n). If y = f (x), we write

yn= f(n)(x) = d

n_y

dxn.

28 2.8 The Derivative as a Function goo.gl/DHdCPZ Example 5. Suppose f (x) = ( 1−cos x sin x x > 0 ax + b x ≤ 0.

Find a and b such that f is continuous and differentiable at x = 0. Solution.



mvhPYZQ-DuY

Example 6. _{Let f (x) = x|x|. Find f}′(x) and f′′(x). Solution.

2.8 The Derivative as a Function goo.gl/DHdCPZ 29











Appendix

Proof of Theorem 4. The goal is to show that lim

x→af (x) = f (a). For x 6= a, we have f (x) − f (a) = f (x) − f (a) x − a · (x − a), so lim

x_→a(f (x) − f (a)) = limx_→a

 f (x) − f(a) x − a · (x − a)  = lim x_→a f (x) − f (a) x − a · limx_→a(x − a) = f′_{(a) · 0 = 0.} Hence lim

x_→af (x) = limx_→a(f (x) − f (a) + f (a)) = limx_→a(f (x) − f (a)) + limx_→af (a)