PRA U STPM MATHEMATICS T PENGGAL 2

CONTENTS

Chapter 1
1 LIMITS AND CONTINUITY
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1.1 Limits 2
1.2 Continuity 6

Chapter 16
2 DIFFERENTIATION
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2.1 Derivatives 17
2.2 Applications of Differentiation 39

Chapter 80
3 INTEGRATION
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3.1 Indefinite Integrals 81
3.2 Definite Integrals 104

Chapter 126
4 DIFFERENTIAL EQUATIONS
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4.1 Differential Equations 127
4.2 First Order Differential Equations with Separable Variables 130
4.3 First Order Linear Differential Equations 134
4.4 Transformations of Differential Equations 136
4.5 Problems Modelled by Differential Equations 140

Chapter 153
5 MACLAURIN SERIES
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5.1 Maclaurin Series 154
5.2 Applications of Maclaurin Series 167

Chapter 172
6 NUMERICAL METHODS
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6.1 Numerical Solution of Equations 173
6.2 Numerical Integration 186

STPM Model Paper (954/2) 193

Answers 195

iii

CHAPTER DIFFERENTIATION

2

Subtopic Learning Outcome
2.1 Derivatives
(a) Identify the derivative of a function as a limit.
2.2 Applications of (b) Find the derivatives of xn (n  Q), ex, In x, sin x, cos x, tan x, sin–1x, cos–1 x,
differentiation
tan–1 x, with constant multiples, sums, differences, products, quotients and
composite.
(c) Perform implicit differentiation.
(d) Find the first derivatives of functions defined parametrically.

(a) Determine where a function is increasing, decreasing, concave upward and
concave downward.

(b) Determine the stationary points, extremum points and points of inflexion.
(c) Sketch the graphs of functions, including asymptotes parallel to the coordinate

axes.
(d) Find the equations of tangents and normals to curves, including parametric

curves.
(e) Solve problems concerning rates of change, including related rates.
(f) Solve optimisation problems.

Bilingual Keywords

asymptote – asimptot
concave downward – cekung ke bawah
curve – lengkung
decreasing – menyusut
derivative – terbitan
differentiation – pembezaan
extremum point – titik ekstremum
implicit function – fungsi tersirat
increasing – menokok
limit – had
point of inflexion – titik lengkok balas
stationary point – titik pegun

Mathematics Term 2  STPM  Chapter 2 Differentiation

2.1 Derivatives

Derivative of a function Differentiation Feynman's
INFO Differentiation
VIDEO

Let A(x, y) be a fixed point on the curve y = f(x) and B be a y y = f(x )
neighbouring point with coordinates (x + x, y + y),
where x (read as ‘delta-ex’) denotes a small increase in x and B (x + δx, y + δy )
y denotes a corresponding small increase in y. T

In Figure 2.1, AC = x, BC = y. 2

Gradient of chord AB is BC = dxy  . A(x, y) C
AC
Figure 2.1
As the point B is moved along the curve towards the fixed 0 x
point A, x → 0 and the direction of the chord AB approaches
closer and closer to the direction of AT, the tangent to the
curve at A.

Thus,

gradient of curve at A = lim 0  dxy (read as ‘the limit as x tends to 0’)
δx →

= dy  (read as ‘dee’ y by ‘dee’ x)
dx

If y = f(x), then dy = f(x) defined by f(x) = lim 0 f(x + x) – f(x)
dx δx → x

This formal definition of a derivative can be used to differentiate any function.

This process is known as differentiation with respect to x (abbreviated to w.r.t.) from first principles.

dy or f(x) is the ‘derivative of f(x)’, the ‘differential coefficient’ or ‘derived function’ of f(x) w.r.t. x.
dx
dy
Note: dy and dx do not have any meaning in themselves; in particular we cannot think of dx as dy ÷ dx.

Example 1

Find the derivatives of the following functions with respect to x, from the first principles.
1
(a) f(x) = x2 (b) f(x) = x

d f(x + x) – f(x)
dx δx
Solution:  4(a)
Using the definition f(x) = lim 0 
δx →

 4 ddx (x2) =
lim 0 (x + x)2 – x2
δx → x

 4 =
lim x2 + 2x x + (x)2 – x2
δx → 0 x

17

Mathematics Term 2  STPM  Chapter 2 Differentiation

2xx + (x)2
x
 4 =
lim 0
δx →

= lim 0 (2x + x)
δx →

= 2x

1 _1
1 2  4(b) ddx  
1 = lim x + x x
x δx → x
0

x – (x + x)
(x + x)x (x)
 4 42
= δxli→m 0 –1
x(x + x)
= lim 0
δx →

= – x12

Example 2

If y = 3x2 + x + 1, find dy from the first principles.
dx

3(x + x)2 + (x + x) + 1 – (3x2 + x + 1)
x
Solution:  4dy = lim 0
δx →
dx

 4 = δxli→m 0 3x2 + 6xx + 3(x)2 + x+ x + 1 – 3x2 – x – 1
x

 4 =
lim 0 6x x + 3(x)2 + x
δx → x

= lim 0 [6x + 3 x + 1]
δx →

= 6x + 1

Exercise 2.1

Find the derivatives of the following functions from the first principles.

1. y = x3 2. y = x4 3. y = 5x2
6. y = x2 – x + 3
4. y= 1 5. y = x2 + 5x 9. y = 2x2 + 3
x2
1
7. y = 4x3 8. y= x3

10. y = 2x2 – 3x + 1

18

Mathematics Term 2  STPM  Chapter 2 Differentiation

Differentiation of standard functions

Derivative of a constant y
Consider the function y = c, where c is a constant.

From the derived definition dy = lim 0 f(x + x) – f(x)
dx δx → δx

As f(x + x) = c = f(x), c y=c

∴ f(x + x) – f(x) = c–c
δx δx
=0 x
0
Hence, dy = lim 00 2
dx δx → Figure 2.2

= 0

Geometrically, y = c represents a straight line parallel to the x-axis.

Therefore, its gradient is zero.

d  (c) = 0, where c is a constant
dx

Derivative of af(x) where a is a constant

From the derived definition,

d [a f(x)] = lim 0 a f(x + x) – a f(x)
dx δx → δx

a[f(x + x) – f(x)]
x
 4 =
lim 0
δx →

lim f(x + x) – f(x)]
x
δx → 0
 4 =
lim 0 a
δx →

= a d [f(x)]
dx

Hence, d [a f(x)] = a ddx [f(x)]
dx

Derivative of xn

Consider the following differentiation from first principles.

d  x 2 = lim 0 (x + x)2 – x2
dx δx → δx

= lim 0 (2x + x)
δx →

= 2x

= 2x2 – 1

19

Mathematics Term 2  STPM  Chapter 2 Differentiation

d  x 3 = lim 0 (x + x)3 – x3
dx δx → δx

= lim 0 [3x2 + 3x x + (x)2]
δx →

= 3x2

= 3x3 – 1

d  x –1 = lim 0 (x + x)–1 – x–1
dx δx → δx

= lim 0 – 1
δx → x(x + x)

2 = – x12
= –x–2
= –1(x–1 – 1)

Hence, d  xn = nx n – 1, n  R.
dx

Example 3

Differentiate with respect to x.
(a) x–3 (b) x
(c) –5x4 (d) x36
Solution:
(a) Let y = x–3
ddxy = –3x–3 – 1 = –3x–4 = – x34
(b) Let y = x  = x—12
ddxy = 1
1  x—21 – 1 = 1  x– —12 = 2x
2 2

(c) Let y = –5x4
ddxy = –5(4x4 – 1) = –20x3

(d) Let y = 3 = 3x – 6
ddxy = x6
3[– 6x–6 – 1] = –18x–7 = – 1x87

Exercise 2.2

Find the derivatives of the following functions with respect to x.

1. 5 2. 3x 3. – 4x 4. x5 5. x–4
6. x—31 7. x–1
8. x3 9. 1 10. 3x2
11. 5x–3 12. 7x10 x
5 7 –2
13. 4x4 14. 8  x3 15. x8

20

Mathematics Term 2  STPM  Chapter 2 Differentiation

Derivative of ex

Consider the exponential function f(x) = ax in Figure 2.3 where a is a positive constant. For any value of a,
a0 = 1. Hence, the curve passes through the point A(0, 1).

The gradient at point A is given by

f(0) = lim 0 f(0 + x) – f(0)
δx → x

= lim 0 f(x) – f(0) y
δx → δx y = ax

= lim ax – a0 2
δx → 0 δx

= lim 0 ax – 1 A (0, 1)
δx → δx
0
x
Figure 2.3
For f(x) = ax

d (ax) = lim 0 ax + x – ax
dx δx → δx

ax – 1
x
1 2 =
lim 0 ax
δx →

= ax lim 0 ax – 1
δx → x

= ax f(0) f(0) = δxli→m 0 a δx – 1
δx

Hence, d (ax) = ax f(0)
dx

Now there must be a value of a for which f(0) = 1, i.e. the gradient of the graph at (0, 1) is 1. We call this
value e.

Thus d (ex) = ex
dx

e is an irrational number. e ≈ 2.718.
ex is known as the exponential function.

Derivative of ln x

Let y = f(x), where f(x) is any function of x,

hence 1 2dy = lim 0 δy
δx → δx
dx

= lim 0 1
δx →
1 δx 2
δy

21

Mathematics Term 2  STPM  Chapter 2 Differentiation

But y → 0 when x → 0,

1 2 dy = 1
dx
lim δx
δx → 0 δy

dy = 1
dx dx
dy

When ln x = y, From the definition of logarithm i.e. if loga y = x, then y = ax.
x = ey

2 Differentiate both sides w.r.t. y,

dx = d (ey) = ey = x
dy dy

dy = 1 = 1
dx dx x

dy

Hence, d  (ln x) = 1
dx x

Derivative of ax

Let y = ax
Taking log of both sides to base e,
ln y = ln ax
= x ln a

Differentiate both sides w.r.t. y, dx
d dy
dy (ln y) = ln a

1 = dx ln a
y dy

dx = y 1 a
dy ln

ddxy == y ln a
ax ln a

Hence, d  (ax) = ax ln a
dx

Example 4

Differentiate each of the following with respect to x. (b) 10x
(a) 3x

Solution: (a) Let y = 3x (b) y = 10x

Hence, dy = 3x ln 3 Hence, dy = 10x ln 10
dx dx

22

Mathematics Term 2  STPM  Chapter 2 Differentiation

Differentiation of trigonometric functions

Derivative of sin x
Let y = sin x

dy = lim 0 sin (x + dx) – sin x
dx δx → dx

 = δxli→m 0 2 cos (x + d2x ) + x s in (x + dx) – x  Using the formula
dx 2 sin A – sin B
= 2 cos –A–+–2– B– s in –A–––2–B–
1 2 = δxli→m 0 2 + d2x dx
cos  x dx sin 2 2

1 2 = δx li→m 0 co s x + d2x s i n d2xd2x

I n the limit, as dx → 0, cos 1 x + d2 x 2 → cos x and sind2 xd2x → 1

Hence d ( sin x) = cos x, where x is in radians.
dx

Derivative of cos x

Let y = cos x

dy = lim 0 cos (x + dx) – cos x
dx δx → dx

 = δxli→m 0 –2 sin (x + d2x ) + x sin (x + dx) – x   Using the formula
dx 2 cos A – cos B
= –2 sin –A–+–2– B– s in –A–––2–B–

x + d2x dx
1 2 –2 sin  sin 2
= lim
δx → 0 dx

1 2 x+ dx sin dx
= lim –sin 2 2
δx → 0
dx
2

the limit, as dx → 0, sin 1x + dx 2 → sin x and sin dx →1
In 2 2

dx
2

Hence d   (cos x) = –sin x, where x is in radians.
dx

23

Mathematics Term 2  STPM  Chapter 2 Differentiation

Derivative of tan x
Let y = tan x

dy tan (x + dx) – tan
  dx =   dx x  
lim 0
δx →

sin (x + dx) sin x 1
cos (x + dx) cos x dx
  = δxli→m 0  –

sin (x + dx) cos x – cos (x + dx) sin x
cos x cos  (x + dx) · dx
  = δxli→m 0

2 = lim 0 sin [(x + dx) – x]   Using the formula
δx → cos  x cos  (x + dx) · dx sin A cos B – cos A sin B = sin (A – B)

= lim 0 sin dx
δx → cos  x cos  (x + dx) · dx

= lim 0 cos  x 1 + dx) · sin dx
δx → cos  (x dx

In the limit, as dx → 0, cos (x + dx) → cos x and sin dx → 1.
dx

Hence, d   (tan x) = 1 = sec2 x
dx cos2 x

Derivative of arcsin x (sin–1 x)

Let y = sin–1 x
x = sin y

Differentiating w.r.t. y, dx = cos y
dy

\  dy = 1 y
dx cos

but cos2 y = 1 – sin2 y = 1 – x2

Hence,  dy = 1
dx (1 – x2)

i.e.  d  (sin–1 x) = 1
dx (1 – x2)

Note: y = sin–1 x ⇒ –   π <y< π  , since y is a principal value angle.
2 2
For this range of values of y, cos x > 0

\ cos x = (1 – x2)   [not –  (1 – x2) ]

24

Derivative of arc cos x (cos–1 x) Mathematics Term 2  STPM  Chapter 2 Differentiation

Let y = cos–1 x 2
x = cos y
dx
dy = –sin y

dy = – 1
dx sin y

but sin2 y = 1 – cos2 y
= 1 – x2
dy 1
\  dx = – (1 – x2)

Hence,   d  (cos–1 x) = –  1   
dx (1 – x2)

Derivative of arc tan x (tan–1 x)

Let y = tan–1 x
x = tan y
dx
dy = sec2 y

dy = 1
dx sec2 y
1
= 1 + tan2 y

= 1 1 x2
+

Hence,   d  (tan–1 x) = 1 1 x2  
dx +

Rules of differentiation

Differentiation of sums and differences of functions

Consider two functions of x, p(x) and q(x), and let f(x) = p(x) + q(x).

From the derived definition,

d [f(x)] = lim [p(x + x) + q(x + x)] – [p(x) + q(x)]
dx δx → δx
0

= lim [p(x + x) – p(x)] + [q(x + x) – q(x)]
δx → δx
0

= lim p(x + x) – p(x) + lim q(x + x) – q(x)
δx δx
δx → 0 δx → 0

= d [p(x)] + d [q(x)]
dx dx

Hence, d [f(x)] = d [p(x)] + d [q(x)]
dx dx dx

25

ANSWERS

1 Limits and Continuity (c) (d) y
y

Exercise 1.1

1. (a) 6 (b) 4 1
(c) 3 (d) –3

2. (a) 1 (b) 37 0 x –1 0 x
4
1
(c) –   2 (c) 6 continuous continuous

3. (a) 1 (b) 3 (e) y (f) y

(c) 1 (d) 13
2

4. (a) 3 (b) 1 (c) 4 x
15 ) 52 x
(d) ∞ (e) 4 (f –1 0 01
–2
5. (a) 1 (b) –1 not continuous
5 continuous
4. (a) 1, –1; Not continuous
6. (a) –1 (b) 1 (c) does not exist (b) y

7. (a) 9 (b) –10 (c) –1
(d) 5 (e) does not exist

8. (a) 2 (b) 2 (c) 2 1
(d) 2 (e) –3 (f) does not exist 5x

Exercise 1.2 –1
1.
(c) does not exist
f(x ) 5. (a) does not exist;  f is not continuous
(b) y
–_2π 0 _π x
2 5


continuous

2. f(x)

2 0 x
–1 x
1
23 4 5
0 1 2 3x 6. Yes
– –41

y
continuous

3. (a) (b) y 1
2
y
– 10
2 1

0 π 2π x 0 πx –1


continuous not continuous

195

  Mathematics Term 2  STPM  Answers

7. (b) right 2 Differentiation
f(x )

2 Exercise 2.1 2. 4x3
1. 3x 2
–3 0 5 x 3. 10x 4. –  2
–2 x3
5. 2x + 5 6. 2x – 1
7. 12x 2 3
8. –  x4
not continuous 9. 4x
10. 4x – 3
8. k = – 3  ;  Continuous
4 Exercise 2.2
9. c = 1 1. 0
3. –4 2. 3
10. (a) 2 4. 5x 4
(b) x– —23
f(x ) 5. –4x–5 6. 1
3
1 3
7. –  x2 8. 2   x

9. – 21 x– —23 10. 2 x– —13
3

0 2 45 x 11. –15x–4 12. 70x 9
13. – x55 21
14. 8 x 2

continuous 16
x9
1 5.

STPM Practice 1

1. (a) 1 (b) 3 Exercise 2.3 1 22. 1 x ln 1
2 1. 2x ln 2 5 5

2. (a) (i) x lim  f(x) ≠ x lim  f(x) 3. 10x + 3 4. 4x 3 – 18x2
→ c– → c+

(ii) x lim  f(x) = x lim  f(x) ≠ f(c) 5. 6x – 5 6. 2x – 4
→ c– → c+ x2

(iii) x lim  f(x) = x lim  f(x) but f(c) is not defined 7. –  2 – 1 8. ex + cos x
→ c– → c+ x3 x2

(b) {k: –∞ < k < 0 or 0 < k < 3 or 3 < k < ∞}
2 2
3. a = –32,  b = 3 9. 2 10. 8x + 2 + 9
x x2 x4
4. (a) 1, 0, 2, 1 (b) Yes, No
11. – 32x – 2x 12. – 21x
5. (a) 8 (b) –2√ 7 
27 (b) a = ±√ 6  1 3. 10x – 3 + 8 1 + 4
x2 x3 14. x2 x3
6. (a) b = 3
3
7. (a) x lim  f(x) = x lim  f(x) = 1, lim  f(x) exists. 15. –  2x 16. 2 cos x – 3 sin x
→ 0– → 0+ x→0

(b) lim  f(x) = 1 ≠ f(0) = 2 1 7. 13 43 18. 13
x→0

f(x) is discontinuous at x = 0

f(x) is continuous in the interval (–∞, 0) < (0, ∞) Exercise 2.4 2. 24x – 5
1. 2x (2x 2 + 1)
8. a = 1,  b = –1 3. 8x 3 – 30x2 + 1 4. 12 – 1 – 2x
x4 x2
9. x lim  f(x) = –6 ≠ x lim  f(x) = 6 5. 3x 2 – 4x – 1 6. 5(cos x – x sin x)
→ 3– → 3+ 7. 2x(2 sin x + x cos x)
f(x) is discontinuous at x = 3 9. 1 + ln x
8. (cos x) ln x–+1)sinx x
1 0. (a) 2 (b) –   1 10. ex (x 2 + 2x
2
1 3. (a) 3 (b) 2
8 11. ex (tan x – cos x + sec2 x + sin x)

1 4. No, not continuous at 1 1 2. sin x (1 + sec2 x)

15. (a) lim  f(x) = 8, lim  f(x) = 2 + c, c = 6 1 3. x(2 cos x – x sin x)
→ 0– → 0+
x x 14. 2x (sin x + cos x) + x2 (cos x – sin x)
(b) (i) f(x) continuous at x = 0 only when c = 6
15. cos 2x
(ii) f(x) continuous at x , 0. (quadratic)

(iii) f(x) continuous at x . 0. (exponential function) Exercise 2.5

1. –6x 3)2 2. 2
(2x 2 – (x + 1)2

196