chapter 8 differentiation

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

CHAPTER 8 DIFFERENTIATION

Focus on Exam 8

1 (a) Let y = (x2 + 3)e-2x

dydx

= (x2 + 3)(-2e-2x) + e-2x(2x)

= 2e-2x(-x2 - 3 + x)

(b) Let u = x and y = sin3 x

= x12 y = sin3 u

dudx

= 12

x- 1

2 dydu

= 3 sin2 u (-cos u)

= 12 x

= -3 sin2 u cos u

Hence, dydx

= dydu

× dudx

= -3 sin2 u cos u × 12 x

= -3 sin2 x cos x

2 x

2 (a) Let y = ln (x3 e-3x)

dydx

= x3(-3e-3x) + e-3x(3x2)

x3e-3x

ddx

(x3e-3x)

= e-3x 3x2(-x + 1)

x3e-3x

= 3(-x + 1)x

Copy back x3e-3x.

(b) Let u = 5x

log5 u = x

ln uln 5

= x

ln u = x ln 5

1u

dudx

= ln 5

dudx

= u ln 5

Chap-08-FWS.indd 1 10/19/2012 10:37:43 AM


ACE AHEAD Mathematics (T) Second Term2

dudx

= 5x ln 5

∴ ddx

(5x) = 5x ln 5

Let y = 5x

1 + 5x2

dydx

= (1 + 5x2)(5x ln 5) - (5x)(10x)(1 + 5x2)2

= 5x [ln 5 (1 + 5x2) - 10x](1 + 5x2)2

3 f (x) = e-2x sin 2x

f ′(x) = e-2x (2 cos 2x) + sin 2x (-2e-2x)

f ′(x) = 2e-2x cos 2x - 2 sin 2x e-2x

f ″(x) = 2e-2x - 2 sin 2x + cos 2x (- 4e-2x) - 2 sin 2x (-2e-2x) + e-2x (-4 cos 2x)

f ″(x) = -4e-2x sin 2x - 4 cos 2x e-2x + 4 sin 2x e-2x - 4e-2x cos 2x

f ″(x) = -8e-2x cos 2x

When x = π6

, f ″(x) = -8e-

π3 cos π

3

= -8e-

π3 1

2 = -4e

-π3

4 ey = x + 12x - 3

y = ln x + 12x - 3

= ln (x + 1) - ln (2x - 3)

dydx

= 1x + 1

- 22x - 3

At the x-axis, y = 0.

e0 = x + 12x - 3

1 = x + 12x - 3

2x - 3 = x + 1 x = 4

The gradient of the tangent at the point (4, 0) = 14 + 1

- 22(4) - 3

= - 15

Hence, the equation of the tangent at the point (4, 0) is

y - 0 = -15

(x - 4)

5y = -x + 4

Chap-08-FWS.indd 2 10/19/2012 10:37:43 AM


Fully Worked Solution 3

5 x2 - xy + y2 = 7

When x = 3, 32 -3y + y2 = 7 y2 - 3y + 2 = 0 (y - 1)(y - 2) = 0 y = 1 or 2 x2 - xy + y2 = 7

Differentiating implicitly with respect to x,

2x - x dydx

+ y(-1) + 2y dydx

= 0

(-x + 2y) dydx

= -2x + y

dydx

= -2x + y-x + 2y

The gradient of the tangent at the point (3, 1) is -2(3) + 1-3 + 2(1)

= 5.

The gradient of the tangent at the point (3, 2) is -2(3) + 2-3 + 2(2)

= -4.

6 2y = ln (xy)

2dydx

= x

dydx

+ y(1)

xy

2xy dydx

= x dydx

+ y

(2xy - x) dydx

= y

dydx

= y

2xy - x

At the point P(e2, 1), dydx

= 12(e2)(1) - e2

= 1e2

Therefore, the gradient of the tangent is 1e2

.

Hence, the equation of the tangent at the point P(e2, 1) is

y - 1 = 1e2

(x - e2)

e2y - e2 = x - e2

e2y = x

7 x = e 4t = e2 t

dxdt

= 2 12 t e2 t

dxdt

= e2 t

t

Chap-08-FWS.indd 3 10/19/2012 10:37:43 AM



y = e6t

y = (e6t)12

y = e3t

dydt

= 3e3t

x = e2 t

ln x = 2 t

(ln x)2 = 4t

t = 14

(ln x)2

dydx

=

dydtdxdt

= 3e3t

e2 t

t

= 3e3t t

e2 t e3t = e

34

(ln x)2

= 3(12 ln x)e34

(ln x)2

x e2 t = x

= 32

ln xx

e34

(ln x)2 t =

12

ln x

8 x = e2t - 2

dxdt

= 2e2t

dydx

=

dydtdxdt

= et + 12e2t

y = et + t

dydt

= et + 1

When t = ln 2, x = e2 ln 2 - 2 = eln 22

- 2 = 22 - 2 = 2 alogax = x

When t = ln 2, y = eln 2 + 2 = 2 + 2 = 4

Chap-08-FWS.indd 4 10/19/2012 10:37:43 AM



When t = ln 2, dydx

= eln 2 + 12e2 ln 2

= 2 + 12(2)2

= 38

Hence, the equation of the tangent at the point where t = ln 2 is

y - 4 = 38

(x - 2)

8y - 32 = 3x - 6 8y = 3x + 26

9 x = - cos2 2θ

dxdθ

= -2 cos 2θ (-2 sin 2θ)

= 4 cos 2θ sin 2θ y = sin2 2θ

dydθ

= 2 sin 2θ (2 cos 2θ)

= 4 sin 2θ cos 2θ

∴ dydx

=

dydθdxdθ

= 4 sin 2θ cos 2θ4 cos 2θ sin 2θ

= 1

The gradient of the tangent is 1. Hence, the gradient of the normal is -1.

When θ = π8

, x = - cos2 π4

= - 12

2

= -12

and y = sin2 π4 = 1

22

= 12

Hence, the equation of the normal is

y - 12

= -13x - - 12

y - 12

= -x - 12

y = -x

Chap-08-FWS.indd 5 10/19/2012 10:37:44 AM



10 y = e2x - 6x + 7

= (e2x - 6x + 7)12

dydx

= 12

(e2x - 6x + 7)- 1

2 (2e2x - 6)

= 2e2x - 62 e2x - 6x + 7

= 2e2x - 62y

= e2x - 3

y

y dydx

= e2x - 3

y d2ydx2

+ dydx

dydx

= 2e2x

y d2ydx2 + dy

dx2

= 2e2x [Shown]

11 y = ex ln x

dydx

= ex 1x + ln x ex

dydx

= ex 1x + y

x dydx

= ex + xy …

x d2ydx2

+ dydx

(1) = ex + x dydx

+ y(1)

x d2ydx2

+ (1 - x) dydx

- y = ex From ,

ex = x dydx

- xy

x d2ydx2

+ (1 - x) dydx

- y = x dydx

- xy

x d2ydx2

+ (1 - 2x) dydx

+ (x - 1) y = 0 [Shown]

12 y = cos xx

xy = cos x

x dydx

+ y(1) = -sin x

x dydx

+ y = -sin x

x d2ydx2

+ dydx

(1) + dydx

= -cos x

x d2ydx2

+ 2 dydx

= -xy

x d2ydx2 + 2

dydx

+ xy = 0 [Shown]

Chap-08-FWS.indd 6 10/19/2012 10:37:44 AM



13 y = cos x

= cos12 x

dydx

= 12

(cos x)- 1

2 (-sin x)

= -sin x

2 cos x

= -sin x2y

2y dydx

= -sin x

2y d2ydx2

+ dydx

2 dydx = -cos x

2y d2ydx2

+ 2dydx

2

+ cos x = 0

2y d2ydx2

+ 2dydx

2

+ y2 = 0 [Shown]

14 y = e-2x sin x

dydx

= e-2x cos x - 2 sin x e-2x

dydx

= e-2x cos x - 2y e-2x sin x = y

d2ydx2 = -e-2x sin x - 2 cos x e-2x - 2

dydx

d2ydx2 = -y - 2dy

dx + 2y - 2

dydx

cos x e-2x =

dydx

+ 2y

d2ydx2 + 4

dydx

+ 5y = 0 [Shown]

15 y = ln (1 - cos x)

dydx

= sin x1 - cos x

…

d2ydx2

= (1 - cos x)(cos x) - sin x sin x

(1 - cos x)2

= cos x - cos2 x - sin2 x(1 - cos x)2

= cos x - (cos2 x + sin2 x)

(1 - cos x)2

= cos x - 1(1 - cos x)2

= - 1 - cos x(1 - cos x)2

e-2x sin x = y

Chap-08-FWS.indd 7 10/19/2012 10:37:44 AM



= -11 - cos x

= 1

cos x - 1

But from , 1cos x - 1

= - 1sin x

dydx.

∴ d2y

dx2 = - 1sin x

dydx

sin x d2ydx2 = -dy

dx

sin x d2ydx2 + dy

dx = 0 [Shown]

16 y = esin x

dydx

= cos x esin x

dydx

= y cos x

d2ydx2 = y(-sin x) + cos x dy

dx

d2ydx2 = -y sin x + cos x dy

dx

d2ydx2 = -y sin x + 1

y dydx

dydx

cos x = 1y dydx

d2ydx2 = -y sin x + 1

y dy

dx2

y d2ydx2 = -y2 sin x + dy

dx2

y d2ydx2 = -y2 ln y + dy

dx2

y d2ydx2 + y2 ln y - dy

dx2

= 0 [Shown]

17 y = ln (sin x + cos x)

dydx

= cos x - sin xsin x + cos x

dydx

2

+ 1 = cos x - sin xsin x + cos x

2

+ 1

= (cos x - sin x)2 + (sin x + cos x)2

(sin x + cos x)2

= cos2 x - 2 sin x cos x + sin2 x + sin2 x + 2 sin x cos x + cos2 x

(sin x + cos x)2

esin x = y

sin x = ln y

Chap-08-FWS.indd 8 10/19/2012 10:37:44 AM



= cos2 x + sin2 x + sin2 x + cos2 x(sin x + cos x)2

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

sin2 x + cos2 x = 1

= 2(sin x + cos x)2

[Shown]

d2ydx2 =

(sin x + cos x)(-sin x - cos x) - (cos x - sin x)(cos x - sin x)(sin x + cos x)2

= -sin2 x - 2 sin x cos x - cos2 x - (cos2 x - 2 sin x cos x + sin2 x)

(sin x + cos x)2

= -2 sin2 x - 2 cos2 x(sin x + cos x)2

= -2(sin2 x + cos2 x)

(sin x + cos x)2

= -2(1)

(sin x + cos x)2

= - 3dydx

2

+ 1 ∴ d

2ydx2 + dy

dx2 + 1 = 0 [Shown]

18 (a) y = x2

(x + 3)(x - 1)

= x2

x2 + 2x - 3

As y → ± ∞, the denominator of x2

(x + 3)(x - 1) → 0

(x + 3)(x - 1) → 0 x → -3 or 1

Therefore, x = -3 and x = 1 are vertical asymptotes.

limx → ±∞

y = limx → ±∞

x2

x2 + 2x - 3

= lim

x→± ∞

x2

x2

x2

x2 + 2x

x2 - 3

x2

= limx→± ∞

1

1 + 2x - 3

x2 = 1

1 + 0 + 0

= 1

Therefore, y = 1 is the horizontal asymptote.

Chap-08-FWS.indd 9 10/19/2012 10:37:44 AM



(b) y = x2

x2 + 2x - 3

dydx

= (x2 + 2x - 3)(2x) - x2(2x + 2)

(x2 + 2x - 3)2

= 2x3 + 4x2 - 6x - 2x3- 2x2

(x2 + 2x - 3)2

= 2x2 - 6x(x2 + 2x - 3)2

d2ydx2 = (x

2 + 2x - 3)2(4x - 6) - (2x2 - 6x) 2(x2 + 2x - 3)(2x + 2)(x2 + 2x - 3)4

= 2(x2 + 2x - 3)[(x2 + 2x - 3)(2x - 3) - (2x2 - 6x)(2x + 2)](x2 + 2x - 3)4

= 2[(x2 + 2x - 3)(2x - 3) - (2x2 - 6x)(2x + 2)]

(x2 + 2x - 3)3

When dydx

= 0, 2x2 - 6x(x2 + 2x - 3)2

= 0

2x2 - 6x = 0

2x(x - 3) = 0

x = 0 or 3

When x = 0, y = 0 and

d2ydx2 =

2[(-3)(-3) - 0](-3)3

= -23 (<0)

Therefore, (0, 0) is a turning point and it is a local maximum point.

When x = 3, y = 96(2)

= 34

and

d2ydx2 =

2[(32 + 2 × 3 - 3)(2 × 3 - 3) - 0](32 + 2 × 3 - 3)3 = 6(>0)

Therefore, 3, 34 is a turning point and it is a local minimum point.

(c) When y = 0, x = 0.

Hence, the graph of y = x2

(x + 3)(x - 1)

= x2

x2 + 2x - 3 is as shown.

Chap-08-FWS.indd 10 10/19/2012 10:37:44 AM



x

Y

−3 O

31

1

34

,� �

19 (a) y = 4(x - 3)2 - 1x - 3

x = 3 is the vertical asymptote.

(b) When x = 0, y = 4(-3)2 - 1(-3)

= 36 13

.

Thus, the graph cuts the y-axis at 0, 36 13.

When y = 0, 4(x - 3)2 - 1x - 3

= 0

4(x - 3)2 = 1x - 3

(x - 3)3 = 14

x - 3 = 1

413

x = 1

413

+ 3

x = 3.63

Thus, the graph cuts the x-axis at (3.63, 0).

(c) y = 4(x - 3)2 - 1x - 3

= 4(x - 3)2 - (x - 3)-1

dydx

= 8(x - 3)1(1) + (x - 3)-2(1)

= 8(x - 3) + 1(x - 3)2

d2ydx2 = 8 - 2(x - 3)-3(1)

= 8 - 2(x - 3)3

Chap-08-FWS.indd 11 10/19/2012 10:37:45 AM



When dydx

= 0,

8(x - 3) + 1(x - 3)2

= 0

8(x - 3) = - 1(x - 3)2

(x - 3)3 = -18

x - 3 = - 12

x = 2 12

When x = 212

,

y = 452 - 32

- 1

52 - 3 = 1 + 2 = 3

d2ydx2 = 8 - 2

52 - 33

= 8 - (-16) = 24 (> 0)

Therefore, the turning point is 212

, 3 and it is a local minimum point.

(d) When d2ydx2

= 0,

8 - 2(x - 3)3

= 0

2(x - 3)3

= 8

(x - 3)3 = 14

x = 1

413

+ 3

x = 3.63

From (b), when x = 3.63, y = 0.

d3ydx3

= 6(x - 3)-4(1)

= 6(x - 3)4

When x = 3.63, d3ydx3

= 6(3.63 - 3)4

= 38.1 (i.e. ≠ 0)

Hence, (3.63, 0) is a point of inflexion.

Chap-08-FWS.indd 12 10/19/2012 10:37:45 AM



(e) The graph of y = 4(x - 3)2 - 1x - 3

is as shown below.

x

y

3 3.63O

2 12

, 3� �

3613

20 (a) The x-axis is the axis of symmetry.

(b) y 2 = x2(4 - x)

y2 ≥ 0

x2(4 - x) ≥ 0

Since x2 ≥ 0, x2(4 - x) ≥ 0 only if 4 - x ≥ 0 i.e. x ≤ 4.

Hence, the set of values of x where the graph does not exist is {x : x > 4}.

(c) y2 = x2(4 - x)

= 4x2 - x3

2y dydx

= 8x - 3x2

dydx

= 8x - 3x2

2y

dydx

= 8x - 3x2

2(± x 4 - x )

dydx

= x(8 - 3x)

± 2x 4 - x

dydx

= 8 - 3x± 2 4 - x

When dydx

= 0,

8 - 3x± 2 4 - x

= 0

8 - 3x = 0

x = 83

Chap-08-FWS.indd 13 10/19/2012 10:37:45 AM



When x = 83

, y = ± 83

4 - 83

= ±3.08

Hence, 223

, 3.08 and 223

, -3.08 are turning points (whose tangents are horizontal).

When dydx

= ∞,

2 4 - x = 0 x = 4

When x = 4, y = ±4 4 - 4 = 0 Hence, (4, 0) is also a turning point where tangent is vertical.

(d) The graph of y 2 = x2(4 - x) is as shown below.

x

y

O 4

2 23

, −3.08� �

2 23

, 3.08� �

21 (a) y = 1 - e2x

1 + e2x

dydx

= (1 + e2x)(-2e2x) - (1 - e2x)(2e2x)

(1 + e2x)2

dydx

= -2e2x[1 + e2x + (1 - e2x)]

(1 + e2x)2

dydx

= -4e2x

(1 + e2x)2

Since e2x > 0 and (1 + e2x)2 > 0, thus dydx

= -4e2x

(1 + e2x)2 < 0 [Shown]

(b) y = 1 - e2x

1 + e2x

y + ye2x = 1- e2x

e2x(1 + y) = 1 - y

e2x = 1 - y1 + y

2x = ln 1 - y1 + y

x = 12

ln 1 - y1 + y [Shown]

Chap-08-FWS.indd 14 10/19/2012 10:37:46 AM



dydx

= -4e2x

(1 + e2x)2

= -4e212 ln 1 - y

1 + y

3(1 + e)214 ln 2 1 - y

1 + y2

=

-41 - y1 + y

1 + 1 - y1 + y

2 alogax = x

= -41 - y

1 + y1 + y + 1 - y

1 + y 2

= -41 - y

1 + y4

(1 + y)2

= -(1 - y)(1 + y)

= y 2 - 1 [Shown]

d2ydx2

= 2ydydx

Since dydx

< 0, d2ydx2

< 0 if y > 0 and d2ydx2

> 0 if y < 0 [Shown]

(c) limx→∞ 1 - e2x

1 + e2x = -1 and limx → -∞1 - e2x

1 + e2x = 1

(d) When y = 0, 1 - e2x

1 + e2x = 0

1 - e2x = 0

e2x = 1

2x = ln 1

2x = 0

x = 0

Thus, (0, 0) is a point of inflexion.

Hence, the graph of y = 1 - e2x

1 + e2x is as shown below.

y

xO

1

−1

Chap-08-FWS.indd 15 10/19/2012 10:37:46 AM



22 (a)

k cm

6 cm

(2k + 6) cm

Q

CRSD

x cm

BA

P

∆CQR and ∆CBS are similar triangles.

k cm

B

Q

CRS

x cm

[(2k + 6) − 6] cm

Thus, RCSC

= QR

BS

RC(2k + 6) - 6

= xk

RC2k

= xk

RC = 2x

Thus, DR = DC - RC

= 2k + 6 - 2x

(b) Area of PQRD,

L = DR × QR

L = (2k + 6 - 2x)(x)

L = (2k + 6)x - 2x2 [Shown]

(c) When L has a stationary value,

dLdx

= 0

2k + 6 - 4x = 0

4x = 2k + 6

x = 2k + 64

x = 2(k + 3)4

x = k + 32

d2Ldx2

= -4 (negative)

Thus, L has a maximum value.

Chap-08-FWS.indd 16 10/19/2012 10:37:47 AM



Hence, the maximum value of L is

L = (2k + 6)k + 32 - 2k + 3

2 2

= 2(k + 3)k + 32 - 2(k + 3)2

4

= (k + 3)2 - (k + 3)2

2

= (k + 3)2

2

23 In ∆QMC, sin x = MCr

MC = r sin x

∴ AC = 2MC

= 2r sin x

r cmr c

m

O

MCA

B

x

x x

r cm

In ∆OMC, cos x = OMr

OM = r cos x

Area of ∆ABC,

L = 12

× AC × BM

L = 12

× AC × (BO + OM)

L = 12

× (2r sin x) × (r + r cos x)

L = r 2 sin x + r 2 sin x cos x

L = r 2 sin x + 12

r 2(2 sin x cos x)

L = r 2 sin x + 12

r 2 sin 2x

L = 12

(2r 2 sin x + r 2 sin 2x)

L = r2

2 (2 sin x + sin 2x) [Shown]

dLdr

= r2

2 (2 cos x + 2 cos 2x)

Chap-08-FWS.indd 17 10/19/2012 10:37:48 AM



When L has a stationary value,

dLdx

= 0

r2

2 (2 cos x + 2 cos 2x) = 0

cos x + cos 2x = 0 cos x + 2 cos2 x - 1 = 0 2 cos2 x + cos x - 1 = 0 (2 cos x - 1)(cos x + 1) = 0

cos x = 12

or cos x = -1

x = π3

x = π (not accepted)

d2Ldx2

= r2

2 (-2 sin x - 4 sin 2x)

When x = π3

,

d2Ldx2

= r 2

2 -2 sin π3

- 4 sin 2π3

= -2.60r 2 (< 0)

Hence, L is a maximum.

Lmax

= r 2

2 2 sin

π3

+ sin 2π3

= r2

2 2 × 3

2 + 3

2 = 3 3

4 r2 [Shown]

24 In ∆ORQ, cos a = ORr

OR = r cos a QM = MP = OR = r cos a

r cmr cm

M

R O

PQ

a a

In ∆ORQ, sin a = QRr

QR = r sin a Therefore, the perimeter of ORQP,

y = OR + RQ + QM + MP + PO y = r cos a + r sin a + r cos a + r cos a + r y = r + r sin a + 3r cos a y = r(1 + sin a + 3 cos a) [Shown]

Chap-08-FWS.indd 18 10/19/2012 10:37:48 AM



dyda

= r(cos a - 3 sin a)

When y has a stationary value,

dydr

= 0

r(cos a - 3 sin a) = 0 cos a - 3 sin a = 0 cos a = 3 sin a

13

= sin acos a

tan a = 13

a = tan-1 13 rad [Shown]

d2yda2

= r(-sin a - 3 cos a)

Since sin a > 0 and cos a > 0, d2yda2

< 0.

Thus, y is a maximum.

ymax

= r(1 + sin a + 3 cos a)

ymax

= r1 + 1

10 + 3 × 3

10 y

max = r1 + 10

10 + 3 × 3 10

10 y

max = (1 + 10 )r [Shown]

√10

3

1

a

sin a = 1

10, cos a =

3

10

25 (a) dVdt

= Change in volume

Change in time

= 12

3

- 13

24

= - 7

192 m3 hour-1

(b) V = x3

dVdx

= 3x2

dxdV

= 1

3x2

dxdt

= dxdV

× dVdt

= 1

3x2 × - 7

192 = 1

3(0.7)2 × - 7

192 = -0.0248 m hour-1

∴ Rate of decrease = 0.0248 m hour-1

Chap-08-FWS.indd 19 10/19/2012 10:37:49 AM



26 AB = x2 + 32

= (x2 + 9)12

d(AB)dx

= 12

(x2 + 9)-1

2 (2x)

= xx2 + 9

d(AB)

dt =

d(AB)dx

× dxdt

= xx2 + 9

× 2

= 442 + 9

× 2

= 1.6 units s-1

27 (a) (x + 2 − r ) cm

r cm

r cm

N

O

P

Q

x cm

R(x + 2) cm

∆RNO and ∆RQP are similar triangles.

Thus, NO QP

= NR QR

rx

= x + 2 - rx + 2

r(x + 2) = x(x + 2 - r) rx + 2r = x2 + 2x - rx 2rx + 2r = x2 + 2x r(2x + 2) = x2 + 2x

r = x2 + 2x

2x + 2 [Shown]

(b) r = x2 + 2x

2x + 2

drdx

= (2x + 2)(2x + 2) - (x2 + 2x)(2)

(2x + 2)2

drdx

= 4x2 + 8x + 4 - 2x2 - 4x(2x + 2)2

drdx

= 2x2 + 4x + 4(2x + 2)2

drdx

= 2(x2 + 2x + 2)

[2(x + 1)]2

drdx

= x2 + 2x + 22(x + 1)2

Chap-08-FWS.indd 20 10/19/2012 10:37:49 AM



dxdt

= dxdr

× drdt

= 2(x + 1)2

x2 + 2x + 2 × (-0.4)

= 2(4 + 1)2

42 + 2(4) + 2 × (-0.4)

= 5026

× -0.4

= -0.769 cm s-1

(c) δrδx

≈ drdx

δr ≈ drdx

× δx

= x2 + 2x + 22(x + 1)2 × (4.002 - 4)

= 42 + 2(4) + 22(4 + 1)2 × 0.002

= 0.00104 cm

28 y = xex + 1

dydx

= x ex + 1 + ex + 1 …

= ex + 1(x + 1)

δy ≈ dydx

× δ x

= ex + 1(x + 1) d x x changes from 1 to 1.01. So, δ x = 1.01 - 1.

ynew

= yoriginal

+ δ y

1.01e2.01 = 1(e1 + 1) + [e1 + 1(1 + 1)](1.01 - 1)

The value of y when x = 1.01.

The value of y when x = 1.

The value of dydx

when x = 1.

1.01e2.01 = e2 + 2e2(0.01)

e2.01 = 7.3891 + 2(7.3891)(0.01)

1.01 ∴ e2.01 = 7.46

29 y = cosx

x xy = cos x

x dydx

+ y(1) = -sin x

x dydx

+ y = -sin x

Chap-08-FWS.indd 21 10/19/2012 10:37:49 AM



x d2ydx2

+ dydx

(1) + dydx

= -cos x

x d2ydx2

+ 2 dydx

= -xy

x d2ydx2 + 2

dydx

+ xy = 0 [Shown]

30 y = x ln (x + 1)

dydx

= x 1x + 1 + ln (x + 1)(1)

= xx + 1

+ ln (x + 1)

δ y = dydx

× δ x

= ( xx + 1

+ ln (x + 1)) δ x

ynew

= yoriginal

+ δ y x changes from 1 to 1.01. So, δ x = 1.01 - 1.

1.01 ln (1.01 + 1) = 1 ln (1 + 1) + 3 11 + 1

+ ln (1 + 1) (1.01 - 1)

The value of y when x = 1.01.

The value of y when x = 1.

The value of dydx

when x = 1.

1.01(ln 2.01) = 0.70508 ∴ ln 2.01 = 0.698

31 (a) f (t) = 4ekt - 14ekt + 1

f (0) = 4e0 - 14e0 + 1

= 35

(b) f ′(t) = (4ekt + 1)(4kekt) - (4ekt - 1)(4kekt)

(4ekt + 1)2

f ′(t) = (16ke2kt + 4kekt - 16ke2kt + 4kekt)

(4ekt + 1)2

f ′(t) = 8kekt

(4ekt + 1)2

Since k is a positive integer, f ′(t) > 0.

(c) LHS = k{1 - [f (t)]2}

= k {1 - 34ekt - 14ekt + 1

2} = k {(4ekt + 1)2 - (4ekt - 1)2

(4ekt + 1)2 }

Chap-08-FWS.indd 22 10/19/2012 10:37:50 AM



= k {16e2kt + 8ekt + 1 - (16e2kt - 8ekt + 1)(4ekt + 1)2 }

= 16kekt

(4ekt + 1)2

= 23 8kekt

(4ekt + 1)2 = 2f ′(t) = RHS

k{1 - [f(t)]2} = 2f ′(t) k - k[f (t)]2 = 2f ′(t) -2k[f(t)] f ′(t) = 2f ″(t) -k[f (t)] f ′(t) = f ″(t) f ″(t) = -k[f(t)] f ′(t) Since k and f ′(t) are both positive, f ″(t) < 0.

(d) limt→∞

f (t) = limt→∞

4ekt - 14ekt + 1

= limt→∞

4ekt

ekt - 1

ekt

4ekt

ekt + 1ekt

= limt→∞

4 - 1ekt

4 + 1ekt

= 4 - 04 + 0

= 1

(e) lim t→∞

4ekt - 14ekt + 1 = lim

t→∞4 - 1ekt

4 + 1ekt

= 4 - 04 + 0

= 1

When t = 0, f (0) = 4(1) - 14(1) + 1

= 35

Therefore, the graph of f(t) intersects the f(t)-axis at the point 0, 35.

When f (t) = 0, 4ekt - 14ekt + 1 = 0

4ekt - 1 = 0

ekt = 14

kt = ln 14

t = 1k ln 1

4

Chap-08-FWS.indd 23 10/19/2012 10:37:50 AM



t

f(t )

O

f(t ) =4e(kt) − 1

4e(kt) + 1

35

−1

1

41—1

k— ln

Key Point:

If f ″(t) = -k[f (t)] f ′(t), since k > 0 and f ′(t) > 0, f ″(t) > 0 only when f (t) = 0. Therefore, the point of

inflexion is on the t-axis, i.e. 1k

ln 14

, 0.

32 y = x1 + x2

dydx

= (1 + x2)(1) - x(2x)

(1 + x2)2

dydx

= 1 - x2

(1 + x2)2

dydx

= 1 - x2

xy

2

dydx

= (1 - x2) y2

x2

x2 dydx

= (1 - x2) y2 [Shown]

33 y = sin x - cos xsin x + cos x

(sin x + cos x)y = sin x - cos x

(sin x + cos x) dydx

+ y (cos x - sin x) = cos x + sin x

(sin x + cos x)dydx

- 1 + y(cos x - sin x) = 0

(sin x + cos x)dydx

- 1 - y(sin x - cos x) = 0

dydx

- 1 - ysin x - cos xsin x + cos x = 0

dydx

- 1 - y(y) = 0

Chap-08-FWS.indd 24 10/19/2012 10:37:50 AM



dydx

- 1 - y2 = 0

d2ydx2

- 2ydydx

= 0

d2ydx2 = 2y

dydx

[Shown]

34 y = x3

x2 - 1

dydx

= (x2 - 1)(3x2) - x3(2x)

(x2 - 1)2

= 3x4 - 3x2 - 2x4

(x2 - 1)2

= x4 - 3x2

(x2 - 1)2

= x2(x2 - 3)(x2 - 1)2

d2ydx2 =

(x2 - 1)2(4x3 - 6x) - (x4 - 3x2)(2)(x2 - 1)1(2x)(x2 - 1)4

= (x2 - 1)2(2x)(2x2 - 3) - 4x(x4 - 3x2)(x2 - 1)

(x2 - 1)4

= 2x(x2 - 1)[(x2 - 1)(2x2 - 3) - 2(x4 - 3x2)]

(x2 - 1)4

= 2x[2x4 - 5x2 + 3 - 2x4 + 6x2](x2 - 1)3

= 2x(x2 + 3)(x2 - 1)3

d3ydx3 = (x

2 - 1)3(6x2 + 6) - (2x3 + 6x)(3)(x2 - 1)2(2x)(x2 - 1)6

= 6(x2 - 1)3(x2 + 1) - 6x(2x3 + 6x)(x2 - 1)2

(x2 - 1)6

= 6(x2 - 1)2[(x2 - 1)(x2 + 1) - x(2x3 - 6x)]

(x2 - 1)6

= 6(x4 - 1 - 2x4 - 6x2)(x2 - 1)4

= 6(-x4 - 6x2 - 1)(x2 - 1)4

When dydx

= 0

x2(x2 - 3) = 0

x = 0 or ± 3

When x = 0, y = 0 and d2ydx2 =

2(0)(02 + 3)(02 - 1)3

= 0

Chap-08-FWS.indd 25 10/19/2012 10:37:50 AM



When x = 0, d3ydx3 =

6(-04 - 6(0)2 - 1)(02 - 1)4

= -6.

Since d3ydx3 ≠ 0, then (0, 0) is a point of reflextion.

When x = 3, y = 3 3

3 - 1

= - 3 3

2 and

d2ydx2 = 2 3(3 + 3)

(3 - 1)3

= 32 3.

Since d2ydx2 > 0, then 3, 3

2 3 is a minimum point.

When x = - 3, y = (- 3)3

3 - 1

= - 3 3

2 and

d2ydx2 =

2(- 3)(3 + 3)(3 - 1)3

= -32 3.

Since d2ydx2 < 0, then - 3, -3

2 3 is a maximum point.

When the denominator of y = x3

x2 - 1 is 0, x2 - 1 = 0 ⇒ x = ±1

Hence, x = -1 and x = 1 are asymptotes.

The graph of y = x3

x2 - 1 is as shown below.

y

√3 , 3√23 ��

√3 , 3√23 ��

1−1 Ox

−−

Chap-08-FWS.indd 26 10/19/2012 10:37:51 AM



x3 = k (x2 - 1)

x3

x2 - 1 = k

By sketching the straight lines y = k on the above graph and as k varies, we obtain the following results.

Value of k Number of real roots

k > 32 3 3

k = 32 3 2

-32 3 < k < 3

2 3 1

k = -32 3 2

k < -32 3 3

35 y = xx2 - 1

dydx

= (x2 - 1)(1) - x(2x)

(x2 - 1)2

= -x2 - 1(x2 - 1)2

= -(x2 + 1)(x2 - 1)2

(that is <0)

Since dydx

<0 for all real values of x, then the gradient of the curve is always decreasing.

d2ydx2 = -(x2 - 1)2(2x) + (x2 + 1)(2)(x2 - 1)(2x)

(x2 - 1)4

= -2x(x2 - 1)[x2 - 1 - 2(x2 + 1)]

(x2 - 1)4

= -2x(-x2 - 3)(x2 - 1)3

= 2x(x2 + 3)(x2 - 1)3

When d2ydx2 = 0,

2x(x2 + 3)(x2 - 1)3

= 0

x = 0

When x = 0, y = 002 - 1

= 0

Chap-08-FWS.indd 27 10/19/2012 10:37:51 AM



d3ydx3 = (x

2 - 1)3(6x2 + 6) - (2x)(x2 + 3)(3)(x2 - 1)2(2x)(x2 - 1)6

= 6(x2 - 1)3(x2 + 1) - (12x2)(x2 + 3)(x2 - 1)2

(x2 - 1)6

= 6(x2 - 1)2[(x2 - 1)(x2 + 1) - (2x2)(x2 + 3)](x2 - 1)6

= 6(x2 - 1)2(x4 - 1 - 2x4 - 6x2)(x2 - 1)6

= 6(-x4 - 1 - 6x2)(x2 - 1)4

When x = 0, d3ydx3 =

6[-04 - 1 - 6(0)2](02 - 1)4

= -6 (that is ≠ 0)

Since d2ydx2 = 0 and

d3ydx3 ≠ 0 when x = 0, then (0, 0) is the point of inflexion.

When the curve concaves upwards,

d2ydx2 > 0

2x(x2 + 3)(x2 - 1)3

> 0

2x(x2 + 3)

[(x + 1)(x - 1)]3 > 0

2x(x2 + 3)(x + 1)3(x - 1)3

> 0

−

+

+

++

+

x

x > 0

− − +(x − 1)3 > 0

+

−

−

−

+ + +x2 + 3 > 0

(x + 1)3 > 0

−0−1 +− +1

Hence, the intervals for which the curve concaves upwards are -1 < x < 0 or x > 1.

The curve y = xx2 - 1

is as shown below.

Ox

y

−1 1

Chap-08-FWS.indd 28 10/19/2012 10:37:52 AM



36 (a) x = t - 2t y = 2t + 1

t

dxdt

= 1 + 2t2

dydt

= 2 - 1t2

dydx

=

dydtdxdt

= 2 - 1

t2

1 + 2t2

= 2t2 - 1t2 + 2

t2 + 2 2

2t2 - 1 2t2 + 4 - 5

∴ dydx

= 2 - 5t2 + 2

[Shown]

Let m = dydx

m = 2 - 5t2 + 2

(m - 2) = - 5t2 + 2

(m - 2)(t2 + 2) = -5

mt2 + 2m - 2t2 - 4 = -5

(m - 2)t2 = -1 - 2m

t2 = -1 - 2mm - 2

t2 = 1 + 2m2 - m

t2 > 0

1 + 2m2 - m

> 0

+

−

+

−+

+

x

1 + 2m > 0

2 − m > 0

+−− −212

Hence, -12

< m < 2, that is, - 12

< dydx

< 2 [Shown]

The question states that t ≠ 0.So, we write t2 > 0 and not t2 ≥ 0.

Chap-08-FWS.indd 29 10/19/2012 10:37:52 AM



(b) dydx

= 13

2 - 5t2 + 2

= 13

2 - 13

= 5t2 + 2

53

= 5t2 + 2

t2 + 2 = 3

t2 = 1

t = ±1

When t = 1, x = 1 - 21

= -1

and

y = 2(1) + 11

= 3

When t = -1, x = (-1) - 2(-1)

= 1

and

y = 2(-1) + 1(-1)

= -3

Hence, the coordinates of the required points are (-1, 3) and (1, -3).

Chap-08-FWS.indd 30 10/19/2012 10:37:52 AM

chapter 8 differentiation

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