EXTRA EXERCISE FROM JOHN BIRDS BOOK

384 Engineering Mathematics

Now try the following exercise f (x)
B

Exercise 148 Further problems on A C f (x2)
functional notation D
f (x1) x2 x
1. If f (x) = 6x2 − 2x + 1 find f (0), f (1), f (2), E
f (−1) and f (−3). B f (x) ϭ x 2
[1, 5, 21, 9, 61] 0 x1

2. If f (x) = 2x2 + 5x − 7 find f (1), f (2), f (−1), Figure 42.2
f (2) − f (−1). f (x)
[0, 11, −10, 21] 10

3. Given f (x) = 3x3 + 2x2 − 3x + 2 prove that 8

f (1) = 1 f (2)
7

4. If f (x) = −x2 + 3x + 6 find f (2), f (2 + a), 6
f (2 + a) − f (2) and f (2 + a) − f (2)
a

[8, −a2 − a + 8, −a2 − a, − a − 1] 4C
2 AD

42.3 The gradient of a curve 0 1 1.5 2 3x
Figure 42.3
(a) If a tangent is drawn at a point P on a curve, then
the gradient of this tangent is said to be the gra- (c) For the curve f (x) = x2 shown in Fig. 42.3:
dient of the curve at P. In Fig. 42.1, the gradient
of the curve at P is equal to the gradient of the (i) the gradient of chord AB
tangent PQ.
= f (3) − f (1) = 9 − 1 = 4
f (x) 3 − 1 2

Q

Section 8 (ii) the gradient of chord AC

P = f (2) − f (1) = 4 − 1 = 3
2 − 1 1

0 x (iii) the gradient of chord AD
Figure 42.1
f (1.5) − f (1) 2.25 − 1
= 1.5 − 1 = 0.5 = 2.5
(b) For the curve shown in Fig. 42.2, let the points
(iv) if E is the point on the curve (1.1, f (1.1))
A and B have co-ordinates (x1, y1) and (x2, y2), then the gradient of chord AE
respectively. In functional notation, y1 = f (x1) and
y2 = f (x2) as shown.

The gradient of the chord AB = f (1.1) − f (1)
1.1 − 1

= BC = BD − CD = 1.21 − 1 = 2.1
AC ED 0.1
= f (x2) − f (x1)
(x2 − x1) (v) if F is the point on the curve (1.01, f (1.01))
then the gradient of chord AF

Introduction to differentiation 385

f (1.01) − f (1) Hence δy = f (x +δx) − f (x)
= 1.01 − 1 δx δx
= 1.0201 − 1 = 2.01
δy
0.01 As δx approaches zero, approaches a limiting

Thus as point B moves closer and closer to point A the δx
gradient of the chord approaches nearer and nearer to the value and the gradient of the chord approaches
value 2. This is called the limiting value of the gradient
of the chord AB and when B coincides with A the chord the gradient of the tangent at A.
becomes the tangent to the curve.
(ii) When determining the gradient of a tangent to
Now try the following exercise a curve there are two notations used. The gradi-
ent of the curve at A in Fig. 42.4 can either be
Exercise 149 A further problem on the written as:
gradient of a curve
δy f (x + δx) − f (x)
1. Plot the curve f (x) = 4x2 − 1 for values of x limit or limit
from x = −1 to x = +4. Label the co-ordinates δx→0 δx δx→0 δx
(3, f (3)) and (1, f (1)) as J and K, respec-
tively. Join points J and K to form the chord dy δy
JK. Determine the gradient of chord JK. By In Leibniz notation, = limit
moving J nearer and nearer to K determine
the gradient of the tangent of the curve at K. dx δx→0 δx
[16, 8]
In functional notation,

f (x) = limit f (x + δx)−f (x)
δx→0 δx

dy
(iii) is the same as f (x) and is called the differ-

dx
ential coefficient or the derivative. The process
of finding the differential coefficient is called
differentiation.

Summarising, the differential coefficient,

42.4 Differentiation from first dy = f (x) = limit δy
principles dx δx→0 δx

(i) In Fig. 42.4, A and B are two points very close = limit f (x+δx)−f (x)
together on a curve, δx (delta x) and δy (delta y) δx→0 δx
representing small increments in the x and y
directions, respectively. Problem 3. Differentiate from first principles Section 8
f (x) = x2 and determine the value of the gradient of
y
the curve at x = 2

B(x ϩd x, y ϩd y) To ‘differentiate from first principles’ means ‘to find
f (x)’ by using the expression

dy f (x) = limit f (x + δx) − f (x)
f (xϩd x)
A(x,y) δx→0 δx

f(x) dx f (x) = x2

0x Substituting (x + δx) for x gives
f (x + δx) = (x + δx)2 = x2 + 2xδx + δx2, hence

Figure 42.4 f (x) = limit (x2 + 2xδx + δx2) − (x2)
δx
Gradient of chord AB = δy δx→0
δx 2xδx + δx2 = limit{2x + δx}
= limit δx δx→0
However, δy = f (x + δx) − f (x)
δx→0

386 Engineering Mathematics

As δx → 0, [2x + δx] → [2x + 0]. Thus f (x) = 2x, i.e. result could have been determined by inspection. ‘Find-
the differential coefficient of x2 is 2x. At x = 2, the ing the derivative’ means ‘finding the gradient’, hence,
in general, for any horizontal line if y = k (where k is a
gradient of the curve, f (x) = 2(2) = 4 constant) then dy = 0.

Problem 4. Find the differential coefficient of dx
y = 5x
Problem 6. Differentiate from first principles
By definition, dy = f (x) f (x) = 2x3
dx

= limit f (x + δx) − f (x) Substituting (x + δx) for x gives
δx→0 δx f (x + δx) = 2(x + δx)3
= 2(x + δx)(x2 + 2xδx + δx2)
The function being differentiated is y = f (x) = 5x. Sub- = 2(x3 + 3x2δx + 3xδx2 + δx3)
stituting (x + δx) for x gives: = 2x3 + 6x2δx + 6xδx2 + 2δx3

f (x + δx) = 5(x + δx) = 5x + 5δx. Hence

dy = f (x) = limit (5x + 5δx) − (5x)
dx δx→0 δx
dy f (x + δx) − f (x)
= f (x) = limit
= lim it 5δx = limit{5} dx δx→0 δx

δx→0 δx δx→0 (2x3 + 6x2δx + 6xδx2 + 2δx3) − (2x3)

= limit δx
δx→0
Since the term δx does not appear in [5] the limiting
value as δx → 0 of [5] is 5. Thus dy = 5, i.e. the differen- = limit 6x2δx + 6xδx2 + 2δx3

dx δx→0 δx
tial coefficient of 5x is 5. The equation y = 5x represents
= limit{6x2 + 6xδx + 2δx2}
a straight line of gradient 5 (see Chapter 28). The ‘differ-
dy δx→0

ential coefficient’ (i.e. or f (x)) means ‘the gradient Hence f (x) = 6x2, i.e. the differential coefficient of 2x3
dx is 6x2.

Section 8 of the curve’, and since the slope of the line y = 5x is 5 Problem 7. Find the differential coefficient of
y = 4x2 + 5x − 3 and determine the gradient of the
this result can be obtained by inspection. Hence, in gen-
curve at x = −3
eral, if y = kx (where k is a constant), then the gradient
dy

of the line is k and or f (x) = k.
dx

Problem 5. Find the derivative of y = 8

y = f (x) = 8. Since there are no x-values in the orig- y = f (x) = 4x2 + 5x − 3
inal equation, substituting (x + δx) for x still gives f (x + δx) = 4(x + δx)2 + 5(x + δx) − 3
f (x + δx) = 8. Hence
= 4(x2 + 2xδx + δx2) + 5x + 5δx − 3
dy f (x + δx) − f (x) = 4x2 + 8xδx + 4δx2 + 5x + 5δx − 3
= f (x) = limit
dx δx→0 δx

= limit 8 − 8 = 0 dy f (x + δx) − f (x)
δx→0 δx = f (x) = limit
dx δx→0 δx

Thus, when y = 8, dy = 0 ⎧ (4x2 + 8xδx + 4δx2 + 5x + 5δx − 3) ⎫
dx ⎪⎪⎨ − (4x2 + 5x − 3) ⎪⎪⎬

The equation y = 8 represents a straight horizontal line = limit ⎩⎪⎪ δx ⎭⎪⎪
and the gradient of a horizontal line is zero, hence the
δx→0

Introduction to differentiation 387

= limit 8xδx + 4δx2 + 5δx 42.5 Differentiation of y = axn by the
δx→0 δx general rule

= limit{8x + 4δx + 5} From differentiation by first principles, a general rule
for differentiating axn emerges where a and n are any
δx→0 constants. This rule is:

i.e. dy = f (x) = 8x + 5 if y = axn then dy = anxn–1
dx dx

At x = −3, the gradient of the curve or, if f (x) = axn then f (x) = anxn–1
= dy = f (x) = 8(−3) + 5 = −19
dx (Each of the results obtained in worked problems 3 to 7
may be deduced by using this general rule).
Now try the following exercise When differentiating, results can be expressed in a
number of ways.
Exercise 150 Further problems on
differentiation from first For example:
principles
(i) if y = 3x2 then dy = 6x,
In Problems 1 to 12, differentiate from first dx
principles.
(ii) if f (x) = 3x2 then f (x) = 6x,
1. y = x [1] (iii) the differential coefficient of 3x2 is 6x,
(iv) the derivative of 3x2 is 6x, and
2. y = 7x [7] (v) d (3x2) = 6x

dx

3. y = 4x2 [8x] Problem 8. Using the general rule, differentiate
4. y = 5x3 [15x2]
5. y = −2x2 + 3x − 12 [−4x + 3] the following with respect to x:
√
(a) y = 5x7 (b) y = 3 x (c) y = 4
x2

6. y = 23 [0] (a) Comparing y = 5x7 with y = axn shows that a = 5

7. f (x) = 9x [9] and n = 7. Using the general rule, Section 8

8. f (x) = 2x 2 dy = anxn−1 = (5)(7)x7−1 = 35x6
3 3 dx

9. f (x) = 9x2 [18x] √1 1
(b) y = 3 x = 3x 2 . Hence a = 3 and n =
2

dy = anxn−1 = (3) 1 x 1 −1
2
10. f (x) = −7x3 [−21x2] dx 2
11. f (x) = x2 + 15x − 4 [2x + 15]
= 3 x− 1 = 3 = √3
2 2x
2 1

12. f (x) = 4 [0] 2x 2

13. Determine d (4x3) from first principles (c) y = 4 = 4x−2. Hence a = 4 and n = −2
dx x2
[12x2]
dy = anxn−1 = (4)( − 2)x−2−1
14. Find d (3x2 + 5) from first principles dx
dx
[6x] = −8x−3 = − 8
x3

388 Engineering Mathematics

Problem 9. Find√the differential coefficient of Now try the following exercise
+ 4 x5 + 7
y = 2 x3 − 4
5 x3
Exercise 151 Further problems on
y = 2 x3 − 4 + 4 x5 + 7 differentiation of y = axn
5 x3 by the general rule

i.e. y = 2 x3 − 4x−3 + 4x5/2 + 7 In Problems 1 to 8, determine the differential
5 coefficients with respect to the variable.

dy 2 1. y = 7x4 [28x3]
= √1
dx 5 (3)x3−1 − (4)( − 3)x−3−1 2. y = √
x 2x
3√
+ (4) 5 x(5/2)−1 + 0 √
2 3. y = t3 t
2
= 6 x2 + 12x−4 + 10x3/2
5 1 3
4. y = 6 + x3 − x4
dy 6 x2 12
i.e. dx = 5 + x4 + 10 x3 = − 1 +1 11
√ + √ −
5. y 3x xx 3 2 x3 x2

1 6. y = 5 − √1 + 2 − 10 + √7
Problem 10. If f (t) = 5t + √ find f (t) x2 x7 x3 2 x9

t3 7. y = 3(t − 2)2

[6t − 12]

1 1 3 8. y = (x + 1)3 [3x2 + 6x + 3]
f (t) = 5t + √ = 5t + 3 2
= 5t1 + t− 9. Using the general rule for axn check the
t3 t2 results of Problems 1 to 12 of Exercise 150,
page 387.
Hence f (t) = (5)(1)t1−1 + −3 t− 3 −1
i.e. 2
2 10. Differentiate f (x) = 6x2 − 3x + 5 and find the

5t0 3 − 5 gradient of the curve at (a) x = −1, and
2
= − 2 t (b) x = 2. [12x − 3 (a) −15 (b) 21]

Section 8 33 11. Find the differential coefficient of
f (t) = 5 − 5 =5− √ y = 2x3 + 3x2 − 4x − 1 and determine
2t 2 2 t5
the gradient of the curve at x = 2.
[6x2 + 6x − 4, 32]

Problem 11. Differentiate y = (x + 2)2 with 12. Determine the derivative of
x y = −2x3 + 4x + 7 and determine the gradi-

respect to x ent of the curve at x = −1.5
[−6x2 + 4, −9.5]
y = (x + 2)2 = x2 + 4x + 4
xx 42.6 Differentiation of sine and
cosine functions
x2 4x 4
=++ Figure 42.5(a) shows a graph of y = sin θ. The gradient
is continually changing as the curve moves from O to
x xx
i.e. y = x + 4 + 4x–1 dy
A to B to C to D. The gradient, given by , may be
Hence dy = 1 + 0 + (4)( − 1)x−1−1
dx dθ

= 1 − 4x−2 = 1 − 4
x2

Introduction to differentiation 389

y y

ϩ A ϩ y ϭ cos q
y ϭ sin u (a)
(a) ππ
0 B D 0 2 3π 2π q radians
π π 3π 2π u radians Ϫ 2
Ϫ 22
DЈ
0Ј C
2π u radians
dy d (sin u) ϭ cos u dy
ϩ dx du ϩ dq
(b)
AЈ CЈ (b) π π 3π 2π q radians
0 0 22
Ϫ ππ 3π
2 2 Ϫ

d (cos q) ϭ Ϫsin q
dq

BЈ

Figure 42.6

Figure 42.5 + 3π , and so on) then the graph shown in Fig. 42.6(b)
2
plotted in a corresponding position below y = sin θ, as
shown in Fig. 42.5(b). would result. This latter graph therefore represents the

(i) At 0, the gradient is positive and is at its steepest. curve of –sin θ.
Hence 0 is the maximum positive value. Thus, if y = cos θ, dy = −sin θ
dθ
(ii) Between 0 and A the gradient is positive but is
decreasing in value until at A the gradient is zero, It may also be shown that:
shown as A . dy

(iii) Between A and B the gradient is negative but is if y = cos a θ, = −a sin aθ
increasing in value until at B the gradient is at its dθ
steepest. Hence B is a maximum negative value. (where a is a constant)

(iv) If the gradient of y = sin θ is further investigated and if y = cos(aθ + α), dy = −a sin(aθ + α)
between B and C and C and D then the resulting dθ
dy
graph of is seen to be a cosine wave. (where a and α are constants).
dθ
Problem 12. Differentiate the following with Section 8
Hence the rate of change of sin θ is cos θ, i.e. respect to the variable: (a) y = 2 sin 5θ
if y = sin θ then dy = cos θ
(b) f (t) = 3 cos 2t
dθ
It may also be shown that: (a) y = 2 sin 5θ
dy = (2)(5) cos 5θ = 10 cos 5θ
dy dθ
if y = sin a θ, = a cos aθ
(b) f (t) = 3 cos 2t
dθ
(where a is a constant) f (t) = (3)(−2) sin 2t = −6 sin 2t

and if y = sin(aθ + α), dy = a cos(aθ + α)
dθ
Problem 13. Find the differential coefficient of
(where a and α are constants). y = 7 sin 2x − 3 cos 4x

If a similar exercise is followed for y = cos θ then the y = 7 sin 2x − 3 cos 4x
dy dy = (7)(2) cos 2x − (3)( − 4) sin 4x
dx
graphs of Fig. 42.6 result, showing to be a graph of
dθ = 14 cos 2x + 12 sin 4x

sin θ, but displaced by π radians. If each point on the

curve y = sin θ (as shown in Fig. 42.5(a)) were to be
π π 3π

made negative, (i.e. + is made − , − is made
2 22

390 Engineering Mathematics

Problem 14. Differentiate the following with 2. Given f (θ) = 2 sin 3θ − 5 cos 2θ, find f (θ)
respect to the variable: [6 cos 3θ + 10 sin 2θ]
(a) f (θ) = 5 sin(100πθ − 0.40)
(b) f (t) = 2 cos(5t + 0.20) 3. An alternating current is given by i = 5 sin 100t
amperes, where t is the time in seconds.
(a) If f (θ) = 5 sin(100πθ − 0.40) Determine the rate of change of current when
f (θ) = 5[100π cos(100πθ − 0.40)] t = 0.01 seconds.
= 500π cos(100πθ − 0.40) [270.2 A/s]

(b) If f (t) = 2 cos (5t + 0.20) 4. v = 50 sin 40t volts represents an alternating
f (t) = 2[−5 sin(5t + 0.20)]
= −10 sin(5t + 0.20) voltage where t is the time in seconds. At a time
of 20 × 10−3 seconds, find the rate of change
Problem 15. An alternating voltage is given by:
v = 100 sin 200t volts, where t is the time in of voltage. [1393.4 V/s]
seconds. Calculate the rate of change of voltage
when (a) t = 0.005 s and (b) t = 0.01 s 5. If f (t) = 3 sin(4t + 0.12) − 2 cos(3t − 0.72)
determine f (t)
v = 100 sin 200t volts. The rate of change of v is given [12 cos(4t + 0.12) + 6 sin (3t − 0.72)]
dv
42.7 Differentiation of eax and ln ax
by .
dt A graph of y = ex is shown in Fig. 42.7(a). The gradient
dy
dv = (100)(200) cos 200t = 20 000 cos 200t
dt of the curve at any point is given by and is continually
(a) When t = 0.005 s, dx

dv = 20 000 cos(200)(0.005) = 20 000 cos 1 changing. By drawing tangents to the curve at many
dt points on the curve and measuring the gradient of the
cos 1 means ‘the cosine of 1 radian’ (make sure
your calculator is on radians — not degrees). dy
tangents, values of for corresponding values of x
dv
Hence = 10 806 volts per second dx
may be obtained. These values are shown graphically
dt
Section 8 (b) When t = 0.01 s, y
20
dv = 20 000 cos(200)(0.01) = 20 000 cos 2.
dt 15 y ϭ e x
Hence dv = −8323 volts per second 10

dt 5

Now try the following exercise (a) Ϫ3 Ϫ2 Ϫ1 0 1 2 3 x

Exercise 152 Further problems on the dy
differentiation of sine and dx 20
cosine functions
15 dy
1. Differentiate with respect to x: (a) y = 4 sin 3x dx
(b) y = 2 cos 6x ϭ ex
[(a) 12 cos 3x (b) −12 sin 6x]
10

5

(b) Ϫ3 Ϫ2 Ϫ1 0 1 2 3 x
Figure 42.7

Introduction to differentiation 391

dy (Note that in the latter expression ‘a’ does not appear in
in Fig. 42.7(b). The graph of against x is identical to dy
dx
the original graph of y = ex . It follows that: the term).
dx
if y = ex, then dy = ex dy 1
dx
Thus if y = ln 4x, then =
It may also be shown that dx x

Problem 16. Differentiate the following with

if y = eax, then dy = aeax respect to the variable: (a) y = 3e2x 4
dx (b) f (t) = 3e5t

Therefore if y = 2e6x, then dy = (2)(6e6x) = 12e6x (a) If y = 3e2x then dy = (3)(2e2x) = 6e2x
dx dx

A graph of y = ln x is shown in Fig. 42.8(a). The (b) If f (t) = 4 = 4 e−5t, then
dy 3e5t 3

gradient of the curve at any point is given by and is = 4 (−5e−5t) = − 20 e−5t = − 20
dx 3 3 3e5t

continually changing. By drawing tangents to the curve

at many points on the curve and measuring the gradient f (t)

dy Problem 17. Differentiate y = 5 ln 3x
of the tangents, values of for corresponding values of

dx
x may be obtained. These values are shown graphically

dy
in Fig. 42.8(b). The graph of against x is the graph
dx If y = 5 ln 3x, then dy = (5) 1 = 5
dy 1 dx x x
of = .
dx x
It follows that: if y = ln x, then dy = 1
dx x
Now try the following exercise

y Exercise 153 Further problems on the
2 differentiation of eax and ln ax
1
0 y ϭ ln x 1. Differentiate with respect to x:
Ϫ1 1 2 3 4 5 6x
Ϫ2 (a) y = 5e3x 2
(a) (b) y = 7e2x
4
dy (a) 15e3x (b) − 7e2x Section 8
dx
dy 1 2. Given f (θ) = 5 ln 2θ − 4 ln 3θ, determine f (θ)
2 dx x 5−4=1
1.5 θθθ
1.0
0.5 ϭ

3. If f (t) = 4 ln t + 2, evaluate f (t) when t = 0.25
[16]

(b) 0 1 2 3 4 5 6x dy
Figure 42.8 4. Evaluate when x = 1, given

dx

y = 3e4x − 5 + 8 ln 5x. Give the answer
2e3x

correct to 3 significant figures. [664]

It may also be shown that
dy 1

if y = ln ax, then =
dx x

Chapter 43

Methods of differentiation

43.1 Differentiation of common Thus dy = (12)(−3)x−3−1
functions dx

= −36x−4 = − 36
x4
The standard derivatives summarised below were
derived in Chapter 42 and are true for all real values of x. Problem 2. Differentiate: (a) y = 6 (b) y = 6x

y or f (x) dy (a) y = 6 may be written as y = 6x0, i.e. in the general
axn of f (x) rule a = 6 and n = 0.
sin ax
cos ax dx Hence dy = (6)(0)x0−1 = 0
eax axn−1 dx

ln ax a cos ax In general, the differential coefficient of a con-
stant is always zero.
−a sin ax (b) Since y = 6x, in the general rule a = 6 and n = 1
aeax
1 Hence dy = (6)(1)x1−1 = 6x0 = 6
x dx

The differential coefficient of a sum or difference is In general, the differential coefficient of kx, where
the sum or difference of the differential coefficients of k is a constant, is always k.
the separate terms.
Thus, if f (x) = p(x) + q(x) − r(x), (where f , p, q Problem 3. Find the derivatives of:
and r are functions), then f (x) = p (x) + q (x) − r (x) √5
Differentiation of common functions is demonstrated
in the following worked problems. (a) y = 3 x (b) y = √
3 x4
Problem 1. Find the differential coefficients of:

(a) y = 12x3 (b) y = 12 √
x3 (a) y = 3 x is rewritten in the standard differential

If y = axn then dy = anxn−1 form as y = 3x1/2
dx In the general rule, a = 3 and n = 1

2

(a) Since y = 12x3, a = 12 and n = 3 thus Thus dy = (3) 1 x 1 −1 = 3 x− 1
2 2

dy = (12) (3)x3−1 = 36x2 dx 2 2
dx
= 3 = 3
12 2x1/2 √
(b) y = x3 is rewritten in the standard axn form as 2x

y = 12x−3 and in the general rule a = 12 and (b) y = 5 = 5 = 5x−4/3
√3 x4 x4/3
n = −3

Methods of differentiation 393

In the general rule, a = 5 and n = − 4 dy 1 6
3 (c) When y = 6 ln 2x then = 6 =
dx x x
Thus dy = (5) − 4 x(−4/3)−1
dx 3

= −20 x−7/3 = −20 = −20 Problem 7. Find the gradient of the curve
3 3x7/3 3√3 x7 y = 3x4 − 2x2 + 5x − 2 at the points (0, −2)

and (1, 4)

Problem 4. Differentiate:

y = 5x4 + 4x − 1 + √1 −3 with respect to x The gradient of a curve at a given point is
2x2 x given by the corresponding value of the deriva-
tive. Thus, since y = 3x4 − 2x2 + 5x − 2 then the
y = 5x4 + 4x − 1 + 1 −3 is rewritten as gradient = dy = 12x3 − 4x + 5
2x2 √
dx
x At the point (0, −2), x = 0.
Thus the gradient = 12(0)3 − 4(0) + 5 = 5
y = 5x4 + 4x − 1 x−2 + x−1/2 − 3
2 At the point (1, 4), x = 1.
Thus the gradient = 12(1)3 − 4(1) + 5 = 13
When differentiating a sum, each term is differentiated
in turn. Problem 8. Determine the co-ordinates of the
point on the graph y = 3x2 − 7x + 2 where the
Thus dy = (5)(4)x4−1 + (4)(1)x1−1 − 1 (−2)x−2−1 gradient is −1
dx 2

+ (1) − 1 x(−1/2)−1 − 0
2

= 20x3 + 4 + x−3 − 1 x−3/2 The gradient of the curve is given by the derivative.
2
When y = 3x2 − 7x + 2 then dy = 6x − 7
i.e. dy = 20x3 + 4 − 1 − 1 dx
dx x3 √
2 x3 Since the gradient is −1 then 6x − 7 = −1, from which,
x=1
Problem 5. Find the differential coefficients of: When x = 1, y = 3(1)2 − 7(1) + 2 = −2
(a) y = 3 sin 4x (b) f (t) = 2 cos 3t with respect Hence the gradient is −1 at the point (1, −2)
to the variable

(a) When y = 3 sin 4x then dy = (3)(4 cos 4x) Now try the following exercise Section 8
dx
= 12 cos 4x Exercise 154 Further problems on
differentiating common
(b) When f (t) = 2 cos 3t then functions
f (t) = (2)(−3 sin 3t) = −6 sin 3t

Problem 6. Determine the derivatives of: In Problems 1 to 6 find the differential coefficients
of the given functions with respect to the variable.

(a) y = 3e5x (b) f (θ) = 2 (c) y = 6 ln 2x 1. (a) 5x5 (b) 2.4x3.5 1
e3θ (c)
x

(a) When y = 3e5x then dy = (3)(5)e5x = 15e5x (a) 25x4 (b) 8.4x2.5 (c) − 1
dx x2

(b) f (θ) = 2 = 2e−3θ, thus −4
e3θ 2. (a) x2 (b) 6 (c) 2x

f (θ) = (2)(−3)e−3θ = −6e−3θ = −6 8
e3θ (a) x3 (b) 0 (c) 2

394 Engineering Mathematics

3. √ √ (c) √4 x Problem 9. Find the differential coefficient of:
(a) 2 x (b) 3 3 x5 y = 3x2 sin 2x

(a) √1 (b) 5√3 x2 (c) − √2 3x2 sin 2x is a product of two terms 3x2 and sin 2x. Let
x x3 u = 3x2 and v = sin 2x

4. −3 (b) (x − 1)2 (c) 2 sin 3x Using the product rule:
(a) √
3x

−3 dy u dv + v du
(a) √3 x4 (b) 2(x − 1) (c) 6 cos 3x = dx dx
dx
↓ ↓ ↓↓
3 dy = (3x2)(2 cos 2x) + (sin 2x)(6x)
5. (a) −4 cos 2x (b) 2e6x (c) e5x gives: dx
i.e.
(a) 8 sin 2x (b) 12e6x −15 dy = 6x2 cos 2x + 6x sin 2x
(c) e5x dx

ex − e−x 1 − √ = 6x(x cos 2x + sin 2x)
x
6. (a) 4 ln 9x (b) (c)
2x Note that the differential coefficient of a product is
not obtained by merely differentiating each term and
4 ex + e−x (c) −1 + √1 multiplying the two answers together. The product rule
(a) (b) x2 2 x3 formula must be used when differentiating products.

x 2

7. Find the gradient of the curve

y = 2t4 + 3t3 − t + 4 at the points (0, 4) Problem 10. Find the √rate of change of y with
respect to x given: y = 3 x ln 2x
and (1, 8). [−1, 16]

8. Find the co-ordinates of the point on graph dy
y = 5x2 − 3x + 1 where the gradient is 2. The rate of change of y with respect to x is given by .
13 √ dx
, 3x 3x1/2
24 y = ln 2x = ln 2x, which is a product.

9. (a) Differentiate Let u = 3x1/2 and v = ln 2x

y = 2 +2 ln 2θ − 2(cos 5θ + 3 sin 2θ) − 2 Then dy = u dv + v du
θ2 e3θ dx dx dx
↓ ↓↓ ↓
(b) Evaluate dy when θ = π , correct to 4
dθ 2 = (3x1/2) 1 + (ln 2x) 3 1 x(1/2)−1
x2
Section 8 significant figures.
⎡ ⎤
⎢⎣ (a) −4 + 2 + 10 sin 5θ ⎦⎥ = 3x(1/2)−1 + (ln 2x) 3 x−1/2
θ3 θ 2

6
− 12 cos 2θ + e3θ (b) 22.30
= 3x−1/2 1 + 1 ln 2x
ds 2

10. Evaluate , correct to 3 significant fig- dy = √3 1
dt i.e. 1 + ln 2x
when t = π √ dx x 2
ures, given s = 3 sin t −3+ t
6
[3.29]

Problem 11. Differentiate: y = x3 cos 3x ln x

43.2 Differentiation of a product Let u = x3 cos 3x (i.e. a product) and v = ln x

When y = uv, and u and v are both functions of x, Then dy = dv + du
where u v
dy dv du dx dx dx
then = u + v
dx dx dx du = (x3)(−3 sin 3x) + (cos 3x)(3x2)
dx
This is known as the product rule.

Methods of differentiation 395

and dv = 1 1 5. et ln t cos t 1 + ln t cos t − ln t sin t
Hence dx x x et t
dy = (x3 cos 3x)
i.e. dx di
6. Evaluate , correct to 4 significant figure,
+ (ln x)[−3x3 sin 3x + 3x2 cos 3x]
dt

= x2 cos 3x + 3x2 ln x(cos 3x − x sin 3x) when t = 0.1, and i = 15t sin 3t

dy = x2{cos 3x + 3 ln x(cos 3x − x sin 3x)} [8.732]
dx
dz
Problem 12. Determine the rate of change of 7. Evaluate , correct to 4 significant figures,
voltage, given v = 5t sin 2t volts, when t = 0.2 s
dt
when t = 0.5, given that z = 2e3t sin 2t

[32.31]

Rate of change of voltage 43.3 Differentiation of a quotient

= dv = (5t)(2 cos 2t) + ( sin 2t)(5) u
dt When y = , and u and v are both functions of x
= 10t cos 2t + 5 sin 2t v
du dv
When t = 0.2, v −u
then dy = dx dx
dv = 10(0.2) cos 2(0.2) + 5 sin 2(0.2) dx
dt v2

= 2 cos 0.4 + 5 sin 0.4 This is known as the quotient rule.

Problem 13. Find the differential coefficient of:

(where cos 0.4 means the cosine of 0.4 radians = y = 4 sin 5x
0.92106) 5x4

Hence dv = 2(0.92106) + 5(0.38942) 4 sin 5x is a quotient. Let u=4 sin 5x and v = 5x4
dt 5x4

= 1.8421 + 1.9471 = 3.7892 (Note that v is always the denominator and u the
numerator)
i.e. the rate of change of voltage when t = 0.2 s is
3.79 volts/s, correct to 3 significant figures. du dv Section 8

dy = v −u
dx dx dx

Now try the following exercise v2

where du
and = (4)(5) cos 5x = 20 cos 5x
Hence
dx
i.e.
Exercise 155 Further problems on dv = (5)(4)x3 = 20x3
differentiating products dx

In Problems 1 to 5 differentiate the given products dy = (5x4)(20 cos 5x) − (4 sin 5x)(20x3)
with respect to the variable. dx (5x4)2

1. 2x3 cos 3x [6x2(cos 3x − x sin 3x)] 100x4 cos 5x − 80x3 sin 5x
√ = 25x8
√ 1 + 3 ln 3x
2. x3 ln 3x x 2 = 20x3[5x cos 5x − 4 sin 5x]
25x8
3. e3t sin 4t [e3t (4 cos 4t + 3 sin 4t)]
4. e4θ ln 3θ dy = 4 − 4 sin 5x)
e4θ 1 + 4 ln 3θ dx 5x5 (5x cos 5x
θ
Note that the differential coefficient is not obtained
by merely differentiating each term in turn and then

396 Engineering Mathematics

dividing the numerator by the denominator. The Let u = te2t and v = 2 cos t then
quotient formula must be used when differentiating
quotients. du = (t)(2e2t) + (e2t)(1) and dv = −2 sin t
dt dt

du dv
v −u
Problem 14. Determine the differential Hence dy = dx dx
coefficient of: y = tan ax dx
v2

= (2 cos t)[2te2t + e2t] − (te2t)(−2 sin t)
(2 cos t)2
y = tan ax = sin ax . Differentiation of tan ax is thus
cos ax 4te2t cos t + 2e2t cos t + 2te2t sin t
= 4 cos2 t
treated as a quotient with u = sin ax and v = cos ax

du dv 2e2t [2t cos t + cos t + t sin t]
v −u = 4 cos2 t
dy = dx dx
dx
v2 dy e2t
i.e. dx = 2 cos 2t (2t cos t + cos t + t sin t)
(cos ax)(a cos ax) − (sin ax)(−a sin ax)
= (cos ax)2

= a cos2 ax + a sin2 ax Problem 17. Determine the g√radient of the curve
(cos ax)2 5x √ 3

a(cos2 ax + sin2 ax) y = 2x2 + 4 at the point 3,
= cos2 ax 2

= a ax since cos2 ax + sin2 ax = 1 Let y = 5x and v = 2x2 + 4
cos2

(see Chapter 26) du dv

Hence dy = a sec2ax since sec2ax = 1 ax dy = v −u = (2x2 + 4)(5) − (5x)(4x)
dx cos2 dx dx dx (2x2 + 4)2
v2

(see Chapter 22) 10x2 + 20 − 20x2 20 − 10x2
(2x2 + 4)2 (2x2 + 4)2
= =

Problem 15. Find the derivative of: y = sec ax √
3
At the point √ 3, √
, x = 3,
y = sec ax = 1 (i.e. a quotient), Let u = 1 and 2
cos ax
Section 8 dy 20 − 10(√3)2
v = cos ax hence the gradient = = √

du dv dx [2( 3)2 + 4]2
v −u = 20 − 30 = − 1
dy = dx dx 100 10
dx
v2

= (cos ax)(0) − (1)(−a sin ax)
(cos ax)2

= a sin ax = a 1 sin ax Now try the following exercise
cos2 ax cos ax cos ax

i.e. dy = a sec ax tan ax Exercise 156 Further problems on
dx differentiating quotients

Problem 16. Differentiate: y = te2t In Problems 1 to 5, differentiate the quotients with
2 cos t respect to the variable.

The function te2t is a quotient, whose numerator is 2 cos 3x −6
1. x3 x4 (x sin 3x + cos 3x)
2 cos t
a product.

Methods of differentiation 397

2x 2(1 − x2) Using the function of a function rule,
2. x2 + 1
(x2 + 1)2 dy = dy × du = (−3 sin u)(10x) = −30x sin u
√ √ dx du dx
3 θ3 3 θ(3 sin 2θ − 4θ cos 2θ) Rewriting u as 5x2 + 2 gives:
3.
2 sin 2θ 4 sin2 2θ dy = −30x sin(5x2 + 2)
dx
4. ln√2t ⎡ 1 −1 ln 2t ⎤
t ⎢⎣ √2 ⎦⎥ Problem 19. Find the derivative of:
t3 y = (4t3 − 3t)6

2xe4x 2e4x Let u = 4t3 − 3t, then y = u6
5. sin2 x {(1 + 4x) sin x − x cos x} Hence du = 12t2 − 3 and dy = 6u5

sin x dt dt
Using the function of a function rule,
6. Find the gradient of the curve y = 2x 5 at the
x2 − dy = dy × du = (6u5)(12t2 − 3)
point (2, −4) [−18] dx du dx
Rewriting u as (4t3 − 3t)gives:
dy
7. Evaluate at x = 2.5, correct to 3 significant dy = 6(4t3 − 3t)5(12t2 − 3)
dx dt
figures, given y = 2x2 + 3
ln 2x [3.82] = 18(4t2 − 1)(4t3 − 3t)5

43.4 Function of a function Problem 20. Determine the differential
√
It is often easier to make a substitution before differen-
tiating. coefficient of: y = 3x2 + 4x − 1

If y is a function of x then dy = dy × du √
dx du dx y = 3x2 + 4x − 1 = (3x2 + 4x − 1)1/2

This is known as the ‘function of a function’ rule (or Let u = 3x2 + 4x − 1 then y = u1/2
sometimes the chain rule).
For example, if y = (3x − 1)9 then, by making the sub- du dy = 1 u−1/2 = √1 Section 8
stitution u = (3x − 1), y = u9, which is of the ‘standard’ Hence = 6x + 4 and du 2
from. dx 2u
Hence dy = 9u8 and du = 3
Using the function of a function rule,
du dx
Then dy = dy × du = (9u8)(3) = 27u8 dy = dy × du = √1 (6x + 4) = 3x√+ 2
dx du dx 2 u u
dx du dx
Rewriting u as (3x − 1) gives: dy = 27(3x − 1)8 i.e. dy 3x + 2
=√
dx dx 3x2 + 4x − 1
Since y is a function of u, and u is a function of x, then
y is a function of a function of x. Problem 21. Differentiate: y = 3 tan4 3x

Problem 18. Differentiate: y = 3 cos (5x2 + 2) Let u = tan 3x then y = 3u4

Let u = 5x2 + 2 then y = 3 cos u Hence du = 3 sec2 3x, (from Problem 14),
du dy dx

Hence = 10x and = −3 sin u and dy = 12u3
dx du du

398 Engineering Mathematics

Then dy = dy × du = (12u3)(3 sec2 3x) 43.5 Successive differentiation
i.e. dx du dx
When a function y = f (x) is differentiated with respect
= 12(tan 3x)3(3 sec2 3x) dy
dy = 36 tan33x sec23x
dx to x the differential coefficient is written as or f (x).
dx
Problem 22. Find the differential coefficient of:
If the expression is differentiated again, the second dif-
y = (2t3 2 5)4 d2y
−
ferential coefficient is obtained and is written as dx2
2 (pronounced dee two y by dee x squared) or f (x) (pro-
(2t3 − 5)4
nounced f double–dash x). By successive differentiation
d3y d4y

further higher derivatives such as dx3 and dx4 may be
obtained.

y= = 2(2t3 − 5)−4. Let u= (2t3 − 5), then Thus if y = 3x4,

y = 2u−4 dy = 12x3, d2y 36x2,
dx dx2
du = 6t2 dy −8u−5 −8 =
dt du u5
Hence and = = d3y d4y d5y

Then dy = dy × du = −8 (6t2) = −48t2 dx3 = 72x, dx4 = 72 and dx5 = 0
dt du dt u5 (2t3 − 5)5

Now try the following exercise Problem 23. If f (x) = 2x5 − 4x3 + 3x − 5, find
f (x)
Exercise 157 Further problems on the
function of a function f (x) = 2x5 − 4x3 + 3x − 5
f (x) = 10x4 − 12x2 + 3
In Problems 1 to 8, find the differential coefficients f (x) = 40x3−24x = 4x(10x2 − 6)
with respect to the variable.

1. (2x3 − 5x)5 [5(6x2 − 5)(2x3 − 5x)4] Problem 24. If y = cos x − sin x, evaluate x, in

2. 2 sin(3θ − 2) [6 cos(3θ − 2)] the range 0 ≤x ≤ π when d2y is zero
, dx2

2

Section 8 3. 2 cos5 α [−10 cos4 α sin α] Since y = cos x − sin x, dy = − sin x − cos x and
1 d2y dx
5(2 − 3x2) dx2 = − cos x + sin x
4. (x3 − 2x + 1)5 (x3 − 2x + 1)6

5. 5e2t+1 [10e2t+1] d2y
When dx2 is zero, − cos x + sin x = 0,
6. 2 cot(5t2 + 3) [−20t cosec2(5t2 + 3)]
i.e. sin x = cos x or sin x = 1
7. 6 tan(3y + 1) [18 sec2(3y + 1)] cos x

8. 2etan θ [2 sec2 θ etan θ] Hence tan x = 1 and x = tan−1 1 = 45◦ or π rads
π4
9. Differentiate: θ sin θ − π with respect to θ,
in the range 0 ≤ x ≤
2

3
and evaluate, correct to 3 significant figures,
π Problem 25. Given y = 2xe−3x show that
when θ = [1.86]
2 d2y dy
+ 6 + 9y = 0
dx2 dx

Methods of differentiation 399

y = 2xe−3x (i.e. a product) Now try the following exercise

Hence dy = (2x)(−3e−3x) + (e−3x)(2) Exercise 158 Further problems on
dx successive differentiation

= −6xe−3x + 2e−3x 1. If y = 3x4 + 2x3 − 3x + 2 find
d2y d3y
d2y = [(−6x)(−3e−3x ) + (e−3x )(−6)]
dx2 (a) dx2 (b) dx3
[(a) 36x2 + 12x (b) 72x + 12]
+ (−6e−3x)

= 18xe−3x − 6e−3x − 6e−3x 2 t2 1 3 √
5 t3 t t
d2y 18xe−3x − 12e−3x 2. (a) Given f (t) = − + − + 1
dx2 (b)
i.e. = determine f (t)

d2y dy Evaluate f (t) when t = 1.
dx2 6
Substituting values into + + 9y gives: ⎡ 4 12 6 1⎤
dx
⎣ (a) − + + √
5 t5 t3 4 t3 ⎦
(18xe−3x − 12e−3x) + 6(−6xe−3x + 2e−3x)
+ 9(2xe−3x) (b) −4.95

= 18xe−3x − 12e−3x − 36xe−3x In Problems 3 and 4, find the second differen-
+ 12e−3x + 18xe−3x = 0 tial coefficient with respect to the variable.

3. (a) 3 sin 2t + cos t (b) 2 ln 4θ

Thus when y = 2xe−3x, d2y + 6 dy + 9y = 0 (a) −(12 sin 2t + cos t) −2
dx2 dx (b) θ2

4. (a) 2 cos2x (b) (2x − 3)4

Problem 26. Evaluate d2y when θ=0 given: [(a) 4(sin2x − cos2x) (b) 48(2x − 3)2]
dθ2
y = 4 sec 2θ
5. Evaluate f (θ) when θ = 0 given

f (θ) = 2 sec 3θ [18]

Since y = 4 sec 2θ, then

dy = (4)(2) sec 2θ tan 2θ (from Problem 15) 6. Show that the differential equation
dθ
d2y − dy + 4y = 0 is satisfied when Section 8
= 8 sec 2θ tan 2θ (i.e. a product) dx2 4

dx
y = xe2x

d2y = (8 sec 2θ)(2 sec2 2θ) 7. Show that, if P and Q are constants and
dθ2 y = P cos(ln t) + Q sin(ln t), then

+ (tan 2θ)[(8)(2) sec 2θ tan 2θ] d2y dy
dt2 t
= 16 sec3 2θ + 16 sec 2θ tan2 2θ t 2 + + y = 0
dt

When θ = 0,

d2y = 16 sec3 0 + 16 sec 0 tan2 0
dθ2

= 16(1) + 16(1)(0) = 16

Chapter 44

Some applications of
differentiation

44.1 Rates of change (b) When θ = 400◦C,

If a quantity y depends on and varies with a quantity x dl
dy = 0.00005 + (0.0000008)(400)

then the rate of change of y with respect to x is . dθ
dx = 0.00037 m/◦C = 0.37 mm/◦C

Thus, for example, the rate of change of pressure p with Problem 2. The luminous intensity I candelas of
dp a lamp at varying voltage V is given by:
I = 4 × 10−4 V2. Determine the voltage at which
height h is . the light is increasing at a rate of 0.6 candelas per
dh volt

A rate of change with respect to time is usually just The rate of change of light with respect to voltage is
called ‘the rate of change’, the ‘with respect to time’ dI
being assumed. Thus, for example, a rate of change of
given by
di dV
current, i, is and a rate of change of temperature, θ,
Since I = 4 × 10−4 V2, dI = (4 × 10−4)(2)V
dt dV
dθ
is , and so on. = 8 × 10−4 V
dt
When the light is increasing at 0.6 candelas per
Problem 1. The length l metres of a certain metal
rod at temperature θ◦C is given by: volt then +0.6 = 8 × 10−4 V, from which, voltage
l = 1 + 0.00005θ + 0.0000004θ2. Determine the
rate of change of length, in mm/◦C, when the V = 8 0.6 = 0.075 × 10+4 = 750 volts
temperature is (a) 100◦C and (b) 400◦C × 10−4

dl Problem 3. Newtons law of cooling is given by:
The rate of change of length means θ = θ0e−kt, where the excess of temperature at zero
time is θ0◦C and at time t seconds is θ◦C.
dθ
Determine the rate of change of temperature after
Since length l = 1 + 0.00005θ + 0.0000004θ2, 40 s, given that θ0 = 16◦C and k = −0.03

dl dθ
then = 0.00005 + 0.0000008θ The rate of change of temperture is

dθ dt
(a) When θ = 100◦C,
Since θ = θ0e−kt then dθ = (θ0)(−k)e−kt
dl dt
= 0.00005 + (0.0000008)(100)
= −kθ0e−kt
dθ
= 0.00013 m/◦C = 0.13 mm/◦C

Some applications of differentiation 401

When θ0 = 16, k = −0.03 and t = 40 then 3. The voltage across the plates of a capacitor at
any time t seconds is given by v = Ve−t/CR,
dθ = −(−0.03)(16)e−(−0.03)(40) where V , C and R are constants. Given
dt V = 300 volts, C = 0.12 × 10−6 farads and
R = 4 × 106 ohms find (a) the initial rate of
= 0.48e1.2 = 1.594◦C/s change of voltage, and (b) the rate of change
of voltage after 0.5 s.
Problem 4. The displacement s cm of the end [(a) −625 V/s (b) −220.5 V/s]
of a stiff spring at time t seconds is given by:
s = ae−kt sin 2πft. Determine the velocity of the end 4. The pressure p of the atmosphere at height h
of the spring after 1 s, if a = 2, k = 0.9 and f = 5 above ground level is given by p = p0e−h/c,
where p0 is the pressure at ground level and
Velocity v = ds where s = ae−kt sin 2πft (i.e. a product) c is a constant. Determine the rate of change
dt of pressure with height when p0 = 1.013 × 105
Pascals and c = 6.05 × 104 at 1450 metres.
Using the product rule, [−1.635 Pa/m]

ds = (ae−kt)(2πf cos 2πft) 44.2 Velocity and acceleration
dt
When a car moves a distance x metres in a time t seconds
+ (sin 2πft)(−ake−kt) along a straight road, if the velocity v is constant then

When a = 2, k = 0.9, f = 5 and t = 1, x
v = m/s, i.e. the gradient of the distance/time graph
velocity, v = (2e−0.9)(2π5 cos 2π5)
+ (sin 2π5)(−2)(0.9)e−0.9 t
shown in Fig. 44.1 is constant.
= 25.5455 cos 10π − 0.7318 sin 10π
= 25.5455(1) − 0.7318(0)
= 25.55 cm/s

(Note that cos 10π means ‘the cosine of 10π radians’,
not degrees, and cos 10π ≡ cos 2π = 1).

Now try the following exercise Section 8

Exercise 159 Further problems on rates
of change

1. An alternating current, i amperes, is given by Figure 44.1

i = 10 sin 2πft, where f is the frequency in

hertz and t the time in seconds. Determine the If, however, the velocity of the car is not constant then

rate of change of current when t = 20 ms, given the distance/time graph will not be a straight line. It may

that f = 150 Hz. [3000π A/s] be as shown in Fig. 44.2.

2. The luminous intensity, I candelas, of a lamp The average velocity over a small time δt and distance δx
is given by I = 6 × 10−4 V2, where V is the
voltage. Find (a) the rate of change of luminous is given by the gradient of the chord AB, i.e. the average
intensity with voltage when V = 200 volts, and
(b) the voltage at which the light is increasing velocity over time δt is δx As δt → 0, the chord AB
.
at a rate of 0.3 candelas per volt. δt
becomes a tangent, such that at point A, the velocity is
[(a) 0.24 cd/V (b) 250 V]
given by: v = dx
dt
Hence the velocity of the car at any instant is given by

the gradient of the distance/time graph. If an expression

402 Engineering Mathematics

Summarising, if a body moves a distance x metres
in a time t seconds then:

(i) distance x = f (t)

dx
(ii) velocity v = f (t) or , which is the gradient

dt
of the distance/time graph

dv d2x
(iii) acceleration a = = f or dt2 , which is the
dt
gradient of the velocity/time graph.

Figure 44.2 Problem 5. The distance x metres moved by a car

for the distance x is known in terms of time t then the in a time t seconds is given by:
velocity is obtained by differentiating the expression. x = 3t3 − 2t2 + 4t − 1. Determine the velocity and
acceleration when (a) t = 0, and (b) t = 1.5 s

Distance x = 3t3 − 2t2 + 4t − 1 m.
Velocity
v = dx = 9t2 − 4t + 4 m/s
Acceleration dt

a = d2x = 18t − 4 m/s2
dx2

Figure 44.3 (a) When time t = 0,
velocity v = 9(0)2 − 4(0) + 4 = 4 m/s
The acceleration a of the car is defined as the rate and acceleration a = 18(0) − 4 = −4 m/s2

of change of velocity. A velocity/time graph is shown (i.e. a deceleration)

in Fig. 44.3. If δv is the change in v and δt the corre- (b) When time t = 1.5 s,
sponding change in time, then a = δv . As δt → 0, the velocity v = 9(1.5)2 − 4(1.5) + 4 = 18.25 m/s
and acceleration a = 18(1.5) − 4 = 23 m/s2
δt
chord CD becomes a tangent, such that at point C, the Problem 6. Supplies are dropped from a
acceleration is given by: a = dv
Section 8 helicopter and the distance fallen in a time t
dt
Hence the acceleration of the car at any instant is seconds is given by: x = 1 gt 2, where g = 9.8 m/s2.
2
given by the gradient of the velocity/time graph. If Determine the velocity and acceleration of the

an expression for velocity is known in terms of time supplies after it has fallen for 2 seconds

t then the acceleration is obtained by differentiating the Distance x = 1 gt2 = 1 (9.8)t2 = 4.9t2m
Velocity 22
expression. and acceleration
v = dv = 9.8 t m/s
dt

a = dv a = d2x = 9.8 m/s2
dt dx2
Acceleration
However, v = dx d2x When time t = 2 s,
Hence dt = dx2 velocity v = (9.8)(2) = 19.6 m/s
and acceleration a = 9.8 m/s2 (which is acceleration
d dx
a= due to gravity).

dt dt

The acceleration is given by the second differential Problem 7. The distance x metres travelled by a
coefficient of distance x with respect to time t vehicle in time t seconds after the brakes are

Some applications of differentiation 403

applied is given by: x = 20t − 5 t2. Determine Problem 9. The displacement x cm of the slide
3 valve of an engine is given by:
x = 2.2 cos 5πt + 3.6 sin 5πt. Evaluate the velocity
(a) the speed of the vehicle (in km/h) at the instant (in m/s) when time t = 30 ms
the brakes are applied, and (b) the distance the
car travels before it stops Displacement x = 2.2 cos 5πt + 3.6 sin 5πt

(a) Distance, x = 20t − 5 t2 Velocity v = dx = (2.2)(−5π) sin 5πt
3 dt
+ (3.6)(5π) cos 5πt
Hence velocity v = dx = 20 − 10 t
dt 3 = −11π sin 5πt + 18π cos 5πt cm/s
When time t = 30 ms,
At the instant the brakes are applied, time = 0 velocity = −11π sin (5π × 30 × 10−3)
Hence
+ 18π cos (5π × 30 × 10−3)
velocity v = 20 m/s = 20 × 60 × 60 km/h
1000 = −11π sin 0.4712 + 18π cos 0.4712
= −11π sin 27◦ + 18π cos 27◦
= 72 km/h
= −15.69 + 50.39
(Note: changing from m/s to km/h merely involves
multiplying by 3.6). = 34.7 cm/s = 0.347 m/s

(b) When the car finally stops, the velocity is zero, i.e. Now try the following exercise

v = 20 − 10 = 0, from which, 20 = 10 giving Exercise 160 Further problems on velocity
t t, and acceleration
33
t = 6 s. Hence the distance travelled before the car 1. A missile fired from ground level rises
x metres vertically upwards in t seconds and
stops is given by: x = 100t − 25 t2. Find (a) the initial velocity
2
x = 20t − 5 t2 = 20(6) − 5 (6)2 of the missile, (b) the time when the height of
33 the missile is a maximum, (c) the maximum
= 120 − 60 = 60 m height reached, (d) the velocity with which the
missile strikes the ground.
Problem 8. The angular displacement θ radians (a) 100 m/s (b) 4 s Section 8
(c) 200 m (d) −100 m/s
of a flywheel varies with time t seconds and follows
the equation: θ = 9t2 − 2t3. Determine (a) the 2. The distance s metres travelled by a car in t
angular velocity and acceleration of the flywheel seconds after the brakes are applied is given by
when time, t = 1 s, and (b) the time when the s = 25t − 2.5t2. Find (a) the speed of the car
angular acceleration is zero (in km/h) when the brakes are applied, (b) the
distance the car travels before it stops.
(a) Angular displacement θ = 9t2 − 2t3 rad. [(a) 90 km/h (b) 62.5 m]
Angular velocity ω = dθ = 18t − 6t2 rad/s.
dt 3. The equation θ = 10π + 24t − 3t2 gives the
When time t = 1 s, angle θ, in radians, through which a wheel
turns in t seconds. Determine (a) the time the
ω = 18(1) − 6(1)2 = 12 rad/s. wheel takes to come to rest, (b) the angle turned
d2θ through in the last second of movement.
[(a) 4 s (b) 3 rads]
Angular acceleration α = dt2 = 18 − 12t rad/s.
When time t = 1 s, α = 18 − 12(1)

= 6 rad/s2

(b) When the angular acceleration is zero, 18 − 12t = 0,
from which, 18 = 12t, giving time, t = 1.5 s

404 Engineering Mathematics

4. At any time t seconds the distance x metres of a y R

particle moving in a straight line from a fixed Positive P
gradient
point is given by: x = 4t + ln(1 − t). Deter- Negative
gradient
mine (a) the initial velocity and acceleration, Positive
gradient
(b) the velocity and acceleration after 1.5 s, and

(c) the time when the velocity is zero.
(a) 3 m/s; −1 m/s2

(b) 6 m/s; −4 m/s2 (c) 3 s O Q
4 Figure 44.4 x

5. The angular displacement θ of a rotating disc
is given by: θ = 6 sin t , where t is the time in
4
seconds. Determine (a) the angular velocity of

the disc when t is 1.5 s, (b) the angular accel- a valley’. Points such as P and Q are given the general
name of turning points.
eration when t is 5.5 s, and (c) the first time
It is possible to have a turning point, the gradient on
when the angular vel⎡ocity is zero. ⎤ either side of which is the same. Such a point is given
(a) ω = 1.40 rad/s the special name of a point of inflexion, and examples
are shown in Fig. 44.5.
⎣(b) α = −0.37 rad/s2⎦

(c) t = 6.28 s

20t3 23t2 y Maximum
6. x = − + 6t + 5 represents the dis- point
Maximum
32 point
tance, x metres, moved by a body in t seconds.

Determine (a) the velocity and acceleration at

the start, (b) the velocity and acceleration when Points of
inflexion
t = 3 s, (c) the values of t when the body is at

rest, (d) the value of t when the acceleration

is 37 m/s2, and (e) the distance travelled in the

third second. ⎡ −23 m/s2 ⎤
(a)
6 m/s,

⎢⎣⎢⎢((bc)) 117 m/s, 97 m/s2 ⎦⎥⎥⎥ 0 Minimum point x
Figure 44.5
3 s or 2 s
4 5

Section 8 (d) 1 1 s (e) 75 1 m
2 6

44.3 Turning points Maximum and minimum points and points of inflexion
are given the general term of stationary points.
In Fig. 44.4, the gradient (or rate of change) of the
curve changes from positive between O and P to nega- Procedure for finding and distinguishing between
tive between P and Q, and then positive again between stationary points.
Q and R. At point P, the gradient is zero and, as x
increases, the gradient of the curve changes from posi- (i) Given y = f (x), determine dy (i.e. f (x))
tive just before P to negative just after. Such a point is dx
called a maximum point and appears as the ‘crest of
a wave’. At point Q, the gradient is also zero and, as x (ii) Let dy = 0 and solve for the values of x
increases, the gradient of the curve changes from nega- dx
tive just before Q to positive just after. Such a point is
called a minimum point, and appears as the ‘bottom of (iii) Substitute the values of x into the original
equation, y = f (x), to find the corresponding
y-ordinate values. This establishes the co-
ordinates of the stationary points.

To determine the nature of the stationary points:
Either

Some applications of differentiation 405

d2y points, and (b) determining the sign of the second
(iv) Find dx2 and substitute into it the values of x derivative

found in (ii).

If the result is: (a) positive — the point is a Since y = x3 − 3x + 5 then dy = 3x2 − 3
(b) minimum one, dx
(c) negative — the point is a
maximum one, dy
or zero — the point is a point For a maximum or minimum value = 0
of inflexion dx
3x2 − 3 = 0
Hence

from which, 3x2 = 3

(v) Determine the sign of the gradient of the curve and x = ±1
just before and just after the stationary points. If When x = 1, y = (1)3 − 3(1) + 5 = 3
the sign change for the gradient of the curve is:
When x = −1, y = (−1)3 − 3(−1) + 5 = 7

(a) positive to negative — the point is a maxi- Hence (1, 3) and (−1, 7) are the co-ordinates of the
mum one turning points.

(b) negative to positive — the point is a mini- (a) Considering the point (1, 3):
mum one
If x is slightly less than 1, say 0.9, then
(c) positive to positive or negative to negative — dy = 3(0.9)2 − 3, which is negative.
the point is a point of inflexion. dx
If x is slightly more than 1, say 1.1, then
Problem 10. Locate the turning point on the dy = 3(1.1)2 − 3, which is positive.
curve y = 3x2 − 6x and determine its nature by dx

examining the sign of the gradient on either side Since the gradient changes from negative to posi-
tive, the point (1, 3) is a minimum point.

Following the above procedure: Considering the point (−1, 7):

(i) Since y = 3x2 − 6x, dy = 6x − 6 If x is slightly less than −1, say −1.1, then Section 8
dx dy = 3(−1.1)2 − 3, which is positive.
dy dx

(ii) At a turning point, = 0, hence 6x − 6 = 0, If x is slightly more than −1, say −0.9, then
dx dy = 3(−0.9)2 − 3, which is negative.
dx
from which, x = 1.
Since the gradient changes from positive to nega-
(iii) When x = 1, y = 3(1)2 − 6(1) = −3 tive, the point (−1, 7) is a maximum point.

Hence the co-ordinates of the turning point is (b) Since dy = 3x2 − 3, then d2y = 6x
(1, −3) dx dx2

(v) If x is slightly less than 1, say, 0.9, then d2y
dy = 6(0.9) − 6 = −0.6, i.e. negative When x = 1, dx2 is positive, hence (1, 3) is a
dx minimum value.

If x is slightly greater than 1, say, 1.1, then d2y
dy When x = −1, dx2 is negative, hence (−1, 7) is a
maximum value.
= 6(1.1) − 6 = 0.6, i.e. positive
dx Thus the maximum value is 7 and the minimum
value is 3.
Since the gradient of the curve is negative just
before the turning point and positive just after (i.e.
− +), (1, −3) is a minimum point

Problem 11. Find the maximum and minimum It can be seen that the second differential method of
values of the curve y = x3 − 3x + 5 by (a) exam- determining the nature of the turning points is, in this
case, quicker than investigating the gradient.
ining the gradient on either side of the turning

406 Engineering Mathematics

(3)3 (3)2 5
Problem 12. Locate the turning point on the When x = 3, y = − − 6(3) +
32 3
following curve and determine whether it is a
maximum or minimum point: y = 4θ + e−θ = −11 5
6

Since y = 4θ + e−θ then dy = 4 − e−θ = 0 for a maxi- Thus the co-ordinates of the turning points
dθ are (−2, 9) and 3, −11 5

mum or minimum value. 6
Hence 4 = e−θ and 1 = eθ
(iv) Since dy = x2 − x −6 then d2y = 2x −1
4 dx dx2
giving θ = ln 1 = −1.3863
d2y
4 When x = −2, dx2 = 2(−2) − 1 = −5, which is
When θ = −1.3863, negative.

y = 4(−1.3863) + e−(−1.3863) = 5.5452 + 4.0000 Hence (−2, 9) is a maximum point.
= −1.5452
d2y
Thus (−1.3863, −1.5452) are the co-ordinates of the When x = 3, dx2 = 2(3) − 1 = 5, which is
turning point. positive.

d2y = e−θ Hence 3, −11 5 is a minimum point.
dθ2 6

Knowing (−2, 9) is a maximum point (i.e. crest

When θ = −1.3863, d2y = e+1.3863 = 4.0, which is 5
dθ2 of a wave), and 3, −11 is a minimum point
positive, hence
6

(−1.3863, −1.5452) is a minimum point. (i.e. bottom of a valley) and that when x = 0, y = 5
,
3
a sketch may be drawn as shown in Fig. 44.6.

Problem 13. Determine the co-ordinates of the

maximum and minimum values of the graph

y = x3 − x2 − 6x + 5 and distinguish between
32 3
them. Sketch the graph

Section 8 Following the given procedure:

(i) Since y = x3 − x2 − 6x + 5 then
32 3

dy = x2 − x − 6
dx

dy
(ii) At a turning point, = 0.

dx

Hence x2 − x − 6 = 0

i.e. (x + 2)(x − 3) = 0 Figure 44.6

from which x = −2 or x = 3 Problem 14. Determine the turning points on the
curve y = 4 sin x − 3 cos x in the range x = 0 to
(iii) When x = −2 x = 2π radians, and distinguish between them.
Sketch the curve over one cycle
(−2)3 (−2)2 5
y = − − 6(−2) + = 9
32 3

Some applications of differentiation 407

Since y = 4 sin x − 3 cos x then

dy
= 4 cos x + 3 sin x = 0, for a turning point,

dx

from which, 4 cos x = −3 sin x and −4 = sin x
3 cos x
= tan x.

Hence x = tan−1 −4 = 126.87◦ or 306.87◦, since
3

tangent is negative in the second and fourth
quadrants.

Figure 44.7

When x = 126.87◦, Now try the following exercise
When y = 4 sin 126.87◦ − 3 cos 126.87◦ = 5
x = 306.87◦
y = 4 sin 306.87◦ − 3 cos 306.87◦ = −5

126.87◦ = 125.87◦ × π radians Exercise 161 Further problems on turning
180 radians points

= 2.214 rad In Problems 1 to 7, find the turning points and
306.87◦ = 306.87◦ × π distinguish between them.

180 1. y = 3x2 − 4x + 2 22
= 5.356 rad Minimum at ,

33

2. x = θ(6 − θ) [Maximum at (3, 9)]

Hence (2.214, 5) and (5.356, −5) are the co-ordinates 3. y = 4x3 + 3x2 − 60x − 12
of the turning points. Minimum (2, −88)
Maximum (−2.5, 94.25)

d2y = −4 sin x + 3 cos x 4. y = 5x − 2 ln x Section 8
dx2 [Minimum at (0.4000, 3.8326)]

5. y = 2x − ex
[Maximum at (0.6931, −0.6136)]

When x = 2.214 rad, 6. y = t3 − t2 − 2t + 4 ⎤
d2y 2⎡
dx2 = −4 sin 2.214 + 3 cos 2.214, which is negative. Minimum at (1, 2.5)

⎣ − 2 , 4 22 ⎦
Maximum at 3 27

Hence (2.214, 5) is a maximum point. 7. x = 8t + 1 [Minimum at (0.5, 6)]
When x = 5.356 rad, 2t2

8. Determine the maximum and minimum values

d2y on the graph y = 12 cos θ − 5 sin θ in the range
dx2 = −4 sin 5.356 + 3 cos 5.356, which is positive. θ = 0 to θ = 360◦. Sketch the graph over one

Hence (5.356, −5) is a minimum point. cycle showing relevant points.
A sketch of y = 4 sin x − 3 cos x is shown in Fig. 44.7. Maximum of 13 at 337.38◦,

Minimum of −13 at 157.38◦

408 Engineering Mathematics

9. Show that the curve y = 2 (t − 1)3 + 2t(t − 2) upwards to form an open box. Determine the
3 maximum possible volume of the box
2
The squares to be removed from each corner are shown
has a maximum value of and a minimum in Fig. 44.8, having sides x cm. When the sides are
3 bent upwards the dimensions of the box will be: length
(20 − 2x) cm, breadth (12 − 2x) cm and height, x cm.
value of −2.

44.4 Practical problems involving x x
maximum and minimum values x x

There are many practical problems involving maxi- 12 cm x (20 Ϫ 2x)
mum and minimum values which occur in science and x
engineering. Usually, an equation has to be determined (12 Ϫ 2x)
from given data, and rearranged where necessary, so
that it contains only one variable. Some examples are x
demonstrated in Problems 15 to 20.
x
Problem 15. A rectangular area is formed having
a perimeter of 40 cm. Determine the length and 20 cm
breadth of the rectangle if it is to enclose the
maximum possible area Figure 44.8

Let the dimensions of the rectangle be x and y. Then the Volume of box, V = (20 − 2x)(12 − 2x)(x)
perimeter of the rectangle is (2x + 2y). Hence
= 240x − 64x2 + 4x3
dV = 240 − 128x + 12x2 = 0 for a turning point.
dx
Hence 4(60 − 32x + 3x2) = 0, i.e. 3x2 − 32x + 60 = 0
Using the quadratic formula,

2x + 2y = 40, or x + y = 20 (1) 32 ± (−32)2 − 4(3)(60)
x=
Since the rectangle is to enclose the maximum possible
area, a formula for area A must be obtained in terms of 2(3)
one variable only. = 8.239 cm or 2.427 cm.

Area A = xy. From equation (1), x = 20 − y Since the breadth is (12 − 2x) cm then x = 8.239 cm is
not possible and is neglected.
Hence, area A = (20 − y)y = 20y − y2 Hence x = 2.427 cm.

Section 8 dA d2V = −128 + 24x
= 20 − 2y = 0 for a turning point, from which, dx2

dy d2V
y = 10 cm. When x = 2.427, dx2 is negative, giving a maximum
value.
d2A
dy2 = −2, which is negative, giving a maximum point. The dimensions of the box are:
length = 20 − 2(2.427) = 15.146 cm,
When y = 10 cm, x = 10 cm, from equation (1). breadth = 12 − 2(2.427) = 7.146 cm, and
height = 2.427 cm.
Hence the length and breadth of the rectangle are
each 10 cm, i.e. a square gives the maximum possible Maximum volume = (15.146)(7.146)(2.427)
area. When the perimeter of a rectangle is 40 cm, the = 262.7 cm3
maximum possible area is 10 × 10 = 100 cm2.

Problem 16. A rectangular sheet of metal having Problem 17. Determine the height and radius of a
dimensions 20 cm by 12 cm has squares removed cylinder of volume 200 cm3 which has the least
from each of the four corners and the sides bent
surface area

Some applications of differentiation 409

Let the cylinder have radius r and perpendicular P Q
height h. y y

Volume of cylinder, V = πr2h = 200 (1)

Surface area of cylinder, A = 2πrh + 2πr2 x

Least surface area means minimum surface area and a Figure 44.9
formula for the surface area in terms of one variable
only is required.

200 (2) Since the maximum area is required, a formula for area
From equation (1), h = πr2
Hence surface area, A is needed in terms of one variable only.

A = 2πr 200 + 2πr2 From equation (1), x = 100 − 2y
πr2 Hence, area A = xy = (100 − 2y)y = 100y − 2y2
dA = 100 − 4y = 0, for a turning point, from which,
= 400 + 2πr2 = 400r−1 + 2πr2 dy
r y = 25 m.
d2A
dA −400 + 4πr = 0, for a turning point. dy2 = −4, which is negative, giving a maximum value.
=
dr r2 When y = 25 m, x = 50 m from equation (1).

400 Hence the maximum possible area
r2 = xy = (50)(25) = 1250 m2

Hence 4πr =

and r3 = 400 Problem 19. An open rectangular box with
4π square ends is fitted with an overlapping lid which
covers the top and the front face. Determine the
from which, r = 3 100 = 3.169 cm. maximum volume of the box if 6 m2 of metal are
π used in its construction

d2A = 800 + 4π A rectangular box having square ends of side x and
dr2 r3 length y is shown in Fig. 44.10.

d2A Section 8
When r = 3.169 cm, dr2 is positive, giving a minimum
value.
From equation (2), when r = 3.169 cm,

200 x
h = π(3.169)2 = 6.339 cm.

x y

Hence for the least surface area, a cylinder of vol- Figure 44.10
ume 200 cm3 has a radius of 3.169 cm and height of

6.339 cm.

Problem 18. Determine the area of the largest Surface area of box, A, consists of two ends and five
piece of rectangular ground that can be enclosed by faces (since the lid also covers the front face).
100 m of fencing, if part of an existing straight wall
is used as one side Hence A = 2x2 + 5xy = 6 (1)

Let the dimensions of the rectangle be x and y as shown Since it is the maximum volume required, a formula
in Fig. 44.9, where PQ represents the straight wall.
for the volume in terms of one variable only is needed.
Volume of box, V = x2y
From equation (1),

From Fig. 44.9, x + 2y = 100 (1) y = 6 − 2x2 = 6 2x
(2) −
Area of rectangle, A = xy 5x 5x 5 (2)

410 Engineering Mathematics

Hence volume V = x2y = x2 6 − 2x Since the maximum volume is required, a formula for
5x 5 the volume V is needed in terms of one variable only.

6x 2x3 From equation (2), r2 = 144 − h2
=− 4

55 Substituting into equation (1) gives:
dV = 6 − 6x2 = 0 for a maximum or minimum value.
dx 5 5 V = π 144 − h2 h = 144πh − πh3
Hence 6 = 6x2, giving x = 1 m (x = −1 is not possible, 44
and is thus neglected).

d2V = −12x dV = 144π − 3πh2 = 0, for a maximum or minimum
dx2 5 dh 4
value.

d2V Hence 144π = 3πh2 from which,
When x = 1, dx2 is negative, giving a maximum value. ,
4
From equation (2), when x = 1, y = 6 − 2(1) = 4
5(1) 5 5 h = (144)(4) = 13.86 cm.
3
Hence the maximum volume of the box is given by
V = x2y = (1)2 4 = 4 m3 d2V −6πh
dh2 = 4
55

Problem 20. Find the diameter and height of a When h = 13.86, d2V is negative, giving a maximum
cylinder of maximum volume which can be cut dh2
from a sphere of radius 12 cm value.

From equation (2),

A cylinder of radius r and height h is shown enclosed r2 = 144 − h2 = 144 − 13.862 , from which, radius
in a sphere of radius R = 12 cm in Fig. 44.11. 44

r = 9.80 cm

Volume of cylinder, V = πr2h (1) Diameter of cylinder = 2r = 2(9.80) = 19.60 cm.
Hence the cylinder having the maximum volume that
Using the right-angled triangle OPQ shown in can be cut from a sphere of radius 12 cm is one in
Fig. 44.11, which the diameter is 19.60 cm and the height is
13.86 cm.

r2 + h 2

Section 8 2 = R2 by Pythagoras’ theorem,

Now try the following exercise

i.e. r2 + h2 = 144 (2)
4
Exercise 162 Further problems on practical
r maximum and minimum
problems
PQ
1. The speed, v, of a car (in m/s) is related to time
h R ϭ 12 cm t s by the equation v = 3 + 12t − 3t2. Deter-
2 mine the maximum speed of the car in km/h.
[54 km/h]
O
h

2. Determine the maximum area of a rectan-

gular piece of land that can be enclosed by

1200 m of fencing. [90 000 m2]

3. A shell is fired vertically upwards and
its vertical height, x metres, is given

Figure 44.11

Some applications of differentiation 411

by: x = 24t − 3t2, where t is the time in 44.5 Tangents and normals

seconds. Determine the maximum height Tangents
The equation of the tangent to a curve y = f (x) at the
reached. [48 m] point (x1, y1) is given by:

4. A lidless box with square ends is to be made y − y1 = m(x − x1)

from a thin sheet of metal. Determine the dy
where m = dx = gradient of the curve at (x1, y1).
least area of the metal for which the volume
Problem 21. Find the equation of the tangent to
of the box is 3.5 m3. [11.42 m2] the curve y = x2 − x − 2 at the point (1, −2)

5. A closed cylindrical container has a surface
area of 400 cm2. Determine the dimensions

for maximum volume.
[radius = 4.607 cm, height = 9.212 cm]

6. Calculate the height of a cylinder of maxi- dy
mum volume that can be cut from a cone of Gradient, m = = 2x − 1
height 20 cm and base radius 80 cm.
[6.67 cm] dx

At the point (1, −2), x = 1 and m = 2(1) − 1 = 1
Hence the equation of the tangent is:

7. The power developed in a resistor R by a

battery of emf E and internal resistance r is y − y1 = m(x − x1)
i.e. y − −2 = 1(x − 1)
given by P = E2R . Differentiate P with i.e. y + 2 = x − 1
(R + r)2 or y = x − 3
respect to R and show that the power is a

maximum when R = r.

8. Find the height and radius of a closed cylinder The graph of y = x2 − x − 2 is shown in Fig. 46.12. The
of volume 125 cm3 which has the least surface line AB is the tangent to the curve at the point C, i.e.
(1, −2), and the equation of this line is y = x − 3.
area. [height = 5.42 cm, radius = 2.71 cm]

9. Resistance to motion, F, of a moving vehicle, y
2 y ϭ x 2ϪxϪ2
is given by: F = 5 + 100x. Determine the
x 1
minimum value of resistance. [44.72]
Section 8
10. An electrical voltage E is given by: Ϫ2 Ϫ1 0 1 2 3x
E = (15 sin 50πt + 40 cos 50πt) volts, where Ϫ1 C B
t is the time in seconds. Determine the Ϫ2
maximum value of voltage. [42.72 volts] D
Ϫ3 A

Figure 44.12

11. The fuel economy E of a car, in miles per Normals
gallon, is given by:
The normal at any point on a curve is the line that passes
E = 21 + 2.10 × 10−2v2 − 3.80 × 10−6v4 through the point and is at right angles to the tangent.
Hence, in Fig. 44.12, the line CD is the normal.
where v is the speed of the car in miles per
hour. It may be shown that if two lines are at right angles
Determine, correct to 3 significant figures, then the product of their gradients is −1. Thus if m is the
the most economical fuel consumption, and gradient of the tangent, then the gradient of the normal
the speed at which it is achieved.
1
[50.0 miles/gallon, 52.6 miles/hour] is −

m

412 Engineering Mathematics

Hence the equation of the normal at the point (x1, y1) i.e. y + 1 = − 5 (x + 1)
is given by: 53

= 1 − i.e. y + 1 = − 5 x − 5
y − y1 − (x x1) 5 33
m
Multiplying each term by 15 gives:
Problem 22. Find the equation of the normal to
the curve y = x2 − x − 2 at the point (1, −2) 15y + 3 = −25x − 25

m = 1 from Problem 21, hence the equation of the Hence equation of the normal is:

1 15y + 25x + 28 = 0
m
normal is y − y1 = − (x − x1)

i.e. y − −2 = − 1 (x − 1) Now try the following exercise
1

i.e. y + 2 = −x + 1 or y = −x − 1 Exercise 163 Further problems on tangents
and normals
Thus the line CD in Fig. 44.12 has the equation
y = −x − 1 For the following curves, at the points given, find
(a) the equation of the tangent, and (b) the equation
Problem 23. Determine the equations of the of the normal

x3 1. y = 2x2 at the point (1, 2)
tangent and normal to the curve y = at the point [(a) y = 4x − 2 (b) 4y + x = 9]

5
1
(−1, − 5 )

x3 2. y = 3x2 − 2x at the point (2, 8)
Gradient m of curve y = is given by [(a) y = 10x − 12 (b) 10y + x = 82]

5 3. x3 −1, − 1
y = at the point
dy 3x2 22
m= =
[(a) y = 3 x + 1 (b) 6y + 4x + 7 = 0]
dx 5 2

At the point (−1, − 1 ), x = −1 and 4. y = 1 + x − x2 at the point (−2, −5)
5

m = 3(−1)2 = 3 [(a) y = 5x + 5 (b) 5y + x + 27 = 0]
55
Section 8 5. θ = 1 at the point 1
3,
Equation of the tangent is: t3

y − y1 = m(x − x1) (a) 9θ + t = 6

i.e. y − −1 = 3 − −1) (b) θ = 9t − 26 2 or 3θ = 27t − 80
(x 3

55

i.e. y + 1 = 3 + 1)
(x
55

or 5y + 1 = 3x + 3 44.6 Small changes

or 5y − 3x = 2 If y is a function of x, i.e. y = f (x), and the approximate
change in y corresponding to a small change δx in x is
Equation of the normal is: required, then:

1 δy ≈ dy
y − y1 = − m (x − x1) δx dx

i.e. y − −1 = −1 (x − −1) dy
5 3 and δy ≈ · δx or δy ≈ f (x) · δx

5 dx

Some applications of differentiation 413

Problem 24. Given y = 4x2 − x, determine the Hence the percentage change in the time of swing is
approximate change in y if x changes from 1 to 1.02 a decrease of 0.156%

Since y = 4x2 − x, then dy = 8x − 1 Problem 26. A circular template has a radius of
dx 10 cm (±0.02). Determine the possible error in
calculating the area of the template. Find also the
Approximate change in y, percentage error

δy ≈ dy · δx ≈ (8x − 1)δx Area of circular template, A = πr2, hence
dx
dA
When x = 1 and δx = 0.02, δy ≈ [8(1) − 1](0.02) = 2πr
≈ 0.14
dr
[Obviously, in this case, the exact value of δy may Approximate change in area,
be obtained by evaluating y when x = 1.02, i.e.
y = 4(1.02)2 − 1.02 = 3.1416 and then subtracting from δA ≈ dA · δr ≈ (2πr)δr
it the value of y when x = 1, i.e. y = 4(1)2 − 1 = 3, giv- dr

dy When r = 10 cm and δr = 0.02,
ing δy = 3.1416 − 3 = 0.1416. Using δy = · δx above
δA = (2π10)(0.02) ≈ 0.4π cm2
dx
gave 0.14, which shows that the formula gives the i.e. the possible error in calculating the template area
approximate change in y for a small change in x]. is approximately 1.257 cm2.

Percentage error ≈ 0.4π 100% = 0.40%
π(10)2
Problem 25. Th√e time of swing T of a pendulum
is given by T = k l, where k is a constant. Now try the following exercise
Determine the percentage change in the time of
swing if the length of the pendulum l changes from Exercise 164 Further problems on small
32.1 cm to 32.0 cm changes

If T = √ = kl1/2 1. Determine the change in y if x changes from
kl 2.50 to 2.51 when (a) y = 2x − x2 (b) y = 5
x
then dT 1 l−1/2 k [(a) −0.03 (b) −0.008]
=k =√
dl 2 2l

Approximate change in T , 2. The pressure p and volume v of a mass of Section 8
gas are related by the equation pv = 50. If the
δT ≈ dT ≈ √k δl ≈ √k (−0.1) pressure increases from 25.0 to 25.4, deter-
δl mine the approximate change in the volume
dl 2 l 2l of the gas. Find also the percentage change in
the volume of the gas. [−0.032, −1.6%]
(negative since l decreases)

Percentage error

3. Determine the approximate increase in (a) the

= approximate change in T 100% volume, and (b) the surface area of a cube
original value of T
of side x cm if x increases from 20.0 cm to

√k ( − 0.1) 20.05 cm. [(a) 60 cm3 (b) 12 cm2]

= 2 l√ × 100% 4. The radius of a sphere decreases from
kl 6.0 cm to 5.96 cm. Determine the approximate
change in (a) the surface area, and (b) the
= −0.1 100% = −0.1 100% volume.
2l 2(32.1) [(a) −6.03 cm2 (b) −18.10 cm3]

= −0.156%

414 Engineering Mathematics liquid. η is obtained by measuring Q, p, r and
L. If Q can be measured accurate to ±0.5%,
5. The rate of flow of a liquid through a tube is p accurate to ±3%, r accurate to ±2% and
given by Poiseuilles’s equation as: Q = pπr4 L accurate to ±1%, calculate the maximum
8ηL possible percentage error in the value of η.
where Q is the rate of flow, p is the pressure
difference between the ends of the tube, r is [12.5%]
the radius of the tube, L is the length of the
tube and η is the coefficient of viscosity of the

Section 8

Revision Test 12

This Revision test covers the material contained in Chapters 42 to 44. The marks for each question are shown in
brackets at the end of each question.

1. Differentiate the following with respect to the 5. The heat capacity C of a gas varies with absolute
temperature θ as shown:
variable: 5 + √ − 1 (b) s = 4e2θ sin 3θ
(a) y = 2 x3 x2 2 C = 26.50 + 7.20 × 10−3θ − 1.20 × 10−6θ2

(c) y = 3 ln 5t (d) x = √
t2 − 3t + 5
cos 2t Determine the maximum value of C and the tem-

(15) perature at which it occurs. (7)

2. If f (x) = 2.5x2 − 6x + 2 find the co-ordinates at the 6. Determine for the curve y = 2x2 − 3x at the point

point at which the gradient is −1. (5) (2, 2): (a) the equation of the tangent (b) the

3. The displacement s cm of the end of a stiff spring equation of the normal. (7)
at time t seconds is given by: s = ae−kt sin 2π f t.

Determine the velocity and acceleration of the end 7. A rectangular block of metal with a square cross-
section has a total surface area of 250 cm2. Find
of the spring after 2 seconds if a = 3, k = 0.75 and
the maximum volume of the block of metal. (7)
f = 20. (10)

4. Find the co-ordinates of the turning points on

the curve y = 3x3 + 6x2 + 3x − 1 and distinguish

between them. (9)

Section 8

Chapter 45

Differentiation of parametric
equations

45.1 Introduction to parametric (a) Ellipse (b) Parabola
equations
(c) Hyperbola (d) Rectangular hyperbola
Certain mathematical functions can be expressed more
simply by expressing, say, x and y separately in terms (e) Cardioid (f) Astroid
of a third variable. For example, y = r sin θ, x = r cos θ.
Then, any value given to θ will produce a pair of values (g) Cycloid
for x and y, which may be plotted to provide a curve of
y = f (x). Figure 45.1
The third variable, θ, is called a parameter and the
two expressions for y and x are called parametric (a) Ellipse x = a cos θ, y = b sin θ
equations.
The above example of y = r sin θ and x = r cos θ are the (b) Parabola x = at2, y = 2at
parametric equations for a circle. The equation of any
point on a circle, centre at the origin and of radius r is (c) Hyperbola x = a sec θ, y = b tan θ
given by: x2 + y2 = r2, as shown in Chapter 19. (d) Rectangular x = ct, y = c
To show that y = r sin θ and x = r cos θ are suitable t
parametric equations for such a circle: hyperbola

Left hand side of equation
= x2 + y2
= (r cos θ)2 + (r sin θ)2
= r2 cos2 θ + r2 sin2 θ
= r2(cos2θ + sin2 θ)
= r2 = right hand side
(since cos2θ + sin2 θ = 1, as shown in
Chapter 26)

45.2 Some common parametric
equations

The following are some of the most common parametric
equations, and Figure 45.1 shows typical shapes of these
curves.

Differentiation of parametric equations 417

(e) Cardioid x = a(2 cos θ − cos 2θ), (a) y = 3 cos 2t, hence dy = −6 sin 2t
dt
y = a(2 sin θ − sin 2θ)
(f) Astroid x = acos3θ, y = a sin3θ x = 2 sin t, hence dx = 2 cos t
dt
(g) Cycloid x = a(θ − sin θ), y = a (1 − cos θ)
From equation (1),

45.3 Differentiation in parameters dy

When x and y are given in terms of a parameter say θ, dy = dt = −6 sin 2t = −6(2 sin t cos t)
then by the function of a function rule of differentiation dx dx 2 cos t 2 cos t
(from Chapter 43):
dt

from double angles, Chapter 27

dy = dy × dθ dy
dx dθ dx i.e. = −6 sin t
It may be shown that this can be written as:
dx
dy (b) From equation (2),

dy = dθ (1) d dy d
dx dx (2) (−6 sin t)
d2y = dt dx = dt = −6 cos t
dx2 dx
dθ 2 cos t 2 cos t

For the second differential, dt

d2y d dy d dy · dθ d2y
dx2 = dx = dx i.e. dx2 = −3
dx dθ dx

or

d dy Problem 3. The equation of a tangent drawn to a
curve at point (x1, y1) is given by:
d2y = dθ dx
d x2 dx
y − y1 = dy1 (x − x1)
dθ dx1

Problem 1. Given x = 5θ − 1 and y = 2θ(θ − 1), Determine the equation of the tangent drawn to the
dy parabola x = 2t2, y = 4t at the point t.

determine in terms of θ At point t, x1 = 2t2, hence dx1 = 4t
dx dt

x = 5θ − 1, hence dy = 5 and y1 = 4t, hence dy1 = 4 Section 8
dθ dt

y = 2θ(θ − 1) = 2θ2 − 2θ, From equation (1),

hence dy = 4θ − 2 = 2(2θ − 1) dy
dθ
dy = dt = 4 = 1
From equation (1), dx dx 4t t

dy dt

dy dθ 2(2θ−1) 2 (2θ−1) Hence, the equation of the tangent is:
dx dx 5 5
= = or 1
(x
dθ y − 4t = t − 2t2)

Problem 2. The parametric equations of a Problem 4. The parametric equations of a cycloid

function are given by y = 3 cos 2t, x = 2 sin t. are x = 4(θ − sin θ), y = 4(1 − cos θ).
dy d2y
dy d2y
Determine expressions for (a) dx (b) dx2 Determine (a) (b) dx2
dx

418 Engineering Mathematics

(a) x = 4(θ − sin θ) 3. The parametric equations for an ellipse
hence dx = 4 − 4 cos θ = 4(1 − cos θ) are x = 4 cos θ, y = sin θ. Determine (a) dy
dθ dx
y = 4(1 − cos θ), hence dy = 4 sin θ d2y
dθ (b) dx2
From equation (1), (a) − 1 cot θ (b) − 1 cosec3θ
4 16

dy 4. Evaluate dy at θ = π radians for the
dx 6
dy = dθ = 4 sin θ = sin θ
dx dx 4(1 − cos θ) (1 − cos θ) hyperbola whose parametric equations are

dθ x = 3 sec θ, y = 6 tan θ. [4]

(b) From equation (2), 5. The parametric equations for a rectangular
hyperbola are x = 2t, y = 2 . Evaluate dy
d dy d sin θ t dx
when t = 0.40
d2y = dθ dx = dθ 1 − cos θ [−6.25]
dx2 dx 4(1 − cos θ)
The equation of a tangent drawn to a curve at
dθ
point (x1, y1) is given by:

(1 − cos θ)( cos θ) − ( sin θ)( sin θ) y − y1 = dy1 (x − x1)
= (1 − cos θ)2 dx1

4(1 − cos θ) Use this in Problems 6 and 7.

= cos θ − cos2 θ − sin2 θ 6. Determine the equation of the tangent drawn
4(1 − cos θ)3 to the ellipse x = 3 cos θ, y = 2 sin θ at θ = π .
6
− ( cos2 θ + sin2 [y = −1.155x + 4]
4(1 − cos θ)3
= cos θ θ)

= cos θ − 1 7. Determine the equation of the tangent drawn
4(1 − cos θ)3 to the rectangular hyperbola x = 5t, y = 5 at
t
Section 8 = −(1 − cos θ) = −1 t = 2.
4(1 − cos θ)3 4(1 − cos θ)2 y = −1x + 5
4

Now try the following exercise 45.4 Further worked problems on
differentiation of parametric
Exercise 165 Further problems on equations
differentiation of
parametric equations Problem 5. The equation of the normal drawn to
a curve at point (x1, y1) is given by:
1. Given x = 3t − 1 and y = t(t − 1), determine
1
dy 1 − 1) y − y1 = − dy1 (x − x1)
in terms of t. (2t
3 dx1
dx Determine the equation of the normal drawn to the
astroid x = 2 cos3 θ, y = 2 sin3 θ at the point θ = π
2. A parabola has parametric equations: x = t2,
4
dy [2]
y = 2t. Evaluate when t = 0.5

dx

Differentiation of parametric equations 419

x = 2 cos3 θ, hence dx = −6 cos2 θ sin θ (b) From equation (2),
dθ
d dy d
y = 2 sin3 θ, hence dy = 6 sin2 θ cos θ (2 cosec θ)
dθ d2y = dθ dx
dx2 dx = dθ
2 sec θ tan θ

From equation (1), dθ

dy = −2 cosec θ cot θ
2 sec θ tan θ
dy dθ 6 sin2 θ cos θ = − sin θ
dx = dx = −6 cos2 θ sin θ cos θ = − tan θ −1 cos θ
= sin θ sin θ
dθ sin θ
1

When θ = π , dy = − tan π = −1 cos θ cos θ

4 dx 4 cos2 θ
sin θ
x1 = 2 cos3 π = 0.7071 and y1 = 2 sin3 π = 0.7071 =− cos θ
4 4 sin2 θ

Hence, the equation of the normal is: = − cos3 θ = − cot3 θ
sin3 θ

y − 0.7071 = − 1 (x − 0.7071) When θ = 1 rad, d2y = −cot3 1 = − 1 =
−1 dx2 (tan 1)3
−0.2647, correct to 4 significant figures.
i.e. y − 0.7071 = x − 0.7071
i.e. y = x Problem 7. When determining the surface
tension of a liquid, the radius of curvature, ρ, of
Problem 6. The parametric equations for a part of the surface is given by:

hyperbola are x = 2 sec θ, y = 4 tan θ. Evaluate dy 2 3
dy d2y 1+

(a) dx (b) dx2 , correct to 4 significant figures, dx
when θ = 1 radian. ρ = d2y

(a) x = 2 sec θ, hence dx = 2 sec θ tan θ dx2 Section 8
dθ
Find the radius of curvature of the part of the
y = 4 tan θ, hence dy = 4 sec2θ surface having the parametric equations x = 3t2,
dθ y = 6t at the point t = 2.

From equation (1), x = 3t2, hence dx = 6t
dt

dy y = 6t, hence dy = 6
dt
dy = dθ = 4 sec2 θ = 2 sec θ
dx dx 2 sec θ tan θ tan θ dy

dθ From equation (1), dy = dt = 6 = 1
dx dx 6t t
1
2
cos θ = 2 or 2 cosec θ dt
sin θ sin θ
= From equation (2),

cos θ d dy d1 1
= dt t t2
d2y dt dx − 1
dx2 dx 6t 6t3
When θ = 1 rad, dy = 2 = 2.377, correct to = = 6t = −
dx sin 1
dt
4 significant figures.

420 Engineering Mathematics

Hence, radius of curvature, ρ = 1 + dy 2 3 2. Determine the equation of the normal drawn
= dx to the parabola x = 1 t2, y = 1 t at t = 2.
42
d2y
dx2 [y = −2x + 3]
1+ 1 2 3
3. Find the equation of the normal drawn to the
t
1 cycloid x = 2(θ − sin θ), y = 2(1 − cos θ) at
− 6t3 θ = π rad.
2 [y = −x + π]

d2y
4. Determine the value of dx2 , correct to 4 sig-
nificant figures, at θ = π rad for the cardioid
6

1 23 x = 5(2θ − cos 2θ), y = 5(2 sin θ − sin 2θ).
1+
[0.02975]
2
(1.25)3 5. The radius of curvature, ρ, of part of a sur-
When t = 2, ρ = 1 = −1 face when determining the surface tension of
− 6(2)3 a liquid is given by:
48

= −48 (1.25)3 = −67.08 1 + dy 2 3/2
dx
Now try the following exercise
ρ = d2y
Exercise 166 Further problems on dx2
differentiation of
parametric equations Find the radius of curvature (correct to 4 sig-
nificant figures) of the part of the surface
1. A cycloid has parametric equations having parametric equations
(a) x = 3t, y = 3 at the point t = 1
x = 2(θ − sin θ), y = 2(1 − cos θ). Evalu-
t2
ate, at θ = 0.62 rad, correct to 4 significant (b) x = 4 cos3t, y = 4 sin3t at t = π rad

dy d2y 6
figures, (a) (b) dx2 [(a) 13.14 (b) 5.196]
dx

Section 8 [(a) 3.122 (b) −14.43]

The equation of the normal drawn to a curve
at point (x1, y1) is given by:

y − y1 = − 1 (x − x1)
dy1

dx1

Use this in Problems 2 and 3.

Chapter 46

Differentiation of implicit
functions

46.1 Implicit functions A simple rule for differentiating an implicit function is

summarised as:

When an equation can be written in the form y = f (x) d [ f (y)] = d [ f (y)] × dy (1)
it is said to be an explicit function of x. Examples of dx dy dx
explicit functions include

y = 2x3 − 3x + 4, y = 2x ln x Problem 1. Differentiate the following functions
3ex
with respect to x:
and y = (a) 2y4 (b) sin 3t
cos x
(a) Let u = 2y4, then, by the function of a function rule:
In these examples y may be differentiated with respect
du = du dy = d (2y4) × dy
to x by using standard derivatives, the product rule and ×
dx dy dx dy dx
the quotient rule of differentiation respectively.
= 8y3 dy
Sometimes with equations involving, say, y and x, dx

it is impossible to make y the subject of the formula. (b) Let u = sin 3t, then, by the function of a func-

The equation is then called an implicit function and tion rule:
examples of such functions include y3 + 2x2 = y2 − x
and sin y = x2 + 2xy. du = du × dt = d ( sin 3t) × dt

dx dt dx dt dx

46.2 Differentiating implicit = 3 cos 3t dy
functions dx

It is possible to differentiate an implicit function by Problem 2. Differentiate the following functions
using the function of a function rule, which may be
stated as with respect to x:
(a) 4 ln 5y (b) 1 e3θ − 2
du = du × dy 5
dx dy dx
(a) Let u = 4 ln 5y, then, by the function of a function
Thus, to differentiate y3 with respect to x, the substitu- rule:

tion u = y3 is made, from which, du = 3y2. du = du × dy = d (4 ln 5y) × dy
dy dx dy dx dy dx

Hence, d (y3) = (3y2) × dy by the function of a
,
dx dx = 4 dy
y dx
function rule.

422 Engineering Mathematics

(b) Let u = 1 e3θ − 2, then, by the function of a function 46.3 Differentiating implicit
5 functions containing products
and quotients
rule:

du = du × dθ = d 1 e3θ−2 × dθ The product and quotient rules of differentiation must
dx dθ dx dθ 5 dx be applied when differentiating functions containing
products and quotients of two variables.
= 3 e3θ−2 dθ
5 dx For example, d (x2y) = (x2) d ( y) + ( y) d (x2),
dx dx dx
Now try the following exercise. by the product rule

Exercise 167 Further problems on = (x2) dy + y(2x),
differentiating 1
implicit functions dx

by using equation(1)

In Problems 1 and 2 differentiate the given func- = x2 dy + 2xy
dx
tions with respect to x.

√ Problem 3. Determine d (2x3y2)
1. (a) 3y5 (b) 2 cos 4θ (c) k dx
⎡ dθ ⎤
⎢⎢⎣ (a) 15y4 dy (b) −8 sin 4θ dx ⎥⎦⎥
1 dx
(c) √ dk In the product rule of differentiation let u = 2x3 and
v = y2.
2 k dx
Thus d (2x3y2) = (2x3) d ( y2) + ( y2) d (2x3)
5 (b) 3 e2y + 1 (c) 2 tan 3y dx dx dx
2. (a) ln 3t
⎡4 ⎤ = (2x3) dy + ( y2)(6x2)
2 ⎢⎢⎣ ⎦⎥⎥ 2y

(a) 5 dt 3 e2y+1 dy dx
(c) (b) 2 dx
= 4x3y dy + 6x2y2
2t dx dx
6 sec23y dy
dx
dy
3. Differentiate the following with respect to y: = 2x2y 2x + 3y
dx
Section 8 √
(a) 3 sin 2θ (b) 4 x3 2 √ dx ⎤ d 3y
(c) et Problem 4. Find
⎡ dθ
dx 2x
⎢⎣⎢ (a) 6 cos 2θ (b) 6 x dy ⎦⎥⎥
−2 dt dy

(c) et dy In the quotient rule of differentiation let u = 3y and
v = 2x.

4. Differentiate the following with respect to u: dd
(2x) (3y) − (3y) (2x)
22 d 3y dx dx
(a) (3x + 1) (b) 3 sec 2θ (c) √y Thus 2x = (2x)2

dx

⎡ −6 dx ⎤ dy − (3y)(2)
(a) (2x) 3
⎢⎢⎢⎣⎢⎢⎢⎢ (b) (3x + 1)2 du ⎥⎦⎥⎥⎥⎥⎥⎥ dx
6 sec 2θ tan 2θ dθ = 4x2
−1 dy du
dy
6x − 6y
dx 3 dy
(c) = 4x2 = 2x2 x − y
dx
y3 du

Differentiation of implicit functions 423

Problem 5. Differentiate z = x2 + 3x cos 3y with dy
respect to y. An expression for the derivative in terms of x and

dx
y may be obtained by rearranging this latter equation.
Thus:

dz = d (x2) + d dy
(3x cos 3y) 1)
dy dy dy (2y + = 5 − 6x
dx

dx dx from which, dy = 5 − 6x
= 2x + (3x)(−3 sin 3y) + (cos3y) 3 dx 2y + 1

dy dy

= dx − 9x sin 3y + 3 cos 3y dx
2x
dy dy Problem 6. Given 2y2 − 5x4 − 2 − 7y3 = 0,
dy

determine
dx

Now try the following exercise

Each term in turn is differentiated with respect to x:

Exercise 168 Further problems on
differentiating implicit
functions involving products Hence d (2y2) − d (5x4) − d (2) − d (7y3)
and quotients i.e. dx dx dx dx

= d
(0)
dx
1. Determine d (3x2y3)
dx dy − 20x3 − 0 − 21y2 dy = 0
4y
3xy2 3x dy + 2y dx dx
dx
Rearranging gives:

d 2y 2 dy (4y − 21y2) dy = 20x3
2. Find x −y dx
5x2 dx dy 20x3
dx 5x
i.e. dx = (4y − 21y2)
d 3u 3 v − udv
3. Determine 4v2 du

du 4v

4. Given z = 3√y cos 3x find dz dy
dx Problem 7. Determine the values of when

3 cos 3x dy − 9√y sin 3x dx
2√y dx x = 4 given that x2 + y2 = 25.

5. Determine dz given z = 2x3 ln y Differentiating each term in turn with respect to x gives: Section 8
dy
d (x2) + d (y2) = d (25)
2x2 x + 3 ln dx dx dx dx
y
y dy 2x + 2y dy = 0
dx
46.4 Further implicit differentiation i.e. dy = − 2x = − x
Hence dx 2y y
An implicit function such as 3x2 + y2 − 5x + y = 2,
may be differentiated term by term with respect to x. Since x2 + y2 = 25, when x = 4, y = (25 − 42) = ±3
This gives:
Thus when x = 4 and y = ±3, dy = − 4 = 4
d (3x2) + d (y2) − d (5x) + d (y) = d (2) dx ±3 ±
dx dx dx dx dx
3
dy dy
i.e. 6x + 2y − 5 + 1 = 0, x2 + y2 = 25 is the equation of a circle, centre at the
origin and radius 5, as shown in Fig. 46.1. At x = 4, the
dx dx
using equation (1) and standard derivatives. two gradients are shown.
Above, x2 + y2 = 25 was differentiated implic-

itly; actually, the equation could be transposed to

424 Engineering Mathematics

y Gradient (b) When x = 1 and y = 2,
5
x 2 ϩ y 2 ϭ 25 3 ϭ Ϫ 4 dy 4(1) + (2)3 12
3 dx 2[5 − (3)(1)(2)] −2
= = = −6

Ϫ5 0 45 x Problem 9. Find the gradients of the tangents
drawn to the circle x2 + y2 − 2x − 2y = 3 at x = 2.

Ϫ3

Ϫ5 Gradient dy
The gradient of the tangent is given by
ϭ 4
3 dx
Differentiating each term in turn with respect to x gives:

Figure 46.1 d (x2) + d (y2) − d (2x) − d (2y) = d
(3)
dx dx dx dx dx

y = (25 − x2) and differentiated using the function of dy dy
a function rule. This gives i.e. 2x + 2y − 2 − 2 = 0

dx dx

dy = 1 (25 − x2 ) −1 (−2x) = − x Hence dy
2 (2y − 2) = 2 − 2x,

dx 2 (25 − x2) dx

dy 4 4 from which dy = 2 − 2x = 1−x
dx 2y − 2 y−1
and when x = 4, = − = ± as obtained
dx (25 − 42) 3
The value of y when x = 2 is determined from the
above. original equation

Problem 8. Hence (2)2 + y2 − 2(2) − 2y = 3
i.e. 4 + y2 − 4 − 2y = 3
dy or y2 − 2y − 3 = 0
(a) Find in terms of x and y given
Factorising gives: (y + 1)(y − 3) = 0, from which
dx y = −1 or y = 3
4x2 + 2xy3 − 5y2 = 0. When x = 2 and y = −1,
(b) Evalate dy when x = 1 and y = 2.

dx

Section 8 (a) Differentiating each term in turn with respect to x dy = 1−x = 1−2 = −1 = 1
gives: dx y−1 −1 − 1 −2 2

d (4x2) + d (2xy3) − d (5y2) = d When x = 2 and y = 3,
(0)
dx dx dx dx dy = 1 − 2 = −1
dx 3 − 1 2
i.e. 8x + (2x) 3y2 dy + (y3)(2)
dx 1
Hence the gradients of the tangents are ±
− dy = 0
10y 2
dx The circle having√the given equation has its centre at
(1, 1) and radius 5 (see Chapter 19) and is shown in
i.e. 8x + 6xy2 dy + 2y3 − dy = 0 Fig. 46.2 with the two gradients of the tangents.
10y
dx dx

Rearranging gives:

8x + 2y3 = (10y − 6xy2) dy Problem 10. Pressure p and volume v of a gas are
dx related by the law pvγ = k, where γ and k are

and dy = 8x + 2y3 = 4x + y3 constants. Show that the rate of change of pressure
dx 10y − 6xy2 y(5 − 3xy) dp p dv

= −γ
dt v dt

Differentiation of implicit functions 425

y Gradient 2. 2y3 − y + 3x − 2 = 0 3
4 1 − 6y2
x 2 ϩ y 2 Ϫ 2x Ϫ 2y ϭ 3 3 ϭ Ϫ 1
2 2
1
3. Given x2 + y2 =9 evaluate dy when x = √
5
dx √
rϭ 5 −5
and y = 2
2

0 12 4x dy
In Problems 4 to 7, determine

dx

Ϫ1

Gradient 4. x2 + 2x sin 4y = 0 −(x + sin 4y)
4x cos 4y
Ϫ2 ϭ 1
2

Figure 46.2 5. 3y2 + 2xy − 4x2 = 0 4x − y
3y + x

Since pvγ = k, then p= k = kv−γ 6. 2x2y + 3x3 = sin y x(4y + 9x)
vγ cos y − 2x2

dp = dp × dv 7. 3y + 2x ln y = y4 + x 1 − 2 ln y
dt dv dt 3 + (2x/y) − 4y3

by the function of a function rule 8. If 3x2 + 2x2y3 − 5 y2 = 0 evaluate dy when

dp = d (kv−γ ) x = 1 and y = 1 4 dx
dv dv
2 [5]

= −γ k v−γ −1 = −γ k 9. Determine the gradients of the tangents
vγ +1
drawn to the circle x2 + y2 = 16 at the point

dp −γk dv where x = 2. Give the answer correct to 4
dt = vγ+1 × dt
significant figures [±0.5774]

Since k = pvγ 10. Find the gradients of the tangents drawn to the Section 8
i.e. x2 y2
dp −γ(pvγ ) dv −γpvγ dv
= = ellipse + = 2 at the point where x = 2
dt vr+1 dt vγ v1 dt 49
[±1.5]

dp p dv 11. Determine the gradient of the curve
= −γ 3xy + y2 = −2 at the point (1, −2) [−6]

dt v dt

Now try the following exercise

Exercise 169 Further problems on
implicit differentiation

dy
In Problems 1 and 2 determine

dx

1. x2 + y2 + 4x − 3y + 1 = 0 2x + 4
3 − 2y

Chapter 47

Logarithmic differentiation

47.1 Introduction to logarithmic which may be simplified using the above laws of
differentiation logarithms, giving;

With certain functions containing more complicated ln y = ln f (x) + ln g(x) − ln h(x)
products and quotients, differentiation is often made
easier if the logarithm of the function is taken before This latter form of the equation is often easier to
differentiating. This technique, called ‘logarithmic dif- differentiate.
ferentiation’ is achieved with a knowledge of (i) the
laws of logarithms, (ii) the differential coefficients of 47.3 Differentiation of logarithmic
logarithmic functions, and (iii) the differentiation of functions
implicit functions.
The differential coefficient of the logarithmic function
47.2 Laws of logarithms ln x is given by:

Three laws of logarithms may be expressed as: d1
(ln x) =

dx x
More generally, it may be shown that:

(i) log (A × B) = log A + log B d f (x)
[ln f (x)] =
A = log A − log B (1)
(ii) log dx f(x)
B

(iii) log An = n log A For example, if y = ln(3x2 + 2x − 1) then,

In calculus, Napierian logarithms (i.e. logarithms to a dy = 6x + 2 1
base of ‘e’) are invariably used. Thus for two func- dx 3x2 + 2x −
tions f (x) and g(x) the laws of logarithms may be
expressed as: Similarly, if y = ln(sin 3x) then

(i) ln[f (x) · g(x)] = ln f (x) + ln g(x) dy = 3 cos 3x = 3 cot 3x.
(ii) ln f (x) = ln f (x) − ln g(x) dx sin 3x

g(x) As explained in Chapter 46, by using the function of a
(iii) ln[ f (x)]n = n ln f (x) function rule:

d (ln y) = 1 dy (2)
dx y dx

Taking Napierian logarithms of both sides of the equa- Differentiation of an expression such as
tion y = f (x) · g(x) gives : (1 + x)2√(x − 1)
y= √ may be achieved by using
h(x) x (x + 2)

ln y = ln f (x) · g(x) the product and quotient rules of differentiation; how-
h(x)
ever the working would be rather complicated. With

Logarithmic differentiation 427

logarithmic differentiation the following procedure is (ii) ln y = ln(x + 1) + ln(x − 2)3 − ln(x − 3),
adopted: by laws (i) and (ii) of Section 47.2,

(i) Take Napierian logarithms of both sides of the i.e. ln y = ln(x + 1) + 3 ln(x − 2) − ln(x − 3),
by law (iii) of Section 47.2.

equation. √ (iii) Differentiating with respect to x gives:
(x
Thus ln y = ln (1 + x)2 − 1)
√
x (x + 2) 1 dy 1 3 1
y dx = + + − − − 3)
(x 1) (x 2) (x
(1 + x)2(x − 1) 1
= ln 2

x(x + 2) 1 by using equations (1) and (2)
2

(ii) Apply the laws of logarithms. (iv) Rearranging gives:

Thus ln y = ln(1 + x)2 + ln(x − 1) 1 dy 1 + 3 − 1
i.e. 2 =y + − −

− ln x − ln(x + 2) 1 , by laws (i) dx (x 1) (x 2) (x 3)
2

and (ii) of Section 47.2 (v) Substituting for y gives:

ln y = 2ln(1 + x) + 1 ln (x − 1) dy = (x + 1)(x − 2)3 1 +3 − 1
2 dx (x − 3) (x + 1) (x − 2) (x − 3)

− ln x − 1 + 2), by law (iii)
ln(x
2

of Section 47.2 (x − 2)3
1)2(2x −
(iii) Differentiate each term in turn with respect of x Problem 2. Differentiate y = (x + 1)
using equations (1) and (2).
dy
Thus 1 dy = 2 + 1 −1− 1 with respect to x and evalualte when x = 3.
2 2
dx

y dx (1 + x) (x − 1) x (x + 2)

dy Using logarithmic differentiation and following the
(iv) Rearrange the equation to make the subject. above procedure:

dx

Thus dy (1 2 x) + 2(x 1 1) − 1 (x − 2)3
=y + − x 1)2(2x −
(i) Since y =
dx

1 (x + 1)
+
− 2(x 2) (x − 2)3
then ln y = ln (x + 1)2(2x − 1)
(v) Substitute for y in terms of x. Section 8

dy = (1 + x)2 √ − 1) 2 (x − 2) 3
Thus √ ( (x (1 + x) 2
= ln (x + 1)2(2x − 1)
dx x (x + 2)

+ 1 −1− 1 (ii) ln y = ln(x − 2) 3 − ln(x + 1)2 − ln(2x − 1)
2(x − 1) x 2(x + 2) 2

i.e. ln y = 3 ln(x − 2) − 2 ln(x + 1) − ln(2x − 1)
2
Problem 1. Use logarithmic differentiation to
(x + 1)(x − 2)3 (iii) 1 dy = 3 −2− 2
2
differentiate y =
(x − 3) y dx (x − 2) (x + 1) (2x − 1)

Following the above procedure: (iv) dy 2(x 3 2) − (x 2 1) − (2x 2 1)
(x + 1)(x − 2)3 =y − + −

(i) Since y = dx
(x − 3)
(x + 1)(x − 2)3 (v) dy = (x + (x − 2)3 1) 3
dx 1)2(2x − 2(x − 2)
then ln y = ln
(x − 3) 22
− (x + 1) − (2x − 1)

428 Engineering Mathematics

dy (1)3 3 2 2 Using logarithmic differentiation and following the
When x = 3, dx = (4)2(5) −− procedure gives:
245
x3 ln 2x
= ± 1 3 = ± 3 or ± 0.0075 (i) ln y = ln ex sin x
80 5 400

(ii) ln y = ln x3 + ln ( ln 2x) − ln (ex) − ln ( sin x)

Problem 3. Given y = 3√e2θ sec 2θ dy i.e. ln y = 3 ln x + ln ( ln 2x) − x − ln ( sin x)
determine
(θ − 2) dθ
1
(iii) 1 dy = 3 + x − 1 − cos x

y dx x ln 2x sin x

Using logarithmic differentiation and following the (iv) dy = y 3 + 1 − 1 − cot x
procedure: dx x x ln 2x

3e2θ sec 2θ dy x3 ln 2x 3 1
(i) Since y = √ (v) dx = ex sin x + − 1 − cot x
x x ln 2x
(θ − 2)

3e2θ sec 2θ Now try the following exercise.
then ln y = ln √

(θ − 2)

⎧⎫
⎨ 3e2θ sec 2θ ⎬
= ln ⎩
1 ⎭ Exercise 170 Further problems on
differentiating logarithmic
(θ − 2) 2 functions

1

(ii) ln y = ln 3e2θ + ln sec 2θ − ln (θ − 2) 2

i.e. ln y = ln 3 + ln e2θ + ln sec 2θ In Problems 1 to 6, use logarithmic differentiation
to differentiate the given functions with respect to
− 1 ln (θ − 2) the variable.
2

i.e. ln y = ln 3 + 2θ + ln sec 2θ − 1 ln (θ − 2) (x − 2)(x + 1)
2 (x − 1)(x + 3)
1. y =

(iii) Differentiating with respect to θ gives: ⎡ (x − 2)(x + 1) ⎤
⎣⎢⎢ (x − 1)(x + 3)
1 dy = 0 + 2 + 2 sec 2θ tan 2θ − 1 (x 1 2) + (x 1 1) ⎦⎥⎥
y dθ sec 2θ 2 − +
Section 8 (θ 2)
− 1 1
− +
from equations (1) and (2) − (x 1) − (x 3)
(iv) Rearranging gives:
(x + 1)(2x + 1)3
dy 1 2. y = (x − 3)2(x + 2)4
dθ −
=y 2 + 2 tan 2θ − ⎡ (x + 1)(2x + 1)3 1 6 ⎤
⎢⎣⎢ (x − 3)2(x + 2)4 + +
2(θ 2) + ⎦⎥⎥

(x 1) (2x 1)

(v) Substituting for y gives: − (x 2 3) − (x 4 2)
− +

dy 3e2θsec 2θ 1 √
=√ 2 + 2 tan 2θ − 2(θ − 2) 3. y = (2x − 1) (x + 2)
dθ (θ − 2)
(x − 3) (x + 1)3

⎡(2x − √ + 2) 2 + 1 ⎤
1) (x − + ⎥⎥⎦

x3 ⎣⎢⎢(x − 3) (x + 1)3 (2x 1) 2(x 2)
ex
Problem 4. Differentiate y = ln 2x with respect − 1 − 3
sin x − +
to x (x 3) 2(x 1)

Logarithmic differentiation 429

4. y = e2x cos 3x from which, dy = y(1 + ln x)
√ i.e. dx
(x − 4) dy = xx(1 + ln x)
dx
e2x cos 3x 1
√ 2 − 3 tan 3x −
(x − 4) 2(x − 4)

dy
5. y = 3θ sin θ cos θ Problem 6. Evaluate when x = −1 given
3θ sin θ cos θ 1 + cot θ − tan θ dx
θ y = (x + 2)x

6. y = 2x4 tan x 2x4 tan x 4+ 1 Taking Napierian logarithms of both sides of
e2x ln 2x e2x ln 2x x sin x cos x y = (x + 2)x gives:

−2− 1 ln y = ln(x + 2)x = x ln(x + 2), by law (iii)
x ln 2x
of Section 47.2

Differentiating both sides with respect to x gives:

7. Evaluate dy when x = 1 given 1 dy = (x) 1 + [ln(x + 2)](1),
dx y dx x + 2

y= (x + 1)2 √ − 1) 13 by the product rule.
(2x 16

(x + 3)3 Hence dy = y x + ln(x + 2)
dx +
dy x 2

8. Evaluate , correct to 3 significant figures, x
dθ = (x + 2)x +
when θ = π given y = 2e√θ sin θ + ln(x + 2)
x 2
[−6.71]
4 θ5

47.4 Differentiation of [f (x)]x When x = −1, dy = (1)−1 −1 + ln 1
dx 1
= (+1)(−1) = −1

Whenever an expression to be differentiated contains a Problem 7. Determine (a) the differential
√
term raised to a power which is itself a function of the coefficient of y = x (x − 1) and (b) evaluate dy

variable, then logarithmic differentiation must be used. when x = 2. dx Section 8

For (exx+am2p)xle, ,√xth(xe differentiation of expressions such as
xx, − 1) and x3x + 2 can only be achieved

using logarithmic differentiation. √1
(a) y = x (x − 1) = (x − 1) x , since by the laws of
Problem 5. Determine dy given y = xx.
dx √m
indices n am = a n
Taking Napierian logarithms of both sides gives:

Taking Napierian logarithms of both sides of y = xx 11
gives: ln y = ln (x − 1) x = ln (x − 1),
ln y = ln xx = x ln x, by law (iii) of Section 47.2
Differentiating both sides with respect to x gives: x
1 dy = (x) 1 + (ln x)(1), using the product rule by law (iii) of Section 47.2.
y dx x
Differentiating each side with respect to x gives:
i.e. 1 dy = 1 + ln x
y dx 1 dy = 1 1 + [ ln (x − 1)] −1
y dx x x−1 x2

by the product rule.

dy 1 − ln (x − 1)
Hence = y − x2
x(x 1)
dx

430 Engineering Mathematics

i.e. dy √ 1 ln(x − 1) Now try the following exercise
= x (x − 1) x(x − 1) − x2
dx

(b) When x = 2, dy √ 1 − ln (1) Exercise 171 Further problems on differen-
= 2 (1) tiating [f (x)]x type functions
dx 2(1) 4

= ±1 1 −0 1 In Problems 1 to 4, differentiate with respect to x
=±
22
1. y = x2x [2x2x(1 + ln x)]

Problem 8. Differentiate x3x+2 with respect to x. 2. y = (2x − 1)x

(2x − 1)x 2x 1 + ln (2x − 1)
2x −
Let y = x3x+2
Taking Napierian logarithms of both sides gives: √
3. y = x (x + 3)
ln y = ln x3x+2
i.e. ln y = (3x + 2) ln x, by law (iii) of Section 47.2 √ 1 ln (x + 3)
Differentiating each term with respect to x gives: x (x + 3) x(x + 3) − x2

1 dy = (3x + 2) 1 + ( ln x)(3), 4. y = 3x4x+1 3x4x + 1 4 + 1 + 4 ln x
y dx x x

by the product rule. 5. Show that when y = 2xx and x = 1, dy = 2.
dx

dy = y 3x + 2 + 3 ln x 6. d √
dx x Evaluate x (x − 2) when x = 3.
dx
Hence 1

3

= x3x+2 3x + 2 + 3 ln x 7. Show that if y = θθ and θ = 2, dy
x = 6.77,
dθ
correct to 3 significant figures.
= x3x+2 3 + 2 + 3 ln x
x

Section 8

Revision Test 13

This Revision test covers the material contained in Chapters 45 to 47. The marks for each question are shown in
brackets at the end of each question.

1. A cycloid has parametric equations given by: 5. Determine the gradient of the tangents drawn to

x = 5(θ − sin θ) and y = 5(1 − cos θ). Evaluate the hyperbola x2 − y2 = 8 at x = 3. (4)

dy d2y
(a) (b) dx2 when θ = 1.5 radians. Give
dx 6. Use logarithmic differentiation to differentiate
√
answers correct to 3 decimal places. (8) (x + 1)2 (x − 2)
y = with respect to x. (6)
(2x − 1) 3 (x − 3)4
2. Determine the equation of (a) the tangent, and

(b) the normal, drawn to an ellipse x = 4 cos θ, 3eθ sin 2θ
y = sin θ at θ = π √
(8) 7. Differentiate y = θ5 and hence evaluate

3 dy π

dz , correct to 2 decimal places, when θ =
3. Determine expressions for for each of the dθ 3 (9)

dy d√
following functions: 8. Evaluate [ t (2t + 1)] when t = 2, correct to 4

(a) z = 5y2 cos x (b) z = x2 + 4xy − y2 (5) dt

4. If x2 + y2 + 6x + 8y + 1 = 0, find dy in terms of significant figures. (6)

x and y. dx (4)

Section 8

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Section 9

Integral Calculus