CHAPTER 52 INTRODUCTION TO DIFFERENTIATION
EXERCISE 219 Page 606
1. If f(x) = 6x2 – 2x + 1 find f(0), f(1), f(2), f(–1) and f(–3)
f(0) = 6(0) 2 – 2(0) + 1 = 1
f(1) = 6(1) 2 – 2(1) + 1 = 5
f(2) = 6(2) 2 – 2(2) + 1 = 24 – 4 + 1 = 21
f(– 1) = 6(–1) 2 – 2(–1) + 1 = 9
f(– 3) = 6(–3) 2 – 2(–3) + 1 = 54 + 6 + 1 = 61
2. If f(x) = 2x2 + 5x – 7 find f(1), f(2), f(–1), f(2) – f(–1)
f(1) = 2(1) 2 + 5(1) – 7 = 0
f(2) = 2(2) 2 + 5(2) – 7 = 8 + 10 – 7 = 11
f(–1) = 2(–1) 2 + 5(–1) – 7 = – 10
f(2) – f(–1) = 11 – –10 = 21
3. Given f(x) = 3x3 + 2x2 – 3x + 2 prove that f(1) =
1
f(2)
7
f(1) = 3(1)3 + 2(1) 2 − 3(1) + 2 = 4
f(2) = 3(2)3 + 2(2) 2 − 3(2) + 2 = 24 + 8 – 6 + 2 = 28
Hence, f(1) =
1
f(2)
7
4. If f(x) = –x2 + 3x + 6 find f(2), f(2 + a), f(2 + a) – f(2) and
f (2 + a ) − f (2)
a
f(2) = −(2) 2 + 3(2) + 6 = –4 + 6 + 6 = 8
897
© 2014, John Bird
f(2) = −(2 + a ) 2 + 3(2 + a ) + 6 = –(4 + 4a + a 2 ) + 6 + 3a + 6 = −a 2 − a + 8
f(2 + a) – f(2) = (−a 2 − a + 8) − 8 = −a 2 − a
f (2 + a ) − f (2) −a 2 − a
= –a – 1
=
a
a
898
© 2014, John Bird
EXERCISE 220 Page 607
1. Plot the curve f(x) = 4x2 – 1 for values of x from x = –1 to x = +4. Label the coordinates (3, f(3))
and (1, f(1)) as J and K, respectively. Join points J and K to form the chord JK. Determine the
gradient of chord JK. By moving J nearer and nearer to K determine the gradient of the tangent of
the curve at K.
The curve of f(x) = 4 x 2 − 1 is shown below.
Gradient of chord JK =
f (3) − f (1) 35 − 3 32
= 16
= =
3 −1
2
2
If, say, J′ is the point (2, f(2)) then gradient of chord J′K =
f (2) − f (1) 15 − 3 12
= 12
= =
2 −1
1
1
If, say, J′′ is a point (1.1, f(1.1)) then gradient of chord J′′K =
899
f (1.1) − f (1) 3.84 − 3 0.84
= 8.4
= =
1.1 − 1
0.1
0.1
© 2014, John Bird
If, say, J′′′ is a point (1.01, f(1.01)) then gradient of chord J′′′K
=
f (1.01) − f (1) 3.0804 − 3 0.0804
= 8.04
= =
1.01 − 1
0.01
0.01
Thus, as J moves closer and closer to point K the gradient of the chord approaches nearer and nearer
to the value 8. Thus the gradient of the tangent to the curve at K is 8
900
© 2014, John Bird
EXERCISE 221 Page 609
1. Differentiate from first principles: y = x
( x + δ x) − ( x) 
dy
 f ( x + δ x) − f ( x) 
δ x 
= f=
'( x) limit 
=
limit 
=
limit =


 limit {1} = 1
δ x →0 
dx
δx
δx
 δ x →0 
 δ x →0  δ x  δ x →0
2. Differentiate from first principles:
y = 7x
 ( 7 x + 7δ x ) − ( 7 x ) 
dy
 f ( x + δ x) − f ( x) 
 7δ x 
'( x) limit 
limit 
= f=
=
=

 limit
 =
 limit {7} = 7
0
0
0
x
x
x
δ
δ
δ
→
→
→
dx
δx
δx


 δ x  δ x →0


3. Differentiate from first principles: y = 4x2
 4 ( x + δ x )2 − ( 4 x 2 ) 
dy
 f ( x + δ x) − f ( x) 
= f=
=
'( x) limit 
 limit 

δ x →0 
dx
δx
δx
 δ x →0 

 4 ( x 2 + 2 xδ x + δ x 2 ) − ( 4 x 2 ) ) 
 4 x 2 + 8 xδ x + 4δ x 2 − 4 x 2 
=
limit
limit 



δ x →0
δx
δx


 δ x →0 
 8 xδ x + 4δ x 2 
= limit 
=
{8 x + 4δ x} = 8x
 limit
δ x →0 
δx
 δ x →0
4. Differentiate from first principles: y = 5x3
 5 ( x + δ x )3 − ( 5 x 3 ) 
dy
 f ( x + δ x) − f ( x) 
= f=
=
'( x) limit 
 limit


δ x →0 
δx
δx
dx
 δ x →0 

 5 ( x3 + 3 x 2δ x + 3 xδ x 2+δ x 2 ) − ( 5 x 3 ) ) 
 5 x3 + 15 x 2δ x + 15 xδ x 2 + δ x 3 − 5 x 3 
=
limit
limit 



δ x →0
δx
δx


 δ x →0 
15 x 2δ x + 15 xδ x 2 + δ x 3 
= limit 
=
{15 x 2 + 15 xδ x + δ x 2 } = 15x 2
 limit
δ x →0 
δx
 δ x →0
5. Differentiate from first principles: y = –2x2 + 3x – 12
y = f(x) = −2 x 2 + 3 x − 12
901
© 2014, John Bird
f(x + δx) = −2 ( x + δ x ) + 3 ( x + δ x ) − 12 =
−2 ( x 2 + 2 xδ x + δ x 2 ) + 3( x + δ x) − 12
2
= −2 x 2 − 4 xδ x − 2δ x 2 + 3 x + 3δ x − 12
 (−2 x 2 − 4 xδ x − 2δ x 2 + 3 x + 3δ x − 12) − ( −2 x 2 + 3x − 12 ) 
dy
 f ( x + δ x) − f ( x) 
'( x) limit 
limit 
= f=
=


δ x →0 
dx
δx
δx
 δ x →0 

 −4 xδ x − 2δ x 2 + 3δ x 
= limit 
=
{−4 x − 2δ x + 3}
 limit
δ x →0 
δx
 δ x →0
= –4x + 3
6. Differentiate from first principles: y = 23
 ( 23) − ( 23) 
dy
 f ( x + δ x) − f ( x) 
 0 
= f=
'( x) limit 
=
limit 
=
limit   = 0


δ x →0 
dx
δx
δx
 δ x →0 
 δ x →0  δ x 
7. Differentiate from first principles: f(x) = 9x
 ( 9 x + 9δ x ) − ( 9 x ) 
dy
 f ( x + δ x) − f ( x) 
 9δ x 
'( x) limit 
limit 
= f=
=
=

 limit
 =
 limit {9} = 9
δ
x
→
0
δ
x
→
0
δ
x
→
0
dx
δx
δx


 δ x  δ x →0


8. Differentiate from first principles: f(x) =
dy
 f ( x + δ x) − f ( x) 
= f=
=
'( x) limit 

δ x →0 
δx
dx

2x
3
 2
2   2 
2 
  3 x + 3 δ x  −  3 x  
 3δ x
2
  =

=
limit  
limit
limit  



δ x →0
δ x →0
δ x →0  3 
δx


 δx 




2
=
3
9. Differentiate from first principles: f(x) = 9x2
 9 ( x + δ x )2 − ( 9 x 2 ) 
dy
 f ( x + δ x) − f ( x) 
= f=
=
'( x) limit 
 limit


δ x →0 
dx
δx
δx
 δ x →0 

 9 ( x 2 + 2 xδ x + δ x 2 ) − ( 9 x 2 ) ) 
 9 x 2 + 18 xδ x + 9δ x 2 − 9 x 2 
=
limit
limit

 δ x →0 

δ x →0
δx
δx




© 2014, John Bird
902
18 xδ x + 9δ x 2 
= limit 
=
{18 x + 9δ x} = 18x
 limit
δ x →0 
δx
 δ x →0
10. Differentiate from first principles: f(x) = –7x3
y = f(x) = −7x3
f(x + δx) = −7 ( x + δ x ) =
−7 ( x 2 + 2 xδ x + δ x 2 )( x + δ x ) =
−7 ( x 3 + 3x 2δ x + 3xδ x 2 + δ x 3 )
3
= −7 x3 − 21x 2δ x − 21xδ x 2 − 7δ x 3
 (−7 x3 − 21x 2δ x − 21xδ x 2 − 7δ x 3 ) − ( −7 x 3 ) 
dy
 f ( x + δ x) − f ( x) 
'( x) limit 
limit 
= f=
=


δ x →0 
dx
δx
δx
 δ x →0 

 −21x 2δ x − 21xδ x 2 − 7δ x 3 
= limit 
{−21x 2 − 21xδ x − 7δ x 2 }
 = limit
x →0
δ x →0 
δ
δx

= −21x 2
11. Differentiate from first principles: f(x) = x2 + 15x – 4
y = f(x) = x2 + 15x – 4
f(x + δx) = ( x + δ x ) + 15 ( x + δ x ) − 4 =( x 2 + 2 xδ x + δ x 2 ) + 15( x + δ x) − 4
2
= x 2 + 2 xδ x + δ x 2 + 15 x + 15δ x − 4
 ( x 2 + 2 xδ x + δ x 2 + 15 x + 15δ x − 4) − ( x 2 + 15 x − 4 ) 
dy
 f ( x + δ x) − f ( x) 
'( x) limit 
limit
= f=
=
 δ x →0 

δ x →0 
δx
δx
dx



2
 2 xδ x + δ x + 15δ x 
= limit 
=
{2 x + δ x + 15}
 limit
δ x →0 
δx
 δ x →0
= 2x + 15
12. Differentiate from first principles: f(x) = 4
 ( 4) − ( 4) 
dy
 f ( x + δ x) − f ( x) 
 0 
= f=
=
=
'( x) limit 
limit 
limit   = 0


δ x →0 
dx
δx
 δ x →0  δ x  δ x →0  δ x 
903
© 2014, John Bird
13. Determine
d
(4x3) from first principles .
dx
Let y = f(x) = 4x3
and f(x + δx) = 4 ( x + δ x ) = 4 ( x 2 + 2 xδ x + δ x 2 )( x + δ x ) = 4 ( x 3 + 3x 2δ x + 3xδ x 2 + δ x 3 )
3
= 4 x3 + 12 x 2δ x + 12 xδ x 2 + 4δ x 3
 (4 x3 + 12 x 2δ x + 12 xδ x 2 + 4δ x 3 ) − ( 4 x 3 ) 
dy
 f ( x + δ x) − f ( x) 
'( x) limit 
limit 
= f=
=


δ x →0 
dx
δx
δx
 δ x →0 

12 x 2δ x + 12 xδ x 2 + 4δ x 3 
= limit 
=
{12 x 2 + 12 xδ x + 4δ x 2 }
 limit
δ x →0 
δx
 δ x →0
d
( 4 x3 ) = 12x 2
dx
Hence,
14. Find
d
(3x2 + 5) from first principles.
dx
Let y = f(x) = 3 x 2 + 5
and f(x + δx) = 3 ( x + δ x ) + 5 = 3 ( x 2 + 2 xδ x + δ x 2 ) + 5
2
= 3 x 2 + 6 xδ x + 6δ x 2 + 5
 (3 x 2 + 6 xδ x + 6δ x 2 + 5) − ( 3 x 2 + 5 ) 
dy
 f ( x + δ x) − f ( x) 
'( x) limit 
limit
= f=
=



δ x →0 
δx
δx
dx
 δ x →0 

 6 xδ x + 6δ x 2 
= limit 
=
{6 x + 6δ x}
 limit
δ x →0 
δx
 δ x →0
Hence,
d
( 3x 2 + 5) = 6x
dx
904
© 2014, John Bird
EXERCISE 222 Page 611
1. Differentiate y = 7x4 with respect to x.
If y = 7x 4 then
dy
= (7) ( 4 x3 ) = 28x3
dx
2. Differentiate y =
If y =
1
1
d y 1 −1
1
=
x 2
x = x 2 then = =
1
dx 2
2 x
2x 2
3. Differentiate y =
If y =
x with respect to x.
t 3 with respect to x.
3
dy 3 1
3
t2 =
t
t 3 = t 2 then = =
dt 2
2
4. Differentiate y = 6 +
If y = 6 +
then
with respect to x.
dy
1
3
= 0 + ( −3 x −4 ) = −
= 6 + x −3 then
3
dt
x
x4
5. Differentiate y = 3x –
If y = 3x –
1
x3
1 1
+
with respect to x.
x x
1
1 1
1
−
+ = 3 x − 1 + x −1 = 3 x − x 2 + x −1
x x
x2
1
1
dy
1 −3
1
1
 1 − 1 −1 
−
= 3+
= 3 −  − x 2  − x −2 =
3 + x 2 − x −2 = 3 + 3 −
2
dt
2
2 x3 x 2
 2

2x 2 x
6. Differentiate y =
5
1
−
+ 2 with respect to x.
2
x
x7
905
© 2014, John Bird
If y =
7
5
1
dy
−
−
+ 2= 5 x −2 − x 2 + 2 then
=
x2
dx
x7
( 5)( −2 x −3 ) −  −
7 −9 
x 2 +0
 2

=−
10
7
10
7
+ 9 = − +
3
3
x 2 x9
x 2x 2
7. Differentiate y = 3(t – 2)2 with respect to x.
If y = 3 ( t − 2 ) = 3 ( t 2 − 4t + 4 ) = 3t 2 − 12t + 12
2
dy
then =
dt
( 3)( 2t ) − 12 + 0
= 6t – 12
8. Differentiate y = (x + 1)3 with respect to x.
If y = ( x + 1) = ( x 2 + 2 x + 1)( x + 1) = x 3 + x 2 + 2 x 2 + 2 x + x + 1 = x3 + 3 x 2 + 3 x + 1
2
then
dy
= 3x 2 + 3 ( 2 x ) + 3 + 0 = 3x 2 + 6 x + 3
dx
9. Using the general rule for axn check the results of Problems 1 to 12 of Exercise 221, page 609.
1. If y = x,
dy
=1
dx
2. If y = 7x,
dy
=7
dx
3. If y = 4x 2 ,
dy
= 8x
dx
4. If y = 5x 3 ,
dy
= 15x 2
dx
5. If y = – 2x 2 + 3x – 12,
6. If y = 23,
dy
= –4x + 3
dx
dy
=0
dx
7. If f(x) = 9x, f ´(x) = 9
906
© 2014, John Bird
8. If f(x) =
2x
2
, f ´(x) =
3
3
9. If f(x) = 9x 2 , f ´(x) = 18x
10. If f(x) = –7x 3 , f ´(x) = –21x 2
11. If f(x) = x 2 + 15x – 4, f ´(x) = 2x + 15
12. If f(x) = 4, f ´(x) = 0
10. Differentiate f(x) = 6x2 – 3x + 5 and find the gradient of the curve at (a) x = –1, and (b) x = 2
Gradient = f ′(x) = 12x – 3
(a) When x = –1, gradient = 12(–1) – 3 = –15
(b) When x = 2, gradient = 12(2) – 3 = 21
11. Find the differential coefficient of y = 2x3 + 3x2 – 4x – 1 and determine the gradient of the curve
at x = 2
Gradient =
dy
= 6 x2 + 6 x − 4
dx
When x = 2, gradient = 6 ( 2 ) + 6(2) − 4 = 24 + 12 – 4 = 32
2
12. Determine the derivative of y = –2x3 + 4x + 7 and determine the gradient of the curve at x = –1.5
dy
=−2 ( 3 x 2 ) + 4 + 0 = −6 x 2 + 4
dx
Gradient =
dy
=
−6 x 2 + 4 and when x = – 1.5
dx
gradient = −6 ( −1.5 ) + 4 =−6(2.25) + 4 =−13.5 + 4 = –9.5
2
907
© 2014, John Bird
EXERCISE 223 Page 613
1. Differentiate with respect to x:
(a) y = 4 sin 3x
(b) y = 2 cos 6x
(a) If y = 4 sin 3x then
dy
= ( 4 ) (3cos 3 x) = 12 cos 3x
dx
dy
(b) If y = 2 cos 6x then =
dx
( 2 ) (−6sin 6 x)
= –12 sin 6x
2. Given f(θ) = 2 sin 3θ – 5 cos 2θ, find f ´(θ)
If f(θ) = 2 sin 3θ – 5 cos 2θ, f ´(θ) = (2)(3 cos 3θ) – (5)( – 2sin 2θ)
= 6 cos 3θ + 10 sin 2θ
3. An alternating current is given by i = 5 sin 100t amperes, where t is the time in seconds.
Determine the rate of change of current when t = 0.01 seconds.
Rate of change of current =
di
= (5)(100 cos100t ) = 500 cos 100t A/s
dt
and when t = 0.01 s, rate of change of current = 500 cos(100 × 0.01) = 500 cos 1 = 270.2 A/s
(Note that cos 1 means the cosine of 1 radian)
4. v = 50 sin 40t volts represents an alternating voltage where t is the time in seconds. At a time of
20 × 10–3 seconds, find the rate of change of voltage.
dv
Rate of change of=
voltage =
(50)(40
=
cos 40t ) 2000 cos 40t V/s
dt
When t = 20 ×10−3 , rate of change of voltage = 2000 cos(40 × 20 ×10−3 ) = 2000 cos 0.8
= 1393.4 V/s
(Note that cos 0.8 means the cosine of 0.8 radian)
908
© 2014, John Bird
5. If f(t) = 3 sin(4t + 0.12) – 2 cos(3t – 0.72) determine f ´(t)
f ′(t) = (3)[4 cos(4t + 0.12)] – (2)[–3 sin(3t – 0.72)]
= 12 cos(4t + 0.12) + 6 sin(3t – 0.72)
909
© 2014, John Bird
EXERCISE 224 Page 614
1. Differentiate with respect to x: (a) y = 5e3x (b) y =
(a) If y = 5e3 x then
2
7e 2 x
dy
= (5) ( 3e3 x ) = 15e3 x
dx
d y 2
4
4
2
2 −2 x
then
=−
− e −2 x = −
2 e −2 x ) =
e
=
(


dx 7
7
7 e2 x
7 e2 x 7
(b) =
If y
2. Given f(θ) = 5 ln 2θ – 4 ln 3θ, determine f ´(θ)
1
1 1 5 4
If f(θ) = 5 ln 2θ – 4 ln 3θ, f ´(θ) = 5   − 4   =− =
θ
θ  θ  θ θ
3. If f(t) = 4 ln t + 2, evaluate f ´(t) when t = 0.25
If f(t) = 4 ln t + 2 then f ′(t) =
When t = 0.25, f ′(t) =
4. Evaluate
4
4
+0 =
t
t
4
4
= 16
=
t 0.25
dy
5
when x = 1, given y = 3e4x –
+ 8 ln 5x. Give the answer correct to 3 significant
dx
2e3 x
figures.
Since y =3e 4 x −
5
5
+ 8ln 5 x =3e 4 x − e −3 x + 8ln 5 x
3
x
2e
2
dy
5
1
then = (3) ( 4 e 4 x ) −   ( −3e −3 x ) + (8)  
dx
2
x
= 12 e 4 x +
and when x = 1,
15 −3 x 8
15 8
e + = 12 e 4 x +
+
2 e3 x x
2
x
dy
15 8
= 12 e 4 +
+ = 655.178 + 0.3734 + =
8 663.55
dx
2 e3 1
= 664, correct to 3 significant figures
910
© 2014, John Bird