CHAPTER 53 METHODS OF DIFFERENTIATION

EXERCISE 225 Page 617

1
1. Differentiate with respect to x: (a) 5x5 (b) 2.4x3.5 (c)
x

dy
(a) If y = 5x5 then = (5) ( 5 x 4 ) = 25x 4
dx

dy
(b) If y = 2.4x3.5 then = (2.4) ( 3.5 x3.5−1 ) = 8.4x 2.5
dx

1 dy 1
(c) If y = = x −1 then = −1 x −2 = −
x dx x2

−4
2. Differentiate with respect to x: (a) (b) 6 (c) 2x
x2

−4 dy 8
(a) If y = = −4x −2 then ( 4) ( −2 x −3 ) =
=− 8 x −3 or
x 2 dx x3
dy
(b) If y = 6 then =0
dx
dy
(c) If y = 2x then =2
dx

4
3. Differentiate with respect to x: (a) 2 x (b) 3 3 x5 (c)
x

1 dy  1 − 12  −
1
1 1
(a) If y = 2 x = 2 x 2 then = (2)  x =  x= 2 =
2 
1
dx x2 x
5 dy  5 23  2
= (3)
(b) If y = 3 3 x5 = 3 x 3 then =
 x  5 x 3 = 5 3 x2
dx  3 
4 4 −
1
dy  1 − 32  −
3 2 2
(c) If y = = = 4x 2 then =−
(4)  x  =
−2 x − 3 =−
2 =
 2 
1
x x2 dx x2 x3

911 © 2014, John Bird

 −3
4. Differentiate with respect to x: (a) 3
(b) (x – 1)2 (c) 2 sin 3x
x

3 3 −
1
dy  1 − 43  −
4
1 1
(a) If y = − =
− =
−3x 3 then =( −3)  − x  =x 3 = =
 
3 1 4 3
x x3 d x 3 x3 x4

(b) If y = (x – 1)2 = (x – 1)(x – 1) = x 2 − x − x + 1 = x 2 − 2 x + 1

dy
then = 2x – 2 or 2(x – 1)
dx
dy
(c) If y = 2 sin 3x then = (2)(3 cos 3x) = 6 cos 3x
dx

3
5. Differentiate with respect to x: (a) – 4 cos 2x (b) 2e6x (c)
e5 x

dy
(a) If y = –4 cos 2x then = (–4)(– 2 sin 2x) = 8 sin 2x
dx
dy
(b) If y = 2e6x then = (2) ( 6 e6 x ) = 12 e6 x
dx

3 dy 15
(c) If y = = 3e −5 x then (3) ( −5e −5 x ) =
= −15e −5 x = −
e5 x dx e5 x

e x − e− x 1− x
6. Differentiate with respect to x: (a) 4 ln 9x (b) (c)
2 x

dy 1 4
(a) If y = 4 ln 9x then = (4)   =
dx x x

e x − e− x 1 x 1 − x dy 1 x 1 e x + e− x
)   e x +   e− x =
1 1
(b) If y = = e − e then =   e −   ( − e x=
2 2 2 dx 2 2 2 2 2

1− x 1 x 1
−1 −
1
(c) If y = = − = x −1 − x 2 = x −1 − x 2
x x x

dy  1 − 1 −1  1 1 −3 1 1 1 1
then =
− x −2 −  − x 2  = − + x 2=− + 3 =− +
dx  2  x2 2 x2 2 x 2 x 2 2 x3

912 © 2014, John Bird

7. Find the gradient of the curve y = 2t4 + 3t3 – t + 4 at the points (0, 4) and (1, 8)

dy
If y= 2t 4 + 3t 3 − t + 4 , then gradient, = 8t 3 + 9t 2 − 1
dt
At (0, 4), t = 0, hence gradient = 8(0)3 + 9(0) 2 − 1 = – 1

At (1, 8), t = 1, hence gradient = 8(1)3 + 9(1) 2 − 1 = 16

8. Find the coordinates of the point on the graph y = 5x2 – 3x + 1 where the gradient is 2

dy
If y = 5 x 2 − 3 x + 1 , then gradient = = 10 x − 3
dx
1
When the gradient is 2, 10x – 3 = 2 i.e. 10x = 5 and x =
2
2
1 1 1 5 3 3
When x = , y = 5  − 3  +1 = − +1 =
2 2 2 4 2 4

1 3
Hence, the coordinates of the point where the gradient is 2 is  , 
2 4

2 2
9. (a) Differentiate y = + 2 ln 2θ – 2(cos 5θ + 3 sin 2θ) –
θ 2 e3θ

dy π
(b) Evaluate when θ = , correct to 4 significant figures.
dθ 2

2 2
(a) y = + 2 ln 2θ − 2(cos 5θ + 3sin 2θ ) −
θ2 e3θ
= 2θ −2 + 2 ln 2θ − 2 cos 5θ − 6sin 2θ − 2e −3θ
dy 2
Hence, =−4θ −3 + − 2(−5sin 5θ ) − 6(2 cos 2θ ) − 2 ( −3e −3θ )
dθ θ
4 2 6
=− + + 10sin 5θ − 12 cos 2θ +
θ3 θ e3θ
π dy 4 2 5π 2π 6
(b) When θ = , =− + + 10sin − 12 cos + π 
dθ π  π 
3
2 2 2 3 
   2  e 2
2
= –1.032049 + 1.2732395 + 10 + 12 + 0.0538997

913 © 2014, John Bird

 = 22.30, correct to 4 significant figures

ds π
10. Evaluate , correct to 3 significant figures, when t = given s = 3 sin t – 3 + t.
dt 6

1 ds 1 −1 1
t 3sin t − 3 + t 2 then = 3cos t − 0 + t=
If s = 3 sin t – 3 + = 2 3cos t +
dt 2 2 t

π ds π  1
When t ==, 3cos   + = 3.29, correct to 3 significant figures
6 dt 6 2 (π / 6 )

11. A mass m is held by a spring with a stiffness constant k. The potential energy p of the system
1 2
is given by:=p kx − mgx where x is the displacement and g is acceleration due to gravity.
2
dp
The system is in equilibrium if = 0 . Determine the expression for x for system equilibrium.
dx

1 2 dp
=
If p kx − mgx then = kx − mg
2 dx

dp mg
If = 0 then kx – mg = 0 i.e. kx = mg and displacement, x =
dx k

12. The current i flowing in an inductor of inductance 100 mH is given by: i = 5 sin 100t amperes,

di
where t is the time t in seconds. The voltage v across the inductor is given by: v = L volts.
dt

Determine the voltage when t = 10 ms.

di
If i = 5 sin 100t=
then =
(5)(100 cos t ) 500 cos100t
dt

di
Voltage across inductor, v = L = ( L)(500 cos100t )
dt

When L = 100 mH and t = 10 ms, voltage, v = (100 ×10−3 )( 500 cos(100 ×10 ×103 ) )

= 50 cos 1 = 27.0 volts

914 © 2014, John Bird

EXERCISE 226 Page 618

1. Differentiate with respect to x: x sin x

dy
=
If y = x sin x, then ( x )( cos x ) + ( sin x )(1)
dx
= x cos x + sin x

2. Differentiate with respect to x: x 2 e 2 x

dy
If y = x 2 e 2 x ,=
then ( x 2 )( 2 e2 x ) + ( e2 x )( 2 x )
dx
= 2 x 2 e 2 x + 2 x e 2 x or 2 x e 2 x ( x + 1)

3. Differentiate with respect to x: x 2 ln x

( x 2 ) 
dy 1
=
If y = x 2 ln x , then  + ( ln x )( 2 x )
dx x
= x + 2x ln x or x(1 + 2 ln x)

4. Differentiate with respect to x: 2x3 cos 3x

dy
If y = 2 x 3 cos 3 x , then =( 2 x3 )( −3sin 3 x ) + ( cos 3 x )( 6 x 2 )
dx
= −6 x3 sin 3 x + 6 x 2 cos 3 x = 6 x 2 ( cos 3 x − x sin 3 x )

5. Differentiate with respect to x: x3 ln 3x

3 d y  3  1  3 1
If y = =  x 2    + ( ln 3 x )  x 2 
x3 ln 3x = x 2 ln 3 x , then
d x   x  2 
3
−1 3 1 1
3 1
= x2 + x 2 ln 3x =
x 2 + x 2 ln 3x
2 2
1
 3   3 
= x 2 1 + ln 3 x  = x 1 + ln 3 x 
 2   2 

915 © 2014, John Bird

6. Differentiate with respect to x: e3t sin 4t

dy
If y = e3t sin 4=
t , then ( e3t )( 4 cos 4t ) + ( sin 4t )( 3e3t )
dt
= e3t ( 4 cos 4t + 3sin 4t )

7. Differentiate with respect to θ: e4θ ln 3θ

( e4θ ) 
dy 1
If y = e4θ ln 3θ, then
=  + ( ln 3θ )( 4 e )
4θ
dθ θ 
1 
= e 4θ  + 4 ln 3θ 
θ 

8. Differentiate with respect to t: et ln t cos t

 
( et ln t )( − sin t ) + ( cos t ) ( et ) 
dy 1
If y = et ln t cos t , then =  + ( ln t )( e ) 
t
dt  t  
 cos t 
= et − ln t sin t + + cos t ln t 
 t 
 1  
= et  + ln t  cos t − ln t sin t 
 t  

di
9. Evaluate , correct to 4 significant figures, when t = 0.1, and i = 15 t sin 3t
dt

di
Since i = 15t sin=
3t, then (15t )(3cos 3t ) + (sin 3t )(15)
dt
= 45t cos 3t + 15 sin 3t
di
When t = 0.1, = 45(0.1) cos 0.3 + 15 sin 0.3 (note 0.3 is radians)
dt
= 4.2990 + 4.4328
= 8.732, correct to 4 significant figures

dz
10. Evaluate , correct to 4 significant figures, when t = 0.5, given that z = 2e3t sin 2t
dt

916 © 2014, John Bird

 dz
=
Since z = 2e3t sin 2t, then (2 e3t )(2 cos 2t ) + (sin 2t )(6 e3t )
dt

= 4 e3t cos 2t + 6 e3t sin 2t

dz
When t = 0.5, = 4 e1.5 cos1 + 6 e1.5 sin1 (note 1 is 1 radian)
dt

= 9.68587 + 22.62727

= 32.31, correct to 4 significant figures

917 © 2014, John Bird

 EXERCISE 227 Page 620

sin x
1. Differentiate with respect to x:
x

sin x d y ( x )( cos x ) − ( sin x )(1)

If y = , then =
( x)
2
x dx

x cos x − sin x
=
x2

2 cos 3x
2. Differentiate with respect to x:
x3

=
If y =
2 cos 3x
, then
dy (=
x 3 )( −6sin 3x ) − ( 2 cos 3 x )( 3 x 2 ) −6 x 2 ( x sin 3x + cos 3x )
( x3 )
2
x3 dx x6

−6( x sin 3 x + cos 3 x) −6

= or ( x sin 3x + cos 3x )
x4 x4

2x
3. Differentiate with respect to x:
x2 + 1

If y =
2x
,=
then
dy ( x 2 + 1)( 2=) − ( 2 x )( 2 x ) 2 x2 + 2 − 4 x2
=
2 − 2 x2
x +1 ( x 2 + 1) ( x 2 + 1) ( x 2 + 1)
2 2 2
2 dx

2(1 − x 2 )
=
( x 2 + 1)
2

x
4. Differentiate with respect to x:
cos x

x x
1
2 dy
( cos x ) 
1 −1 
2
x 2 −

( x ) ( − sin x )
If y = = , then =
( cos x )
2
cos x cos x dx

cos x
+ x sin x
= 2 x
cos 2 x

918 © 2014, John Bird

 3 θ3
5. Differentiate with respect to θ:
2sin 2θ

( 2sin 2θ ) 
9 1  3
3 θ 2  −  3θ 2  ( 4 cos 2θ )
3 θ3 3θ 2 dy 2   
If y = = , then =
2sin 2θ 2sin 2θ dθ ( 2sin 2θ )
2

3 θ { 3sin 2θ − 4θ cos 2θ }
1 3
9 θ 2 sin 2θ − 12 θ 2 cos 2θ
= =
4sin 2 2θ 4sin 2 2θ

ln 2t
6. Differentiate with respect to t:
t

( t )  1t  − ( ln 2t )  12 t 

1
− −
1
1 −1
2
t 2 − t 2 ln 2t −
1
ln 2t dy 2 = t 2  1 
If y = ,= = 1 − ln 2t 
( t)  2 
2
t dt t t

1 1  1  1 
= 3 
1 − ln 2t  =  1 − ln 2t 
t2  2  t3  2 

2 xe 4 x
7. Differentiate with respect to x:
sin x

2 x e4 x d y ( sin x ) ( 2 x )( 4 e 4 x ) + ( e 4 x )( 2 )  − ( 2 x e 4 x )( cos x )

If y = , then =
( sin x )
2
sin x dx

8 x e 4 x sin x + 2 e 4 x sin x − 2 x e 4 x cos x

=
sin 2 x

=
2 e4 x
sin 2 x
{(1 + 4 x ) sin x − x cos x}

2x
8. Find the gradient of the curve y = at the point (2, –4)
x −5
2

If y =
2x
then =
gradient,
dy ( x 2 − 5) (2)
=
− (2 x)(2 x) 2 x 2 − 10 − 4 x 2 −10 − 2 x 2
=
x −5 ( x 2 − 5) ( x 2 − 5) ( x 2 − 5)
2 2 2
2 dx

−10 − 2(2) 2 −10 − 8 −18

At the point (2, –4), x = 2, hence gradient = = = = –18
( 22 − 5 ) (4 − 5) 2
2
1

919 © 2014, John Bird

 dy 2x2 + 3
9. Evaluate at x = 2.5, correct to 3 significant figures, given y =
dx ln 2 x

( ln 2 x ) (4 x) − (2 x 2 + 3) 
1
2x + 3 
If y =
2
then
dy
= x
( ln 2 x )
2
ln 2 x dx

( ln 5) (10) − [2(2.5)2 + 3] 
1 

=
When x = 2.5,
dy
=  2.5  16.09438 − 6.2 = 3.82, correct to 3
( ln 5)
2
dx 2.59029
significant figures

920 © 2014, John Bird

EXERCISE 228 Page 621

( 2 x − 1)
6
1. Differentiate with respect to x:

dy dy
If y = ( 2 x − 1) then = 6 ( 2 x − 1) ( 2 ) i.e. = 12 ( 2 x − 1)
6 5 5

dx dx

2. Differentiate with respect to x: (2x3 – 5x)5

dy
If y = ( 2 x3 − 5 x ) then 5 ( 2 x3 − 5 x ) ( 6 x 2 − 5)
5 4
=
dx

3. Differentiate with respect to θ: 2 sin(3θ – 2)

= 2 ( cos ( 3θ − 2 ) ) ( 3)
dy dy
If y = 2 sin(3θ – 2) then = 6 cos ( 3θ − 2 )
i.e.
dθ dθ

4. Differentiate with respect to α: 2 cos5 α

dy dy
If y = 2 cos5 α=
then (2) ( 5cos 4 α )( − sin α ) i.e. = −10 cos 4 α sin α
dα dα

1
5. Differentiate with respect to x:
( x − 2 x + 1)5
3

1 dy −5 ( 3 x 2 − 2 )
= ( x3 − 2 x + 1) −5 ( x3 − 2 x + 1) ( 3 x 2 − 2 ) =
−5 −6
If y = then =
( x3 − 2 x + 1) ( x3 − 2 x + 1)
5 6
dx

5 ( 2 − 3x 2 )
=
( x3 − 2 x + 1)
6

6. Differentiate with respect to t: 5e2t+1

dy dy
If y = 5e2t+1 then = (5) ( e 2t +1 )( 2 ) i.e. = 10 e 2t +1
dt dt

921 © 2014, John Bird

7. Differentiate with respect to t: 2 cot(5t2 + 3)

cos x
If y = cot x = then
sin x

d y (sin x)(− sin x) − (cos x)(cos x) − ( sin 2 x + cos 2 x ) −1

= = = = − cosec 2 x
dx sin 2 x sin 2 x sin 2 x

dy
Thus, if y = 2 cot ( 5t 2 + 3) , then ( 2 )  − cos ec 2 ( 5t 2 + 3) (10t ) = −20t cosec2 ( 5t 2 + 3)
=
dt

8. Differentiate with respect to y: 6 tan(3y + 1)

dθ dθ
If θ = 6 tan(3y + 1) =
then (6) ( sec 2 ( 3 y + 1) ) ( 3) =
i.e. 18sec 2 ( 3 y + 1)
dy dy

9. Differentiate with respect to x: 2etan θ

dy
If y = 2 e tan θ , then = ( 2e tan θ ) sec 2 θ = 2sec 2 θ e tan θ
dθ

π
10. Differentiate: θ sin(θ – ) with respect to θ, and evaluate, correct to 3 significant figures, when
3
π
θ=
2

 π π π
(θ ) cos  θ − 
dy
If y = θ sin  θ −  then =  + sin  θ −  (1)
 3 dθ  3  3
π d y π  π π  π π  π π π
When θ = , =   cos  −  + sin  − = cos + sin
2 dθ  2  2 3 2 3 2 6 6
= 1.360 + 0.5 = 1.86, correct to 3
significant figures

922 © 2014, John Bird

11. The extension, x metres, of an undamped vibrating spring after t seconds is given by:

x = 0.54 cos(0.3t – 0.15) + 3.2

dx
Calculate the speed of the spring, given by , when (a) t = 0, (b) t = 2 s
dt

dx
If x = 0.54 cos(0.3t – 0.15) + 3.2 then = (0.54) [– sin(0.3t – 0.15)](0.3) + 0
dt

= –0.162 sin(0.3t – 0.15)

dx
(a) When t = 0, speed of spring, = –0.162 sin(–0.15) = 0.02421 m/s = 24.21 mm/s
dt

dx
(b) When t = 2 s, speed of spring, = –0.162 sin(0.45) = –0.07046 m/s = –70.46 mm/s
dt

923 © 2014, John Bird

EXERCISE 229 Page 622

d2 y d3 y
1. If y = 3x4 + 2x3 – 3x + 2 find (a) (b)
d x2 d x3

dy d2 y
(a) If y = 3x4 + 2x3 – 3x + 2 then = 12 x3 + 6 x 2 − 3 and = 36 x 2 + 12 x
dx d x2

d3 y
(b) = 72 x + 12
d x3

2 2 1 3
2. (a) Given f(t) = t – + – t + 1 determine f ´´(t)
5 t3 t

(b) Evaluate f ´´(t) when t = 1

1
2 2 1 3 2 2 −3
(a) f(t) = t − + − t +=
1 t − t + 3t −1 − t 2 + 1
5 t3 t 5
4 1 −1
f ′(t) = t + 3t −4 − 3t −2 − t 2
5 2
4 1 −3 4 12 6 1
f ′′(t) = − 12t −5 + 6t −3 + t 2 = − + +
5 4 5 t 5 t 3
4 t3
4 12 6 1 4 1
(b) When t = 1, f ′′(t) = − + + = − 12 + 6 + = –4.95
5 (1) (1) 4 1
5 3 3 5 4

t
−
3. The charge q on the plates of a capacitor is given by q = CV e CR , where t is the time, C is the

capacitance and R the resistance. Determine (a) the rate of change of charge, which is given by

dq d2 q
, (b) the rate of change of current, which is given by
dt d t2

−
t dq  1 − CRt  V − t
(a) If q = CV e CR ,= (CV )  − e  = − e CR
dt  CR  R

dq V − t d 2 q  V  1 − t  V − t
(b) If = − e CR , = −  − e CR  = e CR
dt R d t 2  R  CR  CR 2

924 © 2014, John Bird

4. Find the second differential coefficient with respect to the variable:

(a) 3 sin 2t + cos t (b) 2 ln 4θ

dy
(a) If y = 3 sin 2t + cos t then = (3)(2 cos 2t ) − sin=
t 6 cos 2t − sin t
dx
d2 y
and = (6)(– 2 sin 2t) – cos t = – 12 sin 2t – cos t or –(12 sin 2t + cos t)
d x2

dy 1 2
(b) If y = 2 ln 4θ then = (2)  = = 2θ
−1
dx θ  θ
d2 y 2
and = −2θ −2 = −
dx 2 θ2

5. Find the second differential coefficient with respect to the variable:

(a) 2 cos2x (b) (2x – 3)4

dy
(a) If y = 2 cos 2 x , =
4 cos x(− sin x) =
−4sin x cos x
dx
d2 y
and =
(−4sin x)(− sin x) + (cos x)(−4 cos x) =
4sin 2 x − 4 cos 2 x
d x2
= 4 ( sin 2 x − cos 2 x )

dy
(b) If y = ( 2 x − 3) ,
4
=4(2 x − 3)3 (2) =8(2 x − 3)3
dx
d2 y
= 24(2 x − 3) 2 (2) = 48 ( 2 x − 3)
2
and
d x2

6. Evaluate f ´´(θ) when θ = 0 given f(θ) = 2 sec 3θ

If f(θ) = 2 sec 3θ, then f ′(θ) = 6 sec 3θ tan 3θ

and f ′′(θ) = ( 6sec 3θ )( 3sec 2 3θ ) + ( tan 3θ )(18sec 3θ tan 3θ )

= 18sec3 3θ + 18sec 3θ tan 2 3θ

18 18 tan 2 0 18 18(0)
When θ = 0, f ′′(0) = + = + = 18
cos3 0 cos 0 1 1

925 © 2014, John Bird

 d2 y dy
7. Show that the differential equation –4 + 4y = 0 is satisfied when y = xe2x
dx 2 dx

dy
If y = xe2x then =( x)(2 e 2 x ) + (e 2 x )(1) =2 x e 2 x + e 2 x
dx
d2 y
and = (2 x)(2 e 2 x ) + (e 2 x )(2) + 2 e 2 x = 4 x e 2 x + 4 e 2 x
d x2
d2 y dy
–4 + 4y = ( 4 xe 2 x + 4e 2 x ) – 4( 2 x e 2 x + e 2 x ) + 4(xe2x)
dx 2 dx

= 4 xe 2 x + 4e 2 x – 8 x e 2 x − 4 e 2 x + 4xe2x

=0

d2 y dy
8. Show that, if P and Q are constants and y = P cos(ln t) + Q sin(ln t), then t 2 +t +y=0
dt 2 dt

y = P cos(ln t) + Q sin(ln t)

dy P Q − P sin(ln t ) + Q cos(ln t )
=
− sin(ln t ) + cos(ln t ) =
dt t t t

 − P cos(ln t ) Q sin(ln t ) 
d y
2
(t ) 
 t
−
t  − [ − P sin(ln t ) + Q cos(ln t ) ] (1)
=
d t2 t2

− P cos(ln t ) − Q sin(ln t ) + P sin(ln t ) − Q cos(ln t )

=
t2

d2 y dy − P cos(ln t ) − Q sin(ln t ) + P sin(ln t ) − Q cos(ln t )

Hence, t 2 +t + y = (t 2 )
dt 2 dt t2

− P sin(ln t ) + Q cos(ln t )
+ (t) + P cos(ln t) + Q sin(ln t)
t

= − P cos(ln t ) − Q sin(ln t ) + P sin(ln t ) − Q cos(ln t ) − P sin(ln t ) + Q cos(ln t )

+ P cos(ln t) + Q sin(ln t)

=0

926 © 2014, John Bird

9. The displacement s of a mass in a vibrating system is given by: s= (1 + t ) e−ω t where ω is the
d2 s ds
natural frequency of vibration. Show that: + 2ω + ω 2s =
0
dt 2 dt

ds
If s= (1 + t ) e−ω t (1 t ) ( −ω e −ωt ) + (e −ωt )(1) =
=+ −ω e −ωt − ωt e −ωt + e −ωt
dt

d2 y
and +ω 2 e − ωt − [(ωt ) ( −ω e − ωt ) + ( e −ωt ) (ω )] − ω e −ωt
=
d x2
= ω 2 e − ωt + ( ω 2 t e − ωt ) − ( ω e − ωt ) − ω e − ωt

= ω 2 e − ωt + (ω 2t e − ωt ) − 2ω e −ωt

Hence,

d2 s ds
+ 2ω ω 2 e − ωt + (ω 2t e − ωt ) − 2ω e −ωt + 2ω[−ω e −ωt − ωt e −ωt + e −ωt ] + ω 2 [(1 + t ) e −ωt ]
+ ω 2s =
dt 2 dt

= ω 2 e − ωt + ω 2te −ωt − 2ω e −ωt − 2ω 2 e −ωt − 2ω 2te −ωt + 2ω e −ωt + ω 2 e −ωt + ω 2te −ωt

=0

927 © 2014, John Bird