316 Basic Engineering Mathematics

When differentiating, results can be expressed in a

number of ways. For example,

(a) ify= 3 x^2 then

dy

dx

= 6 x

(b) iff(x)= 3 x^2 thenf′(x)= 6 x

(c) the differential coefficient of 3x^2 is 6x

(d) the derivative of 3x^2 is 6x

(e)

d

dx

( 3 x^2 )= 6 x

34.5.1 Revision of some laws of indices

1

xa

=x−a For example,

1

x^2

=x−^2 andx−^5 =

1

x^5

√

x=x

1

(^2) For example,
√
5 = 5
1
(^2) and
16
1
(^2) =
√
16 =±4and
1
√
x

1
x
1
2
=x−
1
2
√a
xb=x
b
a For example,^3
√
x^5 =x
5
(^3) andx
4
(^3) =^3
√
x^4
and
1
√ (^3) x 7 =
1
x
73 =x
−^73
x^0 = 1 For example, 7^0 =1 and 43. 50 = 1
Here are some worked problems to demonstrate the
general rule for differentiatingy=axn.
Problem 4. Differentiate the following with
respect tox:y= 4 x^7
Comparingy= 4 x^7 withy=axnshows thata=4and
n=7. Using the general rule,
dy
dx
=anxn−^1 =( 4 )( 7 )x^7 −^1 = 28 x^6
Problem 5. Differentiate the following with
respect tox:y=
3
x^2
y=
3
x^2
= 3 x−^2 , hencea=3andn=−2 in the general
rule.
dy
dx
=anxn−^1 =( 3 )(− 2 )x−^2 −^1 =− 6 x−^3 =−
6
x^3
Problem 6. Differentiate the following with
respect tox:y= 5
√
x
y= 5
√
x= 5 x
1
(^2) , hence a=5andn=
1
2
in the
general rule.
dy
dx
=anxn−^1 =( 5 )
(
1
2
)
x
1
2 −^1

5
2
x−
1
(^2) =
5
2 x
1
2

5
2
√
x
Problem 7. Differentiatey= 4
y=4 may bewritten asy= 4 x^0 ; i.e., in the general rule
a=4andn=0. Hence,
dy
dx
=( 4 )( 0 )x^0 −^1 = 0
The equationy=4representsastraight horizontal
lineand the gradient of a horizontal line is zero, hence
the result could have been determined on inspection.
In general,the differential coefficient of a constant is
always zero.
Problem 8. Differentiatey= 7 x
Sincey= 7 x,i.e.y= 7 x^1 , in the general rulea=7and
n=1. Hence,
dy
dx
=( 7 )( 1 )x^1 −^1 = 7 x^0 = 7 sincex^0 = 1
The gradient of the line y= 7 x is 7 (from
y=mx+c), hence the result could have been obtained
by inspection.In general,the differential coefficient of
kx,wherekis a constant, is alwaysk.
Problem 9. Find the differential coefficient of
y=
2
3
x^4 −
4
x^3

• 9
  y=
  2
  3
  x^4 −
  4
  x^3


• 
  

  9
  i.e. y=
  2
  3
  x^4 − 4 x−^3 + 9
  dy
  dx

  
  

  (
  2
  3
  )
  ( 4 )x^4 −^1 −( 4 )(− 3 )x−^3 −^1 + 0

  
  

  8
  3
  x^3 + 12 x−^4