316 Basic Engineering Mathematics
When differentiating, results can be expressed in a
number of ways. For example,
(a) if y = 3x 2 then
dy
dx
= 6x
(b) if f (x) = 3x 2 then f (x) = 6x
(c) the differential coefficient of 3x 2 is 6x
(d) the derivative of 3x
2 is 6x
(e)
d
dx
(3x 2 ) = 6x
34.5.1 Revision of some laws of indices
1
x a = x
−a
For example,
1
x 2 = x
−2 and x
−5
=
1
x 5
√ x = x
1
2
For example,
√
5 = 5
1
2 and
16
1
2 =
√
16 = ±4 and
1
√
x
=
1
x
1
2
= x
−
1
2
a
√
x b = x
b
a
For example,
3
√
x 5 = x
5
3 and x
4
3 =
3
√
x 4
and
1
3
√
x 7
=
1
x
7
3
= x
−
7
3
x 0 = 1
For example, 7 0 = 1 and 43.5 0 = 1
Here are some worked problems to demonstrate the
general rule for differentiating y = ax n .
Problem 4. Differentiate the following with
respect to x: y = 4x 7
Comparing y = 4x 7 with y = ax n shows that a = 4 and
n = 7. Using the general rule,
dy
dx
= anx
n−1
= (4)(7)x
7−1
= 28x
6
Problem 5. Differentiate the following with
respect to x: y =
3
x 2
y =
3
x 2 = 3x −2 , hence a = 3 and n = −2 in the general
rule.
dy
dx
= anx
n−1
= (3)(−2)x
−2−1
= −6x
−3
= −
6
x 3
Problem 6. Differentiate the following with
respect to x: y = 5
√
x
y = 5
√
x = 5x
1
2 , hence a = 5 and n =
1
2
in the
general rule.
dy
dx
= anx
n−1
= (5)
1
2
x
1
2 −1
=
5
2
x
−
1
2 =
5
2x
1
2
=
5
2
√ x
Problem 7. Differentiate y = 4
y = 4 may be written as y = 4x
0 ; i.e., in the general rule
a = 4 and n = 0. Hence,
dy
dx
= (4)(0)x
0−1
= 0
The equation y = 4 represents a straight horizontal
line and 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. Differentiate y = 7x
Since y = 7x, i.e. y = 7x
1 , in the general rule a = 7 and
n = 1. Hence,
dy
dx
= (7)(1)x
1−1
= 7x
0
= 7
since x
0
= 1
The gradient of the line y = 7x is 7 (from
y = mx + c), hence the result could have been obtained
by inspection. In general, the differential coefficient of
kx, where k is a constant, is always k.
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
− 4x
−3
+ 9
dy
dx
=
2
3
(4)x
4−1
− (4)(−3)x
−3−1
+ 0
=
8
3
x
3
+ 12x
−4
When differentiating, results can be expressed in a
number of ways. For example,
(a) if y = 3x 2 then
dy
dx
= 6x
(b) if f (x) = 3x 2 then f (x) = 6x
(c) the differential coefficient of 3x 2 is 6x
(d) the derivative of 3x
2 is 6x
(e)
d
dx
(3x 2 ) = 6x
34.5.1 Revision of some laws of indices
1
x a = x
−a
For example,
1
x 2 = x
−2 and x
−5
=
1
x 5
√ x = x
1
2
For example,
√
5 = 5
1
2 and
16
1
2 =
√
16 = ±4 and
1
√
x
=
1
x
1
2
= x
−
1
2
a
√
x b = x
b
a
For example,
3
√
x 5 = x
5
3 and x
4
3 =
3
√
x 4
and
1
3
√
x 7
=
1
x
7
3
= x
−
7
3
x 0 = 1
For example, 7 0 = 1 and 43.5 0 = 1
Here are some worked problems to demonstrate the
general rule for differentiating y = ax n .
Problem 4. Differentiate the following with
respect to x: y = 4x 7
Comparing y = 4x 7 with y = ax n shows that a = 4 and
n = 7. Using the general rule,
dy
dx
= anx
n−1
= (4)(7)x
7−1
= 28x
6
Problem 5. Differentiate the following with
respect to x: y =
3
x 2
y =
3
x 2 = 3x −2 , hence a = 3 and n = −2 in the general
rule.
dy
dx
= anx
n−1
= (3)(−2)x
−2−1
= −6x
−3
= −
6
x 3
Problem 6. Differentiate the following with
respect to x: y = 5
√
x
y = 5
√
x = 5x
1
2 , hence a = 5 and n =
1
2
in the
general rule.
dy
dx
= anx
n−1
= (5)
1
2
x
1
2 −1
=
5
2
x
−
1
2 =
5
2x
1
2
=
5
2
√ x
Problem 7. Differentiate y = 4
y = 4 may be written as y = 4x
0 ; i.e., in the general rule
a = 4 and n = 0. Hence,
dy
dx
= (4)(0)x
0−1
= 0
The equation y = 4 represents a straight horizontal
line and 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. Differentiate y = 7x
Since y = 7x, i.e. y = 7x
1 , in the general rule a = 7 and
n = 1. Hence,
dy
dx
= (7)(1)x
1−1
= 7x
0
= 7
since x
0
= 1
The gradient of the line y = 7x is 7 (from
y = mx + c), hence the result could have been obtained
by inspection. In general, the differential coefficient of
kx, where k is a constant, is always k.
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
− 4x
−3
+ 9
dy
dx
=
2
3
(4)x
4−1
− (4)(−3)x
−3−1
+ 0
=
8
3
x
3
+ 12x
−4