290 DIFFERENTIAL CALCULUS

(a) Since y= 12 x^3 , a= 12 and n= 3 thus
dy
dx

=(12)(3)x^3 −^1 = 36 x^2

(b)y=

12

x^3

is rewritten in the standardaxnform as

y= 12 x−^3 and in the general rulea=12 and

n=−3.

Thus

dy

dx

=(12)(−3)x−^3 −^1 =− 36 x−^4 =−

36

x^4

Problem 3. Differentiate (a)y=6 (b)y= 6 x.

(a)y=6 may be written as y= 6 x^0 , i.e. in the
general rulea=6 andn=0.

Hence

dy

dx

=(6)(0)x^0 −^1 = 0

In general,the differential coefficient of a
constant is always zero.
(b) Sincey= 6 x, in the general rulea=6 andn=1.

Hence

dy

dx

=(6)(1)x^1 −^1 = 6 x^0 = 6

In general, the differential coefficient of kx,

wherekis a constant, is alwaysk.

Problem 4. Find the derivatives of

(a)y= 3

√

x (b)y=

5

√ 3

x^4

.

(a)y= 3

√

xis rewritten in the standard differential

form asy= 3 x

1

(^2).
In the general rule,a=3 andn=
1
2
Thus
dy
dx
=(3)
(
1
2
)
x
1
2 −^1 =
3
2
x−
1
2

3
2 x
1
2

3
2
√
x
(b)y=
5
√ 3
x^4

5
x
4
3
= 5 x−
4
(^3) in the standard differen-
tial form.
In the general rule,a=5 andn=−^43
Thus
dy
dx
=(5)
(
−
4
3
)
x−
4
3 −^1 =
− 20
3
x−
7
3

− 20
3 x
7
3

− 20
3
√ 3
x^7
Problem 5. Differentiate, with respect tox,
y= 5 x^4 + 4 x−
1
2 x^2

• 1
  √
  x
  −3.
  y= 5 x^4 + 4 x−
  1
  2 x^2


• 
  

  1
  √
  x
  −3 is rewritten as
  y= 5 x^4 + 4 x−
  1
  2
  x−^2 +x−
  1
  (^2) − 3
  When differentiating a sum, each term is differenti-
  ated in turn.
  Thus
  dy
  dx
  =(5)(4)x^4 −^1 +(4)(1)x^1 −^1 −
  1
  2
  (−2)x−^2 −^1
  +(1)
  (
  −
  1
  2
  )
  x−
  1
  2 −^1 − 0
  = 20 x^3 + 4 +x−^3 −
  1
  2
  x−
  3
  2
  i.e.
  dy
  dx
  = 20 x^3 + 4 +
  1
  x^3
  −
  1
  2
  √
  x^3
  Problem 6. Find the differential coefficients
  of (a)y=3 sin 4x(b)f(t)=2 cos 3twith respect
  to the variable.
  (a) Wheny=3 sin 4xthen
  dy
  dx
  =(3)(4 cos 4x)
  =12 cos 4x
  (b) When f(t)=2 cos 3tthen
  f′(t)=(2)(−3 sin 3t)=−6 sin 3t
  Problem 7. Determine the derivatives of
  (a)y=3e^5 x(b)f(θ)=
  2
  e^3 θ
  (c)y=6ln2x.
  (a) Wheny=3e^5 xthen
  dy
  dx
  =(3)(5)e^5 x=15e^5 x
  (b) f(θ)=
  2
  e^3 θ
  =2e−^3 θ, thus
  f′(θ)=(2)(−3)e−^30 =−6e−^3 θ=
  − 6
  e^3 θ