Methods of differentiation 291
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=3sin4x(b)f(t)=2cos3twith respect to
the variable.
(a) Wheny=3sin4xthen
dy
dx
=( 3 )(4cos4x)
=12cos4x
(b) When f(t)=2cos3tthen
f′(t)=( 2 )(−3sin3t)=−6sin3t
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
(c) Wheny=6ln2xthen
dy
dx
= 6
(
1
x
)
=
6
x
Problem 8. Find the gradient of the curve
y= 3 x^4 − 2 x^2 + 5 x−2 at the points (0,−2)
and (1, 4).
The gradient of a curve at a given point is given by
the corresponding value of the derivative. Thus, since
y= 3 x^4 − 2 x^2 + 5 x− 2
Then the gradient=
dy
dx
= 12 x^3 − 4 x+ 5
At the point (0,−2),x= 0
Thus the gradient= 12 ( 0 )^3 − 4 ( 0 )+ 5 = 5
At the point (1, 4),x= 1
Thus the gradient= 12 ( 1 )^3 − 4 ( 1 )+ 5 = 13.
Problem 9. Determine the co-ordinates of the
point on the graphy= 3 x^2 − 7 x+2wherethe
gradient is−1.
The gradient of the curve is given by the derivative.
Wheny= 3 x^2 − 7 x+2then
dy
dx
= 6 x− 7
Since the gradient is−1then6x− 7 =−1, from which,
x= 1
Whenx=1,y= 3 ( 1 )^2 − 7 ( 1 )+ 2 =− 2
Hence the gradient is−1 at the point (1,−2).
Now try the following exercise
Exercise 115 Further problemson
differentiating common functions
In Problems 1 to 6 find the differential coeffi-
cients of the given functions with respect to the
variable.
- (a) 5x^5 (b) 2. 4 x^3.^5 (c)
1
x
[
(a) 25 x^4 (b) 8. 4 x^2.^5 (c)−
1
x^2
]
- (a)
− 4
x^2
(b) 6 (c) 2x
[
(a)
8
x^3
(b) 0 (c) 2
]
- (a) 2
√
x(b) 3
√ 3
x^5 (c)
4
√
x
[
(a)
1
√
x
(b) 5
√ 3
x^2 (c)−
2
√
x^3
]
- (a)
− 3
√ (^3) x (b)(x− 1 )
(^2) (c) 2sin3x
⎡
⎢
⎣
(a)
1
√ (^3) x 4 (b) 2(x− 1 )
(c) 6cos3x
⎤
⎥
⎦
- (a)−4cos2x(b) 2e^6 x(c)
3
e^5 x
[
(a) 8sin2x(b) 12e^6 x(c)
− 15
e^5 x
]