Methods of differentiation 291
Thus
d y
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
= 20x 3 + 4 + x −3 −
1
2
x
−
3
2
i.e.
dy
dx
= 20x 3 + 4 +
1
x 3 −
1
2
√
x 3
Problem 6. Find the differential coefficients of
(a) y = 3 sin4x (b) f (t ) = 2 cos3t with respect to
the variable.
(a) When y = 3 sin4x then
d y
dx
= (3)(4 cos 4x)
= 12 cos 4x
(b) When f (t ) = 2 cos 3t then
f (t ) = (2)(−3 sin 3t ) =−6 sin 3t
Problem 7. Determine the derivatives of
(a) y = 3e 5x (b) f (θ) =
2
e 3θ (c) y = 6 ln2x.
(a) When y = 3e 5x then
dy
dx
= (3)(5)e 5x = 15e
5x
(b) f (θ) =
2
e 3θ = 2e −3θ , thus
f (θ) = (2)(−3)e −30 =−6e −3θ =
−6
e 3θ
(c) When y = 6 ln2x then
dy
dx
= 6
1
x
=
6
x
Problem 8. Find the gradient of the curve
y = 3x 4 − 2x 2 + 5x − 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 = 3x 4 − 2x 2 + 5x − 2
Then the gradient =
dy
dx
= 12x 3 − 4x + 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 graph y = 3x 2 − 7x + 2 where the
gradient is −1.
The gradient of the curve is given by the derivative.
When y = 3x 2 − 7x + 2 then
dy
dx
= 6x − 7
Since the gradient is −1 then 6x − 7 =−1, from which,
x = 1
When x = 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 problems on
differentiating common functions
In Problems 1 to 6 find the differential coefficients of the given functions with respect to the
variable.
1. (a) 5x 5 (b) 2.4x 3.5 (c)
1
x
(a) 25x 4 (b) 8.4x 2.5 (c) −
1
x 2
2. (a)
−4
x 2 (b) 6 (c) 2x
(a)
8
x 3 (b) 0 (c) 2
3. (a) 2
√
x (b) 3
3
√
x 5 (c)
4
√
x
(a)
1
√
x
(b) 5
3
√
x 2 (c) −
2
√
x 3
4. (a)
−3
3
√
x
(b) (x − 1) 2 (c) 2 sin 3x
⎡
⎢
⎣
(a)
1
3
√
x 4
(b) 2(x − 1)
(c) 6 cos 3x
⎤
⎥
⎦
5. (a) −4 cos 2x (b) 2e 6x (c)
3
e 5x
(a) 8 sin 2x (b) 12e 6x (c)
−15
e 5x
Thus
d y
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
= 20x 3 + 4 + x −3 −
1
2
x
−
3
2
i.e.
dy
dx
= 20x 3 + 4 +
1
x 3 −
1
2
√
x 3
Problem 6. Find the differential coefficients of
(a) y = 3 sin4x (b) f (t ) = 2 cos3t with respect to
the variable.
(a) When y = 3 sin4x then
d y
dx
= (3)(4 cos 4x)
= 12 cos 4x
(b) When f (t ) = 2 cos 3t then
f (t ) = (2)(−3 sin 3t ) =−6 sin 3t
Problem 7. Determine the derivatives of
(a) y = 3e 5x (b) f (θ) =
2
e 3θ (c) y = 6 ln2x.
(a) When y = 3e 5x then
dy
dx
= (3)(5)e 5x = 15e
5x
(b) f (θ) =
2
e 3θ = 2e −3θ , thus
f (θ) = (2)(−3)e −30 =−6e −3θ =
−6
e 3θ
(c) When y = 6 ln2x then
dy
dx
= 6
1
x
=
6
x
Problem 8. Find the gradient of the curve
y = 3x 4 − 2x 2 + 5x − 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 = 3x 4 − 2x 2 + 5x − 2
Then the gradient =
dy
dx
= 12x 3 − 4x + 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 graph y = 3x 2 − 7x + 2 where the
gradient is −1.
The gradient of the curve is given by the derivative.
When y = 3x 2 − 7x + 2 then
dy
dx
= 6x − 7
Since the gradient is −1 then 6x − 7 =−1, from which,
x = 1
When x = 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 problems on
differentiating common functions
In Problems 1 to 6 find the differential coefficients of the given functions with respect to the
variable.
1. (a) 5x 5 (b) 2.4x 3.5 (c)
1
x
(a) 25x 4 (b) 8.4x 2.5 (c) −
1
x 2
2. (a)
−4
x 2 (b) 6 (c) 2x
(a)
8
x 3 (b) 0 (c) 2
3. (a) 2
√
x (b) 3
3
√
x 5 (c)
4
√
x
(a)
1
√
x
(b) 5
3
√
x 2 (c) −
2
√
x 3
4. (a)
−3
3
√
x
(b) (x − 1) 2 (c) 2 sin 3x
⎡
⎢
⎣
(a)
1
3
√
x 4
(b) 2(x − 1)
(c) 6 cos 3x
⎤
⎥
⎦
5. (a) −4 cos 2x (b) 2e 6x (c)
3
e 5x
(a) 8 sin 2x (b) 12e 6x (c)
−15
e 5x