292 Higher Engineering Mathematics
6. (a) 4 ln 9x (b)
e x − e −x
2
(c)
1 −
√
x
x
⎡
⎢
⎢
⎣
(a)
4
x
(b)
e x + e −x
2
(c)
−1
x 2 +
1
2
√
x 3
⎤
⎥
⎥
⎦
7. Find the gradient of the curve y = 2t 4 +
3t 3 − t + 4 at the points (0, 4) and (1, 8).
[−1, 16]
8. Find the co-ordinates of the point on the
graph y = 5x 2 − 3x + 1 where the gradient
is 2.
1
2 ,
3
4
9. (a) Differentiate y =
2
θ 2 + 2 ln2θ −
2 (cos 5θ + 3 sin 2θ)−
2
e 3θ
(b) Evaluate
dy
dθ
in part (a) when θ =
π
2
,
correct to 4 significant figures.
⎡
⎢
⎢
⎢
⎣
(a)
−4
θ 3 +
2
θ
+ 10 sin 5θ
−12 cos 2θ +
6
e 3θ
(b) 22.30
⎤
⎥
⎥
⎥
⎦
10. Evaluate
ds
dt
, correct to 3 significant figures,
when t =
π
6
given
s = 3 sint − 3 +
√
t.
[3.29]
27.5 Differentiation of a product
When y = uv, and u and v are both functions of x,
then
dy
dx
= u
dv
dx
+ v
du
dx
This is known as the product rule.
Problem 10. Find the differential coefficient of
y = 3x 2 sin 2x.
3x 2 sin 2x is a product of two terms 3x 2 and sin 2x
Let u = 3x 2 and v = sin 2x
Using the product rule:
dy
dx
= u
dv
dx
+
v
du
dx
↓ ↓
↓
↓
gives:
dy
dx
= (3x 2 )(2 cos 2x) + (sin 2x)(6x)
i.e.
dy
dx
= 6x 2 cos 2x + 6x sin 2x
= 6x(xcos 2x +sin 2x)
Note that the differential coefficient of a product is
not obtained by merely differentiating each term and
multiplying the two answers together. The product rule
formula must be used when differentiating products.
Problem 11. Find the rate of change of y with
respect to x given y = 3
√
x ln 2x.
The rate of change of y with respect to x is given by
dy
dx
y = 3
√
x ln 2x = 3x
1
2 ln 2x, which is a product.
Let u = 3x
1
2 and v = ln 2x
Then
dy
dx
= u
dv
dx
+ v
du
dx
↓
↓
↓
↓
=
3x
1
2
1
x
+ (ln 2x)
3
1
2
x
1
2 −1
= 3x
1
2 −1 + (ln 2x)
3
2
x
−
1
2
= 3x
−
1
2
1 +
1
2
ln 2x
i.e.
dy
dx
=
3
√
x
1 +
1
2
ln 2x
Problem 12. Differentiate y = x 3 cos 3x ln x.
Let u = x 3 cos 3x (i.e. a product) and v = ln x
Then
dy
dx
= u
dv
dx
+ v
du
dx
where
du
dx
= (x 3 )(−3 sin 3x) + (cos 3x)(3x 2 )
and
dv
dx
=
1
x
6. (a) 4 ln 9x (b)
e x − e −x
2
(c)
1 −
√
x
x
⎡
⎢
⎢
⎣
(a)
4
x
(b)
e x + e −x
2
(c)
−1
x 2 +
1
2
√
x 3
⎤
⎥
⎥
⎦
7. Find the gradient of the curve y = 2t 4 +
3t 3 − t + 4 at the points (0, 4) and (1, 8).
[−1, 16]
8. Find the co-ordinates of the point on the
graph y = 5x 2 − 3x + 1 where the gradient
is 2.
1
2 ,
3
4
9. (a) Differentiate y =
2
θ 2 + 2 ln2θ −
2 (cos 5θ + 3 sin 2θ)−
2
e 3θ
(b) Evaluate
dy
dθ
in part (a) when θ =
π
2
,
correct to 4 significant figures.
⎡
⎢
⎢
⎢
⎣
(a)
−4
θ 3 +
2
θ
+ 10 sin 5θ
−12 cos 2θ +
6
e 3θ
(b) 22.30
⎤
⎥
⎥
⎥
⎦
10. Evaluate
ds
dt
, correct to 3 significant figures,
when t =
π
6
given
s = 3 sint − 3 +
√
t.
[3.29]
27.5 Differentiation of a product
When y = uv, and u and v are both functions of x,
then
dy
dx
= u
dv
dx
+ v
du
dx
This is known as the product rule.
Problem 10. Find the differential coefficient of
y = 3x 2 sin 2x.
3x 2 sin 2x is a product of two terms 3x 2 and sin 2x
Let u = 3x 2 and v = sin 2x
Using the product rule:
dy
dx
= u
dv
dx
+
v
du
dx
↓ ↓
↓
↓
gives:
dy
dx
= (3x 2 )(2 cos 2x) + (sin 2x)(6x)
i.e.
dy
dx
= 6x 2 cos 2x + 6x sin 2x
= 6x(xcos 2x +sin 2x)
Note that the differential coefficient of a product is
not obtained by merely differentiating each term and
multiplying the two answers together. The product rule
formula must be used when differentiating products.
Problem 11. Find the rate of change of y with
respect to x given y = 3
√
x ln 2x.
The rate of change of y with respect to x is given by
dy
dx
y = 3
√
x ln 2x = 3x
1
2 ln 2x, which is a product.
Let u = 3x
1
2 and v = ln 2x
Then
dy
dx
= u
dv
dx
+ v
du
dx
↓
↓
↓
↓
=
3x
1
2
1
x
+ (ln 2x)
3
1
2
x
1
2 −1
= 3x
1
2 −1 + (ln 2x)
3
2
x
−
1
2
= 3x
−
1
2
1 +
1
2
ln 2x
i.e.
dy
dx
=
3
√
x
1 +
1
2
ln 2x
Problem 12. Differentiate y = x 3 cos 3x ln x.
Let u = x 3 cos 3x (i.e. a product) and v = ln x
Then
dy
dx
= u
dv
dx
+ v
du
dx
where
du
dx
= (x 3 )(−3 sin 3x) + (cos 3x)(3x 2 )
and
dv
dx
=
1
x