294 Higher Engineering Mathematics
Note that the differential coefficient is not obtained by
merely differentiating each term in turn and then dividing the numerator by the denominator. The quotient
formula must be used when differentiating quotients.
Problem 15. Determine the differential
coefficient of y = tan ax.
y = tan ax =
sin ax
cos ax
. Differentiation of tan ax is thus
treated as a quotient with u = sin ax and v = cos ax
dy
dx
=
v
du
dx
− u
dv
dx
v 2
=
(cos ax)(a cos ax) − (sin ax)(−a sin ax)
(cos ax) 2
=
a cos 2 ax + a sin
2 ax
(cos ax) 2
=
a(cos 2 ax + sin
2 ax)
cos 2 ax
=
a
cos 2 ax
, sincecos 2 ax + sin 2 ax = 1
(see Chapter 15)
Hence
dy
dx
= a sec
2 ax since sec
2 ax =
1
cos 2 ax
(see
Chapter 11).
Problem 16. Find the derivative of y = sec ax.
y = sec ax =
1
cos ax
(i.e. a quotient). Let u = 1 and
v = cos ax
dy
dx
=
v
du
dx
− u
dv
dx
v 2
=
(cos ax)(0) − (1)(−a sin ax)
(cos ax) 2
=
a sin ax
cos 2 ax
= a
1
cos ax
sin ax
cos ax
i.e.
dy
dx
= a sec ax tan ax
Problem 17. Differentiate y =
t e 2t
2 cost
The function
t e
2t
2 cost
is a quotient, whose numerator is a
product.
Let u = t e
2t and v = 2 cos t then
du
dt
= (t )(2e 2t ) + (e 2t )(1) and
dv
dt
=−2 sint
Hence
dy
dx
=
v
du
dx
− u
dv
dx
v 2
=
(2 cos t )[2t e 2t + e 2t ] − (t e 2t )(−2 sint )
(2 cos t ) 2
=
4t e 2t cos t + 2e 2t cos t + 2t e 2t sin t
4 cos 2 t
=
2e 2t [2t cos t + cos t + t sin t ]
4 cos 2 t
i.e.
dy
dx
=
e 2t
2 cos 2 t
(2t cos t + cos t +t sin t)
Problem 18. Determine the gradient of the curve
y =
5x
2x 2 + 4
at the point
√
3,
√
3
2
.
Let y = 5x and v = 2x 2 + 4
dy
dx
=
v
du
dx
− u
dv
dx
v 2
=
(2x 2 + 4)(5) − (5x)(4x)
(2x 2 + 4) 2
=
10x 2 + 20 − 20x 2
(2x 2 + 4) 2
=
20 − 10x 2
(2x 2 + 4) 2
At the point
√
3,
√
3
2
, x =
√
3,
hence the gradient =
dy
dx
=
20 − 10(
√
3) 2
[2(
√
3) 2 + 4] 2
=
20 − 30
100
= −
1
10
Now try the following exercise
Exercise 117
Further problems on
differentiating quotients
In Problems 1 to 7, differentiate the quotients with
respect to the variable.
1.
sin x
x
x cos x − sin x
x 2
Note that the differential coefficient is not obtained by
merely differentiating each term in turn and then dividing the numerator by the denominator. The quotient
formula must be used when differentiating quotients.
Problem 15. Determine the differential
coefficient of y = tan ax.
y = tan ax =
sin ax
cos ax
. Differentiation of tan ax is thus
treated as a quotient with u = sin ax and v = cos ax
dy
dx
=
v
du
dx
− u
dv
dx
v 2
=
(cos ax)(a cos ax) − (sin ax)(−a sin ax)
(cos ax) 2
=
a cos 2 ax + a sin
2 ax
(cos ax) 2
=
a(cos 2 ax + sin
2 ax)
cos 2 ax
=
a
cos 2 ax
, sincecos 2 ax + sin 2 ax = 1
(see Chapter 15)
Hence
dy
dx
= a sec
2 ax since sec
2 ax =
1
cos 2 ax
(see
Chapter 11).
Problem 16. Find the derivative of y = sec ax.
y = sec ax =
1
cos ax
(i.e. a quotient). Let u = 1 and
v = cos ax
dy
dx
=
v
du
dx
− u
dv
dx
v 2
=
(cos ax)(0) − (1)(−a sin ax)
(cos ax) 2
=
a sin ax
cos 2 ax
= a
1
cos ax
sin ax
cos ax
i.e.
dy
dx
= a sec ax tan ax
Problem 17. Differentiate y =
t e 2t
2 cost
The function
t e
2t
2 cost
is a quotient, whose numerator is a
product.
Let u = t e
2t and v = 2 cos t then
du
dt
= (t )(2e 2t ) + (e 2t )(1) and
dv
dt
=−2 sint
Hence
dy
dx
=
v
du
dx
− u
dv
dx
v 2
=
(2 cos t )[2t e 2t + e 2t ] − (t e 2t )(−2 sint )
(2 cos t ) 2
=
4t e 2t cos t + 2e 2t cos t + 2t e 2t sin t
4 cos 2 t
=
2e 2t [2t cos t + cos t + t sin t ]
4 cos 2 t
i.e.
dy
dx
=
e 2t
2 cos 2 t
(2t cos t + cos t +t sin t)
Problem 18. Determine the gradient of the curve
y =
5x
2x 2 + 4
at the point
√
3,
√
3
2
.
Let y = 5x and v = 2x 2 + 4
dy
dx
=
v
du
dx
− u
dv
dx
v 2
=
(2x 2 + 4)(5) − (5x)(4x)
(2x 2 + 4) 2
=
10x 2 + 20 − 20x 2
(2x 2 + 4) 2
=
20 − 10x 2
(2x 2 + 4) 2
At the point
√
3,
√
3
2
, x =
√
3,
hence the gradient =
dy
dx
=
20 − 10(
√
3) 2
[2(
√
3) 2 + 4] 2
=
20 − 30
100
= −
1
10
Now try the following exercise
Exercise 117
Further problems on
differentiating quotients
In Problems 1 to 7, differentiate the quotients with
respect to the variable.
1.
sin x
x
x cos x − sin x
x 2