294 Higher Engineering Mathematics
Note that the differential coefficient isnotobtained by
merely differentiating each term in turn and then divid-
ing the numerator by the denominator. The quotient
formulamustbe used when differentiating quotients.
Problem 15. Determine the differential
coefficient ofy=tanax.
y=tanax=
sinax
cosax
. Differentiation of tanax is thus
treated as a quotient withu=sinaxandv=cosax
dy
dx
=
v
du
dx
−u
dv
dx
v^2
=
(cosax)(acosax)−(sinax)(−asinax)
(cosax)^2
=
acos^2 ax+asin^2 ax
(cosax)^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
=asec^2 ax since sec^2 ax=
1
cos^2 ax
(see
Chapter 11).
Problem 16. Find the derivative ofy=secax.
y=secax=
1
cosax
(i.e. a quotient). Letu=1and
v=cosax
dy
dx
=
v
du
dx
−u
dv
dx
v^2
=
(cosax)( 0 )−( 1 )(−asinax)
(cosax)^2
=
asinax
cos^2 ax
=a
(
1
cosax
)(
sinax
cosax
)
i.e.
dy
dx
=asecaxtanax
Problem 17. Differentiatey=
te^2 t
2cost
The function
te^2 t
2cost
is a quotient, whose numerator is a
product.
Letu=te^2 tandv=2costthen
du
dt
=(t)(2e^2 t)+(e^2 t)( 1 )and
dv
dt
=−2sint
Hence
dy
dx
=
v
du
dx
−u
dv
dx
v^2
=
(2cost)[2te^2 t+e^2 t]−(te^2 t)(−2sint)
(2cost)^2
=
4 te^2 tcost+2e^2 tcost+ 2 te^2 tsint
4cos^2 t
=
2e^2 t[2tcost+cost+tsint]
4cos^2 t
i.e.
dy
dx
=
e^2 t
2cos^2 t
(2tcost+cost+tsint)
Problem 18. Determine the gradient of the curve
y=
5 x
2 x^2 + 4
at the point
(
√
3 ,
√
3
2
)
.
Lety= 5 xandv= 2 x^2 + 4
dy
dx
=
v
du
dx
−u
dv
dx
v^2
=
( 2 x^2 + 4 )( 5 )−( 5 x)( 4 x)
( 2 x^2 + 4 )^2
=
10 x^2 + 20 − 20 x^2
( 2 x^2 + 4 )^2
=
20 − 10 x^2
( 2 x^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 problemson
differentiating quotients
In Problems 1 to 7, differentiate the quotients with
respect to the variable.
1.
sinx
x
[
xcosx−sinx
x^2
]