Higher Engineering Mathematics

294 DIFFERENTIAL CALCULUS

Problem 16. Find the derivative ofy=secax.

y=secax=

1

cosax

(i.e. a quotient). Letu=1 and

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

2 cost

The function

te^2 t

2 cost

is a quotient, whose numerator

is a product.

Letu=te^2 tandv=2 costthen

du

dt

=(t)(2e^2 t)+(e^2 t)(1) and

dv

dt

=−2 sint

Hence

dy

dx

=

v

du

dx

−u

dv

dx

v^2

=

(2 cost)[2te^2 t+e^2 t]−(te^2 t)(−2 sint)

(2 cost)^2

=

4 te^2 tcost+2e^2 tcost+ 2 te^2 tsint

4 cos^2 t

=

2e^2 t[2tcost+cost+tsint]

4 cos^2 t

i.e.

dy

dx

=

e^2 t

2 cos^2 t

(2tcost+cost+tsint)

Problem 18. Determine the gradient of the

curvey=

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

=

(2x^2 +4)(5)−(5x)(4x)

(2x^2 +4)^2

=

10 x^2 + 20 − 20 x^2

(2x^2 +4)^2

=

20 − 10 x^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 119 Further problems on differen-

tiating quotients

In Problems 1 to 5, differentiate the quotients

with respect to the variable.

1.

2 cos 3x

x^3

[

− 6

x^4

(xsin 3x+cos 3x)

]

2.

2 x

x^2 + 1

[

2(1−x^2 )

(x^2 +1)^2

]

3.

3

√

θ^3

2 sin 2θ

[

3

√

θ(3 sin 2θ− 4 θcos 2θ)

4 sin^22 θ

]

4.

ln 2t

√

t

⎡

⎢

⎣

1 −

1

2

ln 2t

√

t^3

⎤

⎥

⎦

5.

2 xe^4 x

sinx

[

2e^4 x

sin^2 x

{(1+ 4 x) sinx−xcosx}

]

6. Find the gradient of the curvey=

2 x

x^2 − 5

at

the point (2,−4). [−18]

7. Evaluate

dy

dx

atx= 2 .5, correct to 3 significant

figures, giveny=

2 x^2 + 3

ln 2x

.

[3.82]