Higher Engineering Mathematics

326 DIFFERENTIAL CALCULUS

Whenx=3,

dy

dx

=

√

(1)^3

(4)^2 (5)

(

3

2

−

2

4

−

2

5

)

=±

1

80

(

3

5

)

=±

3

400

or± 0. 0075

Problem 3. Giveny=

3e^2 θsec 2θ

√

(θ−2)

determine

dy

dθ

Using logarithmic differentiation and following the

procedure gives:

(i) Since y=

3e^2 θsec 2θ

√

(θ−2)

then lny=ln

{

3e^2 θsec 2θ

√

(θ−2)

}

=ln

{

3 e^2 θsec 2θ

(θ−2)

1

2

}

(ii) lny=ln 3e^2 θ+ln sec 2θ−ln(θ−2)

1

2

i.e. lny=ln 3+ln e^2 θ+ln sec 2θ

−^12 ln(θ−2)

i.e. lny=ln 3+ 2 θ+ln sec 2θ−^12 ln(θ−2)

(iii) Differentiating with respect toθgives:

1

y

dy

dθ

= 0 + 2 +

2 sec 2θtan 2θ

sec 2θ

−

1

2

(θ−2)

from equations (1) and (2)

(iv) Rearranging gives:

dy

dθ

=y

{

2 +2 tan 2θ−

1

2(θ−2)

}

(v) Substituting forygives:

dy

dθ

=

3e^2 θsec 2θ

√

(θ− 2 )

{

2 +2 tan 2θ−

1

2 (θ− 2 )

}

Problem 4. Differentiatey =

x^3 ln 2x

exsinx

with

respect tox.

Using logarithmic differentiation and following the

procedure gives:

(i) lny=ln

{

x^3 ln 2x

exsinx

}

(ii) lny=lnx^3 +ln(ln 2x)−ln(ex)−ln(sinx)

i.e. lny=3lnx+ln(ln 2x)−x−ln(sinx)

(iii)

1

y

dy

dx

=

3

x

+

1

x

ln 2x

− 1 −

cosx

sinx

(iv)

dy

dx

=y

{

3

x

+

1

xln 2x

− 1 −cotx

}

(v)

dy

dx

=

x^3 ln 2x

exsinx

{

3

x

+

1

xln 2x

− 1 −cotx

}

Now try the following exercise.

Exercise 133 Further problems on differen-

tiating logarithmic functions

In Problems 1 to 6, use logarithmic differenti-

ation to differentiate the given functions with

respect to the variable.

1.y=

(x−2)(x+1)

(x−1)(x+3)

⎡

⎢

⎢

⎣

(x−2)(x+1)

(x−1)(x+3)

{

1

(x−2)

+

1

(x+1)

−

1

(x−1)

−

1

(x+3)

}

⎤

⎥

⎥

⎦

2.y=

(x+1)(2x+1)^3

(x−3)^2 (x+2)^4

⎡

⎢

⎢

⎣

(x+1)(2x+1)^3

(x−3)^2 (x+2)^4

{

1

(x+1)

+

6

(2x+1)

−

2

(x−3)

−

4

(x+2)

}

⎤

⎥

⎥

⎦

3.y=

(2x−1)

√

(x+2)

(x−3)

√

(x+1)^3

⎡

⎢

⎢

⎢

⎣

(2x−1)

√

(x+2)

(x−3)

√

(x+1)^3

{

2

(2x−1)

+

1

2(x+2)

−

1

(x−3)

−

3

2(x+1)

}

⎤

⎥

⎥

⎥

⎦