Higher Engineering Mathematics

328 DIFFERENTIAL CALCULUS

i.e.

dy

dx

=x

√

(x− 1 )

{

1

x(x− 1 )

−

ln(x− 1 )

x^2

}

(b) Whenx=2,

dy

dx

=^2

√

(1)

{

1

2(1)

−

ln (1)

4

}

=± 1

{

1

2

− 0

}

=±

1

2

Problem 8. Differentiatex^3 x+^2 with respect

tox.

Lety=x^3 x+^2
Taking Napierian logarithms of both sides gives:

lny=lnx^3 x+^2

i.e. lny=(3x+2) lnx, by law (iii) of Section 31.2

Differentiating each term with respect toxgives:

1

y

dy

dx

=(3x+2)

(

1

x

)

+(lnx)(3),

by the product rule.

Hence

dy

dx

=y

{

3 x+ 2

x

+3lnx

}

=x^3 x+^2

{

3 x+ 2

x

+3lnx

}

=x^3 x+^2

{

3 +

2

x

+3lnx

}

Now try the following exercise.

Exercise 134 Further problems on differen-

tiating[f(x)]xtype functions

In Problems 1 to 4, differentiate with respect tox

1.y=x^2 x [2x^2 x(1+lnx)]

2.y=(2x−1)x

[

(2x−1)x

{

2 x

2 x− 1

+ln(2x−1)

}]

3.y= x

√

(x+3)

[

√x(x+3)

{

1

x(x+3)

−

ln(x+3)

x^2

}]

4.y= 3 x^4 x+^1

[

3 x^4 x+^1

{

4 +

1

x

+4lnx

}]

5. Show that wheny= 2 xxandx=1,

dy

dx

=2.

6. Evaluate

d

dx

{√x

(x−2)

}

whenx=3.

[

1

3

]

7. Show that ify=θθ and θ=2,

dy

dθ

= 6 .77,

correct to 3 significant figures.