Understanding Engineering Mathematics

So by product rule, with sayu=x^2 ,v=cosx,

dy

dx

=

d(uv)

dx

=

d

dx

(x^2 cosx)

=

vdu

dx

+

udv

dx

=

d(x^2 )

dx

cosx+x^2

d

dx

(cosx)

= 2 xcosx+x^2 (−sinx)

= 2 xcosx−x^2 sinx

(iii) y=

x− 1

x+ 1

We can treat this as a quotienty=

u

v

and use the quotient rule,or

treat as a product:

y=(x− 1 )×

(

1

x+ 1

)

so

dy

dx

= 1 ×

1

x+ 1

+(x− 1 )

(

−

1

(x+ 1 )^2

)

=

x+ 1 −(x− 1 )

(x+ 1 )^2

=

2

(x+ 1 )^2

(iv)y=cos(x^2 + 1 )

This is a function(cos)of a function(x^2 + 1 )(97

➤

), so we can

use the function of a function rule:

dy

dx

=

dy

du

du

dx

(

u=x^2 + 1

y=cosu

)

giving

dy

dx

=−sin(x^2 + 1 )×( 2 x)

=− 2 xsin(x^2 + 1 )

(v)y=ln 3x

Again, the function of a function rule can be used:

dy

dx

=

1

3 x

× 3 =

1

x

Orthis can be simplified using rules of logarithms (131

➤

)

y=ln 3x=ln 3+lnx

Therefore

dy

dx

= 0 +

1

x

as before