9.2.4 To determine the first derivative of a product of 2 polynomial.

Example Exercise 1 y= x ( x3+1)

u=x , v =x3+1

=

= x(3x2)+(x3+1)(1)

=3 x3+x3+1 = 4x3 +1

a y= x ( x4+2)

[ 5x4+2]

y= ( x 5+1)x

[ 6x5+1]

c y= ( x3-1)x

[4x3 -1]

2 y= 2x2 ( x3+1)

u=2x2 , v =x3+1

=

= 2x2(3x2)+(x3+1)(4x)

=6 x4+4x4 +4x = 10x4 +4x

a y= 3x2 ( x3-1)

[ 15x4-6x]

b y= ( x3-1)(5x2)

[ 25x4-10x]

c y= ( x3-1)(-4x2)

[ -20x4+8x]

3 f(x)= (x+1) ( x3+1)

u=x+1 , v =x3+1

=

= (x+1)(3x2)+(x3+1)(1)

=3x3+3x2 +x3+1

= 4x3+3x2+1

a f(x)= (x-1) ( 1+x3)

[4 x3-3x2+1]

b f(x)= (1-x) ( x3+2)

[ -4x3+3x2-2 ]

c f(x)= (2-x) ( x3+3)

[ -4x3+6x2-3]

Differentiation 4

4y= (x+1) ( +1)

u=x+1 , v = +1

=

= (x+1)( )+

( )(1)

=

=

ay= (x-1) ( +1)

[ ]

by= (2x+1) ( +1)

[2x+ ]

cy= (3-2x) ( 2- )

[2x-5 ]

5y= (x2+1) ( +1)

u=x2+1 , v = +1

=

= (x2+1)( )+

( )(2x)

= +

=

ay= (x4+1) ( +1)

[ ]

b y= (2+x2) ( -1)

[

cy= (3-x3) ( +1)

[ - ]

Differentiation 5