The Chemistry Maths Book, Second Edition

4.6 Differentiation by rule 105

EXAMPLE 4.12Product rule

The function

y 1 = 1 (2x 1 + 13 x

2

) 1 sin 1 x

is easily differentiated only by means of the product rule. Let y 1 = 1 uv where

u 1 = 1 (2x 1 + 13 x

2

)andv 1 = 1 sin 1 x. Then

= 1 (2x 1 + 13 x

2

) 1 cos 1 x 1 + 1 (2 1 + 16 x) 1 sin 1 x

0 Exercises 29–32

The quotient rule

By Rule 4 in Table 4.3,

EXAMPLE 4.13Differentiate

Lety 1 = 1 u 2 vwhereu 1 = 1 (2x 1 + 13 x

2

)andv 1 = 1 (5 1 + 17 x

3

). Then

0 Exercises 33–36

The chain rule (function of a function)

The polynomial

y 1 = 1 f(x) 1 = 1 (2x

2

1 − 1 1)

3

=

++−+

• 
  

()()( )()

()

57 26 2 3 21

57

322

32

xxxxx

x

dy

dx

x

d

dx

xx xx

d

dx

=+ + − + +x



()()()()57 23 23 57

3223











()57+

32

x

y

xx

x

=

• 
  

• 
  

23

57

2

3

d

dx

udu

dx

u

d

v dx

v

v

v











=−













2

=+()sinsin()23+ 23 +

22

xx

d

dx

xx

d

dx

xx

dy

dx

u

d

dx

du

dx

=+

v

v