Cambridge International AS and A Level Mathematics Pure Mathematics 1

Differentiation

168

P1^

5

In the limit, as δx → 0,

δ

δ

δ

δ

δ

δ

y

x

y

x

y

u

yu

x

u

→→ → x

d

d

d

duand

d

, d

and so the relationship above becomes

d

d

d

d

d

d

y

x

y

u

u

x

=×.

This is known as the chain rule.

EXAMPLE 5.18 Differentiate y = (x^2 + 1)

(^12)
.
SOLUTION
As you saw earlier, you can break down this expression as follows.
y = u
(^12)
, u = x^2 + 1
Differentiating these gives
d
d
y
u
u
x

==

+

1

2

1

2 1

(^12)
2

• 
  

and

d

d

u

x=2.x^

By the chain rule

d

d

d

d

d

d

y

x

y

u

u

x

x

x

x

x

=×

=

+

×

=

+

1

21

2

1

2

2

! Notice that the answer must be given in terms of the same variables as the

question, in this case x and y. The variable u was your invention and so should

not appear in the answer.

You can see that effectively you have made a substitution, in this case

u = x^2 + 1. This transformed the problem into one that could easily be solved.

Note

Notice that the substitution gave you two functions that you could differentiate.

Some substitutions would not have worked. For example, the substitution u = x^2 ,

would give you

y = (u + 1)

(^12)
and u = x^2.
You would still not be able to differentiate y, so you would have gained nothing.