Download Chain Rule

Lesson-5
Introduction to Calculus
Chain Rule
The chain rule applies to procedures when one of the functions is a
composition of other function. A function of a function can be regarded as
a function within a function, such as, for example,
y = (x2 + 1)8
It is inconvenient, and sometimes impossible, to expand such functions
into polynomials.
The way to tackle the problem is to substitute u for the inner function;
that is, substitute
u = x2 + 1
(5.1)
so that y = u8
(5.2)
As x increases to x + x there will be corresponding increases of u and
y in u and y respectively. or  means multiplication. Since these finite
quantities it is possible to write down:
dy
dy
du


dx
dx
du
or better in different order as
y
y
u
u

1

u
 x (because u
x
)
Both y and u are dependent variables as x  0 and
u  0 , then
y
dy

x
dx
y
dy

u
du
u
du

x
dx
Therefore, in the limit this becomes the chain rule
dy
dy
du


dx
du
dx
(5.3)
Returning to the function y = (x2 + 1)8 and
substitution u = x2 + 1
du
= 2x
dx
and
dy
= 8x7
du
so that
dy
dy
du
= 8u7*2x = 8(x2 + 1)2x = 16x(x2 + 1)7


dx
du
dx
Example 1
Differentiate y = (x2 - 6x + 1)1/2
Solution:
let u = x2 - 6x + 1 so that
du
= 2x - 6
dx
then y =u1/2
dy
= 1/2u-1/2
du
and
dy
dy
du
= 1/2u-1/2(2x – 6) =


dx
du
dx
½(x2 – 6x + 1)-1/2(2x – 6) if we taking 2 in front of
second bracket, we obtain:
(x – 3)(x2 – 6x + 1)-1/2
(because ½ and 2 cancel
each other)
Differentiation of a function of a function may be carried out without
going through this substitution as procedure each time. The method can
be summarised as shown below:
a)
Differentiate the first function as a simple variable.
b)
Differentiate the internal function with respect to x.
c)
dy
= product (multiplication) of (a) and (b).
dx
Very often the chain rule involves more than two functions. In this
case procedure is similar to describe above, but more function are
involved.
EXERCISES:
Differentiate the following with respect to appropriate variable.
1.
(2x + 3)4
2.
(7x2 – 3)6
3.
(-3x1/2 + 4)3
4.
(4t – 3)1/2
5.
(6t2 - 62 )3/4
6.
(Q - 1 )2
t
1
7.
3
x 3
Q
1
8.
t  3t  4
2
2
9.
(3x + 2)6
10.
(5 – x2)3
11.
3(x5 – 8x2 + x)100
12.
(x2 – 2)-3
13.
(2x2 – 3x + 1)-10/3
14.
5x 2  x
15.
4
2x  1
16.
2
1
( x  3x) 2
17.
6
2x  x  1
18.
19.
2( 8x  1 )-1
20.
2
( x 3  1) 2
5
2
3
7x +
3
7x
Answers:
1.
8(2x + 3)3
2.
4.
(4t - 3)-1/2
5.
84x(7x2 – 3)5
3.
-9/2x-1/2(3x1/2 + 4)2
9 (t + 1 / t3) (6t2 - 6/t2)-1/4
6.
(Q - 1/Q)(1 + 1/Q2)
7.
-2/3x(x2 – 3)-4/3
8.
-1/2(2t – 3)(t2 – 3t + 4)-3/2
9.
18 (3x + 2)5
10.
-6x(5 – x2)2
11.
12.
-6x(x2 – 2)-4
13.
300 (x5 – 8x2 +x)99(5x4 - 16x +1)
-10/3(4x - 3)(2x2 3x + 1)-13/3
14. 1/2(10x - 1)(x2 –x)-1/2
15.
½(2x – 1)-3/4
-6(x – 1)(2x2 – x + 1)-2
16.
12/5x2(x3 + 1)-3/5 17.
18.
-2(2x - 3)(x2 - 3x)-3
20.
2/3(7)1/3 x-2/3 = 14/9x-2/3
19.
-8(8x - 1)-3/2