1) The document explains the chain rule for differentiating composite functions. 2) It provides examples of applying the chain rule to functions of the form f(g(x)) where f and g are different functions. 3) The key steps are to let the inner function g(x) equal a variable u, take the derivative of the outer function f(u) with respect to u, and multiply it by the derivative of the inner function g(x) with respect to x.

1. Block 1
The Chain Rule

2. What is to be learned?
• How to differentiate a composite function

3. f(x) = (x + 5)2
= x2
+ 10x + 25
f(x) = (x + 5)7

f/
(x) = 2x + 10

4. y = (2x + 5)7
remember sin(2x + 30) = 0.8
sinA = 0.8,where A = 2x + 30let y= u7
where u = 2x + 5
dy
/du = 7u6 du
/dx = 2
dy
/dx 
dy
/du
du
/dxX = dydu
dudx
dy
/du
du
/dx
dy
/dx=
dy
/dx
= 7u6
2
= 14u6

5. y = (2x + 5)7
remember sin(2x + 30) = 0.8
sinA = 0.8,where A = 2x + 30let y= u7
where u = 2x + 5
dy
/du = 7u6 du
/dx = 2
dy
/dx 
dy
/du
du
/dx
= dydu
dudx
dy
/du
du
/dx
dy
/dx=
dy
/dx
= 7u6
2
= 14u6

6. y = (2x + 5)7
remember sin(2x + 30) = 0.8
sinA = 0.8,where A = 2x + 30let y= u7
where u = 2x + 5
dy
/du = 7u6 du
/dx = 2
dy
/dx 
dy
/du
du
/dx
= dydu
dudx
dy
/du
du
/dx
dy
/dx=
dy
/dx
= 7u6
2
= 14u6
= 14(2x + 5)6
The Chain Rule

7. y = (7x + 2)5
let y= u5
where u = 7x + 2
dy
/du = 5u4 du
/dx = 7
dy
/du
du
/dx
dy
/dx =
dy
/dx
= 5u4
= 35u4
7
= 35(7x + 2)4

8. y = (5 – 2x)4
let y= u4
where u = 5 – 2x
dy
/du = 4u3 du
/dx = -2
dy
/du
du
/dx
dy
/dx =
dy
/dx
= 4u3
= -8u3
-2
= -8(5 – 2x)3

9. y = (3x2
– 4)3
let y= u3
where u = 3x2
– 4
dy
/du = 3u2 du
/dx = 6x
dy
/du
du
/dx
dy
/dx =
dy
/dx
= 3u2
= 18u2
x
6x
= 18(3x2
– 4)2
x
= 18x(3x2
– 4)2

10. The Chain Rule
(Non Trig)
Key to the chain rule
dy
/du
du
/dx
dy
/dx =

11. y = (x2
+ 2)9
let y= u9
where u = x2
+ 2
dy
/du = 9u8 du
/dx = 2x
dy
/du
du
/dx
dy
/dx =
dy
/dx
= 9u8
= 18x u8
2x
= 18x(x2
+ 2)8
from*
*
change to u

12. y = (x3
– 6)4
let y= u4
where u = x3
– 6
dy
/du = 4u3 du
/dx = 3x2
dy
/du
du
/dx
dy
/dx =
dy
/dx
= 4u3
= 12u3
x2
3x2
= 12(x3
– 6)3
x2
= 12x2
(x3
– 6)3
Key Question

13. y = 4
5x + 2( )1
y = 4(5x + 2)-1
let y= 4u-1
where u = 5x + 2
dy
/du = -4u-2 du
/dx
dy
/du
du
/dx
dy
/dx =
dy
/dx
= -4u-2
= -20u-2
5
= -20(5x + 2)-2
= 5
= -20
5x + 2( )2
Nasty

14. y = (7x + 2)5
let y= u5
where u = 7x + 2
dy
/du = 5u4 du
/dx = 7
dy
/du
du
/dx
dy
/dx =
dy
/dx
= 5u4
7
Fast Tracking
derivative of
bit in brackets
=5(7x + 2)4