This document discusses the chain rule for finding derivatives. It explains that the chain rule is needed when taking the derivative of a composition of functions, where an "inside function" is plugged into an "outside function". The chain rule formula is given as the derivative of the outside function multiplied by the derivative of the inside function. Several examples are worked through, applying the chain rule when the power rule alone cannot be used, such as when the base of an exponent is a function rather than a variable. The document also notes that problems may require using multiple derivative rules, like the product rule and chain rule, to fully solve them.

1. 7.7 The Chain Rule
A. What’s a Composition of Functions
B. The Chain Rule
C. When a Problem Uses More than
One Rule

2. A. What’s a composition of
functions?
• You’ll have an “inside function” and an
“outside function” sort of.
• We will call the outside function f and the
inside function g.
• When you plug the function g INTO f, you
get the original composition of functions.
• You will understand these words better if I
show you….

3. f ( x ) = x and g ( x ) = 4 − x.
2
Find f ( g ( x ) ).
This means PLUG g INTO f.
2
Skeleton of f : ( )
Insert g : ( 4 − x )
2
This is a composition of two functions!

5. Consider ( 4 − x ) .
2
You canNOT just use the power rule to bring down the 2
and get 2( 4 − x ) . That would be WRONG.
1
Why can' t we do this? If it were x 2 , we could do it!
The base isn' t just plain x this time. Instead
the base is ( 4 - x ).
This will be a job for.....THE CHAIN RULE!
The first step is identifying an inside function g and an
outside function f such that f(g(x)) = ( 4 - x ) .
2

6. B. The Chain Rule
d
( f ( g ) ) = f ′( g ) ⋅ g ′ means that we CAN use the power rule
dx
in that way AS LONG AS we multiply it by the derivative of
the inside function, which I might refer to as its " tail."
d
dx
( )
( 4 − x ) 2 = 2( 4 − x )1 ⋅ ( − 1)
In this one, the derivative of the inside function (4 - x) is (-1).

7. d
((
dx
2
))
5
x +1 =

8. f ( x ) = x − 7 x + 1. Find f ′( x ).
5

9. d
dx
( )
x − 3x − 4 =
4 3

10. 3
d  1 
 2 
dx  x + 1 

11. d
dx
((
x −x
3
) −1 / 2
)

12. d  1 

You try :
dx  3 ( 2 x + 4 ) 2 
 

13. You try : 27 x + 42
3

14. C. When a problem uses more than
one rule
d
dx
[ ]
( 5 x − 2) ( 9 x + 2) =
4 7

15. 4
d  x 
 
dx  x + 1 

16. dz
[
d 2
(
z + z −1
2
)]
3 5

17. You try :
d
dx
[
( 4 x − 1) ( 5 x + 1)
3 4
]

18. 3
d  x  2
You try :  2 
dx  x + 3 

19. You try : [(
d 3
dx
)
2
x +1 + x 3
] 4