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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
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….
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!
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
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).
d
 ((
dx
    2
      ))
       5
   x +1 =
f ( x ) = x − 7 x + 1. Find f ′( x ).
           5
d
dx
     (            )
         x − 3x − 4 =
          4   3
3
d  1 
    2 
dx  x + 1 
d
dx
  ((
   x −x
    3
          )   −1 / 2
                       )
d         1          
                                 
You try :
          dx  3 ( 2 x + 4 ) 2   
                                
You try : 27 x + 42
        3
C. When a problem uses more than
               one rule
d
dx
   [                   ]
   ( 5 x − 2) ( 9 x + 2) =
             4          7
4
d  x 
          
dx  x + 1 
dz
  [
d 2
      (
   z + z −1
        2
           )]
            3   5
You try :
          d
          dx
               [
             ( 4 x − 1) ( 5 x + 1)
                       3           4
                                       ]
3
         d  x  2
You try :  2       
         dx  x + 3 
You try :  [(
          d 3
          dx
                 )
                 2
             x +1 + x 3
                          ]   4

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Chain Rule Guide

  • 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!
  • 4.
  • 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