3.6
The Chain Rule
Bell Ringer
Solve even #’s
We now have a pretty good list of “shortcuts” to find
derivatives of simple functions.
Of course, many of the functions that we will encounter
are not so simple. What is needed is a way to combine
derivative rules to evaluate more complicated functions.

Consider a simple composite function:
y  6 x  10
y  2  3x  5 
If u  3x  5
then y  2u
y  6 x  10
y  2u
u  3x  5
dy
6
dx
dy
2
du
du
3
dx
6  23
dy dy du


dx du dx

and another:
y  5u  2
where u  3t
y  5  3t   2 y  5u  2
u  3t
y  15t  2
then y  5  3t   2
dy
 15
dt
dy
5
du
du
3
dt
15  5  3
dy dy du


dt du dt

and one more:
y  9x  6x  1
2
y   3 x  1
y  u2
u  3x  1
dy
 18 x  6
dx
dy
 2u
du
du
3
dx
2
If u  3x  1
then y  u 2
y  9x2  6x  1
This pattern is called
the chain rule.
dy
 2  3 x  1
du
dy
 6x  2
du
18x  6   6 x  2   3
dy dy du


dx du dx

Chain Rule:
dy dy du


dx du dx
If f g is the composite of y  f  u  and u  g  x  ,
then:

f
g

  fat u  g x  gat x
example:
f  x   sin x
g  x   x2  4
f   x   cos x
g  x   2x
Find:
f
g  at x  2
g  2  4  4  0
f   0  g   2
cos  0   2  2
1 4  4

We could also do it this way:
f  g  x    sin  x 2  4 
y  sin  x  4 
2
y  sin u
dy
 cos u
du
dy
 cos  x 2  4   2 x
dx
u  x2  4
du
 2x
dx
dy dy du


dx du dx
dy
 cos u  2 x
dx
dy
 cos  22  4   2  2
dx
dy
 cos  0   4
dx
dy
4
dx

Here is a faster way to find the derivative:
y  sin  x 2  4 
d 2
y  cos  x  4    x  4 
dx
2
y  cos  x 2  4   2 x
Differentiate the outside function...
…then the inside function
At x  2, y  4

Another example:
d
cos 2  3 x 
dx
2
d
cos  3 x  
dx
It looks like we need to
use the chain rule again!
d
2 cos  3 x    cos  3 x 
dx
derivative of the
outside function
derivative of the
inside function

Another example:
d
cos 2  3 x 
dx
2
d
cos  3 x  
dx
d
2 cos  3 x    cos  3 x 
dx
d
2 cos  3x    sin  3x    3x 
dx
2cos  3x   sin  3x   3
6cos  3x  sin  3x 
The chain rule can be used
more than once.
(That’s what makes the
“chain” in the “chain rule”!)

Derivative formulas include the chain rule!
d n
n 1 du
u  nu
dx
dx
d
du
sin u  cos u
dx
dx
d
du
cos u   sin u
dx
dx
d
du
2
tan u  sec u
dx
dx
etcetera…
The formulas on the memorization sheet are written with
instead of du . Don’t forget to include the u term!
dx
u

The most common mistake on the chapter 3 test is to
forget to use the chain rule.
Every derivative problem could be thought of as a
chain-rule problem:
d 2
d
x  2 x x  2 x 1  2x
dx
dx
derivative of
outside function
derivative of
inside function
The derivative of x is one.

The chain rule enables us to find the slope of
parametrically defined curves:
dy dy dx


dt dx dt
dy
dt  dy
dx
dx
dt
dx
Divide
sides
Theboth
slope
of aby
parametrized
dt
curve is given by:
dy
dy
 dt
dx
dx
dt

Example:
x  3cos t
y  2sin t
These are the equations for
an ellipse.
dx
dy
 3sin t
 2 cos t
dt
dt
dy 2 cos t
2

  cot t
dx 3sin t
3

Example:
x  3cos t
y  2sin t

4
Now we can find the slope for any value of t :
For example, when t 

4
:
2
2
 
slope   cot    
3
3
4

Don’t forget to use the chain rule!

Homework:
3.6a
3.6 p153 1,15,31,45,61
3.5 p146 21,27,33,43
2.1 p66 9,18,27,36
3.6b
3.6 p153 5,7,21,23,35,37,51,53
2.1 p66 41,44,55
3.6c
3.6 p153 9,13,27,39,43,57
2.2 p76 9,18,27,36