3.6
The Chain Rule
Photo by Vickie Kelly, 2002
Created by Greg Kelly, Hanford High School, Richland, Washington
Revised by Terry Luskin, Dover-Sherborn HS, Dover, Massachusetts
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 
y  2(u )
and u  3x  5,
y  6 x  10
y  2(u )
u  3x  5
dy
6
dx
dy
2
du
du
3
dx
6  23
dy dy du


dx du dx
d
f ( g ( x))  f '  g ( x)  * g '  x 
dx

and one more:
y  9x  6x 1
2
y   3 x  1
when y  u
u  3x 1
2
y  9x2  6x  1
y  u2
u  3x  1
dy
 18 x  6
dx
dy
 2u
du
du
3
dx
2
dy
1
 2  3x  1
du
This pattern is
called the
chain rule.
dy/dx = 2(3x + 1)1 • 3
dy dy du
d


or
f ( g ( x))  f '  g ( x)  * g '  x 
dx du dx
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 g  x  gat x
If f (g(x)) is the composite of y = f (u) and u = g (x), then:
d
d
d
 f ( g ( x)   f at g  x   g at x
dx
dx
dx

Chain Rule:
dy dy du


dx du dx
If f ( g ( x)) is the composite of y  f  u  and u  g  x  ,
then:
f
g   f at g  x   gat x  f  ( g ( x))  g  ( x)
example:
d
f  x   sin x
dx
f   x   cos x
f '  0  cos(0)  1
g  x  x 2  4
g  2  2 2  4  0
g  2  2 2
g  x   2x
Find:
 f ( g ( x)   f   g ( x)   g   x 
at x  2
d
 f ( g (2)   f   g (2)   g   2 
dx
d
 f ( g (2)   f   0   g   2 
dx
d
 f ( g (2)   1  4  4

dx
Here is a way to find the derivative by seeing “layers:”
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,
(keep the inner function unchanged...)
…then multiply by the derivative
of the inner function
Evaluate this general derivative at x  2, to find 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
1
derivative of the
outside power
function
derivative of the
inside trig function

Another example:
d
cos 2  3 x 
dx
2
d

cos  3 x  
dx
d
 2 cos  3 x    cos  3 x 
dx
1
d
 2 cos  3 x     sin  3 x     3 x 
dx
 2cos  3x     sin  3x    3
 6cos  3x  sin  3x 
The chain rule can be
needed more than once.
(That’s what makes the
“chain” in the “chain rule”!)

Each derivative formula will now 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
et cetera…

The most common mistake in differentiating is to
forget to use the chain rule.
Every old familiar derivative problem could be thought of
now as a chain-rule situation:
d
d
2
 x   2 x  x  2x 1  2x
dx
dx
derivative of
outside function
derivative of
inside function
The derivative of x is one.

Don’t forget to use the chain rule!
p