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					3.6

The Chain Rule
Greg Kelly, Hanford High School, Richland, Washington

Photo by Vickie Kelly, 2002

U.S.S. Alabama Mobile, Alabama

Photo by Vickie Kelly, 2002

Greg Kelly, Hanford High School, Richland, Washington

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  6 x  10

y  2u

u  3x  5

y  2  3x  5

If u  3x  5
then y  2u

dy 6 dx

dy 2 du
6  23

du 3 dx

dy dy du   dx du dx



and another:

y  5u  2

y  5  3t   2 y  5u  2
y  15t  2

u  3t

where u  3t
then y  5  3t   2

dy  15 dt

dy 5 du
15  5  3

du 3 dt

dy dy du   dt du dt


and one more:

y  9x  6x  1
2

y  9x2  6x  1

y  u2

u  3x  1

y   3x  1
then y  u 2

2

If u  3x  1

dy  18 x  6 dx

dy  2u du dy  2  3x  1 du dy  6x  2 du

du 3 dx

This pattern is called the chain rule.

18 x  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  
example:

  f at u  g  x   g at x
Find:

f  x   sin x f   x   cos x

g  x   x2  4 g  x   2x f   0  g   2 cos  0    2  2 

 f  g 

at x  2

g  2  4  4  0

1 4  4



We could also do it this way:

f  g  x    sin  x 2  4 

y  sin  x  4 
2

dy  cos  x 2  4   2 x dx dy  cos  22  4   2  2 dx dy  cos  0   4 dx dy 4 dx


y  sin u

u  x2  4

dy  cos u du

du  2x dx

dy dy du   dx du dx dy  cos u  2 x 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

Differentiate the outside function...
…then the inside function

y  cos  x 2  4   2 x

At x  2, y  4



Another example:

d cos2  3x  dx
2 d cos  3x   dx 

It looks like we need to use the chain rule again!

d 2 cos  3x   cos  3x    dx
derivative of the outside function derivative of the inside function


Another example:

d cos2  3x  dx
2 d cos  3x   dx 

d 2 cos  3x   cos  3x    dx d 2cos  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 cos u   sin u dx dx

d du sin u  cos 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 both sidesa parametrized The slope of by curve is given by: dt
dy dy  dt dx dx dt


Example:

x  3cos t

y  2sin t

These are the equations for an ellipse.

dx dy  3sin t  2cos t dt dt

dy 2cos t 2    cot t dx 3sin t 3

Don’t forget to use the chain rule!

p


				
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