Find y´´
y  csc x
y   csc x cot x
y   (csc x)( csc 2 x)  (cot x)( csc x cot x) 
y     csc3 x  csc x cot 2 x) 
y  csc3 x  csc x cot 2 x
y  csc x(csc2 x  cot 2 x)
y  csc x(csc2 x  csc2 x  1)
y  csc(2csc2 x  1)
y  2csc3 x  csc x
State Standard – 5.0 Students know the Chain Rule and its
proof and applications to the calculation of the derivative
of a variety of composite functions.
Objective – To be able to use the Chain Rule to solve
applications.
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
dy
dy
2
6
du
dx
6  23
dy dy du


dx du dx
u  3x  5
du
3
dx
The Chain Rule:
If f and g are both differentiable and F  f g is
the composite function defined by F ( x)  f ( g ( x))
then F is differentiable and F ' is given by the
product F ( x)  f ( g ( x)) g ( x)
In Leibniz notation, if y  f  u  and u  g  x  are both
differentiable functions, then
dy dy du


dx du dx
one more:
y  9x  6x 1
2
y   3 x  1
y u
2
u  3x  1
2
If u  3x  1
then y  u 2
y  9x2  6x  1
du
dy
dy
 18 x  6
3
 2u
dx
du
dx
dy
 2  3 x  1
du
dy
 6x  2
du
This pattern is called
the chain rule.
18x  6   6 x  2   3
dy dy du


dx du dx
The Power Rule and Chain Rule:
If n is any real number and u  g ( x) is
differentiable, then
dy n
n 1 du
(u )  nu
dx
dx
Also,
d
n
n 1
 g ( x)  n  g ( x) g '( x)
dx
Example 1
dy
Find
of
dx
y u
y  (4  3x)
9
u  4  3x
9
du
 3
dx
dy
8
 9u
du
dy du


du dx
9u ( 3)
8
 27u
8
dy
8
 27(4  3x)
dx
Example 2
dy
Find
of
dx
y  (5x  x )
y u
3
7
dy
6
 7u
du
4 7
u  5x  x
3
4
du
 15 x 2  4 x3
dx
dy du
2
3
6

 7u (15 x  4 x )
du dx
dy
3
4 6
2
3
 7(5 x  x ) (15 x  4 x )
dx
Example 3
dy
Find
of
dx
yu
3
dy
 3u 2
du
y  (3x  7 x  5)
4
3
u  3x  7 x  5
du
 12 x 3  7
dx
4
dy du
3
2

 3u (12 x  7)
du dx
dy
4
2
3
 3(3x  7 x  5) (12 x  7)
dx
Example 4
dy
Find
of
dx
y  x 1
2
y  u u
1
u  x 1
2
2
dy 1 12
1
 u 
du 2
2 u
du
 2x
dx
dy du
2/ x
1


 2x 
du dx
2 u
2/ x 2  1
dy

dx
x
x 1
2
Example 5
dy
Find
of
dx
y u
3
dy
4
 3u
du
y
1
 x  5
3
u  x 5
du
1
dx
dy du

 3u 4 1
du dx
 3u
4
dy
3

4
dx ( x  5)
Worksheet