2.4 2nd part:
d
(tan x) 
dx
d
(cot x) 
dx
d
(sec x) 
dx
d
(csc x) 
dx
d
Ex. Prove (cot x)   csc2 x
dx
d2
Ex. Find 2 (csc x) 
x
dx
4
sec
Ex. f ( ) 
, find f ( )
1  sec
2.5 The chain rule
So many rules so far, but are we able to use
d 3
those rules to do ( x  1)9 ?
dx
This is a ‘composite’ function, so we may
break down to two simple functions:
y  u 9 , u  x3  1
This is like a chain:
The chain rule is:
If y depends on u and u depends on x, then y
depends on x, and
dy dy du
( Chain Rule version 1)
 
dx du dx
Trick to memorize: looks like a cancellation.
Other notations:
y  f (u )  u 9 , u  g ( x)  x3  1
Then
dy dy du
   f (u ) g ( x)  f ( g ( x)) g ( x)
dx du dx
Chain Rule version 2:
If y  f ( g ( x)), then y  f ( g ( x)) g ( x)
Version 2 is probably more useful since you
don’t have to add u every time. Another
reason is: we may explain version 2 this
way:
If the derivative of f (x) is f (x) , then the
derivative of f (stuff ) is f (stuff )  (stuff)
Or:
The ‘outer function’ f is differentiated like
always (leaving the inner function as-is),
and then you multiply by the derivative of
the inner function.
Let’s try version 2:
d x
e
dx
2
1
d
cos( x 4  1)
dx
((2  tan x)3 )
In general, the CHAIN RULE and POWER
RULE combine to give us:
Since we know that the derivative of x n is
nx n1, if follows that the derivative of
(stuff) n is n(suff) n1  (stuff)
d
Or [ g ( x)]n  n[ g ( x)]n1 g ( x)
dx
Ex. Find the indicated derivatives and
simplify.
a. g ( x) if g ( x)  5 x sec(4 x)
b. f (e), if f ( x)  2 cot x
d
c.
[3x( x 2  1)3 ]
dx
d ( x 3  7) 4
d.
dx 2 x 3
d
e.
cot(2 x 3  1)
dx
d
1
f.
(3 2
)
dt t  t  1
g. y  sec x 2 , find y
h. y  (2 x  1)5 ( x3  x  1) 4 , find y
Ex2. Find f (x) and find the value(s) of
x where the tangent line is horizontal.
x
a. f ( x) 
(2 x  5) 2
b. f ( x)  x 2  8 x  20