1
Section 2.4: The Chain Rule
Practice HW from Larson Textbook (not to hand in)
p. 122 # 1-23 odd, 33-97 odd
In this section, we want to differentiate different type of function compositions. The rules
displayed are all special cases of the chain rule, which we discuss later in the section.
General Power Rule
d
( g ( x)) n  n ( g ( x)) n1 g ( x)
dx
Example 1: Differentiate y  (3x 2  1) 5 .
Solution:
█
Example 2: Differentiate y  4 5  4 x  x 3 .
Solution:
█
2
Example 3: Differentiate f (t ) 
1
(t 5  2) 4
Solution: Using the exponent law a  n 
.
1
, start by rewriting the function as
an
f (t )  (t 5  2) 4 . Then using the general power rule, we calculate the derivative as
follows.
f (t )  4(t 5  2) 41 (5t 4  0)
(Take derivative using general power rule)
 4(t 5  2) 5 (5t 4 )
(Simplify)
 20t 4 (t 5  2) 5
(Multiply  4 and 5t 4 )

 20t 4
(t  2)
5
5
(Rewrite using exponent law a  n 
1
an
)
█
Example 4: Differentiate y  ( x 2  1) 5 (3x  4) 6 .
Solution:
█
3
Derivative of an Exponential Function for an Arbitrary Base
Base e:
d x
(e )  e x
dx
Arbitrary Base:
d x
(a )  (ln a ) a x
dx
Example 5: Differentiate y  5 x .
Solution:
█
Chain Rule For Exponential Functions
Base e:
d g ( x)
(e
)  e g ( x )  g ( x)  g ( x) e g ( x )
dx
Arbitrary Base:
d g ( x)
(a
)  (ln a ) a g ( x )  g ( x)  (ln a ) g ( x) a x
dx
Example 6: Differentiate y  e 2 x  52 x .
Solution:
█
4
Example 7: Differentiate y  e x
5
2 x1
.
Solution:
█
5
3
Example 8: Differentiate f (t )  t 2 e t .
Solution: Since this function is a product of two functions of t, we need to use the
product rule. When taking the derivative of the second function, we apply the chain rule
applied to the exponential function of base e. This gives
f (t )  t 2 
(3t 2 )e t
3
First Derivative of 2nd
Chain Rule e function
3
f (t )  3t 4 e t  2tet
3
3
 et 
2t
Second Derivative of First
(Simplify)
█
Chain Rule
Used to take the derivative of a composition of two functions.
Chain Rule
d
( f ( g ( x))  f ( g ( x)) g ( x)
dx
Note: The general power rule and chain rule for the exponential function are special cases
of the Chain Rule.
6
Example 9: Differentiate y  (3x 2  1) 5 using the general power rule and the chain rule.
Solution:
█
Chain Rule Applied to the Trigonometric Functions
d
sin( g ( x))  cos( g ( x))  g ( x)
dx
d
cos( g ( x))   sin( g ( x))  g ( x)
dx
d
tan( g ( x))  sec 2 ( g ( x))  g ( x)
dx
d
sec( g ( x))  sec( g ( x))  tan( g ( x))  g ( x)
dx
d
csc( g ( x))   csc( g ( x))  cot( g ( x))  g ( x)
dx
d
cot( g ( x))   csc 2 ( g ( x))  g ( x)
dx
7
Example 10: Differentiate y  sin 5 x .
Solution:
█
Example 11: Differentiate y  cos 2 x  cos x 2 .
Solution:
█
Example 12: Differentiate y  5 tan( x 3  2 x)  sec 3x .
Solution:
8
█
Example 13: Differentiate y  sin (cot x) .
2
Solution:
█
Formulas for the Derivative of Logarithmic Functions
Natural Logarithm:
d
(ln x) 
dx
d
( ln x ) 
dx
Logarithm of an Arbitrary Base:
d
(log a x) 
dx
9
Example 14: Differentiate y  x 3 ln x  log 5 x
Solution:
█
Chain Rule for the Logarithmic Functions
Natural Logarithm:
d
ln( g ( x))  
dx
d
ln(| g ( x) | 
dx
Logarithm of an Arbitrary Base:
d
log a ( g ( x))  
dx
10
Example 15: Differentiate y  2 ln( x 4  2)  2 log 3 ( x 4  2) .
Solution:
█
Example 16: Differentiate y  (ln x) 3  ln x 3 .
Solution:
█
11
Example 17: Differentiate y  ln | x 4  3x 2 | .
Solution: Using the chain rule formula for the natural logarithm of an absolute value
d
ln(| g ( x) |  1 g ( x) , we calculate the derivative as follows:
dx
g ( x)
dy
1
 4
(4 x 3  6 x)
2
dx x  3 x

4x3  6x
x 4  3x 2
█
Example 18: Differentiate y  ln
(3t  5) 7
.
2t  1
Solution: On this one, it is easier to use some properties of logarithms to rewrite the
function before differentiating. We do this as follows:
(3t  5) 7
y  ln
 ln( 3t  5) 7  ln( 2t  1)
2t  1
 7 ln( 3t  5)  ln( 2t  1)
(Use ln prop ln
u
 ln u  ln v)
v
(Use ln prop ln u k  k ln u
We now apply the chain rule for the ln function to differentiate y  7 ln( 3t  5)  ln( 2t  1)
as follows:
dy
1
1
7
(3) 
(2)
dt
3t  5
2t  1
dy
21
2


dt 3t  5 2t  1
█