Section 3–4B
The Chain Rule
If y = f (u) is a differentiable function of u
and u = g(x) is a differentiable function of x
then
dy dy du
=
•
dx du dx
or
If y = f (u) then y′ = f ′(u) • u′
The Chain Rule with the Power Rule
Theorem: If a and n are a real numbers
If y = a • un
then y′ = a • n • u n-1 • u′
or in English
If y = a • (some function of x inside the brackets )n
then
y′ = a • n• (some function of x inside the brackets ) n-1 • the derivative of the function inside the brackets
Example 1
The Chain Rule with the Power Rule
Find y ′ if y = (5x − 2)3
if we let u = 5x − 2 then
y = (u) 3 where u = 5x − 2 and u′ = 5
y ′ = n(u)n−1 • u′
n−1
n 6
4u74
8 u}′
}
y ′ = 3 (5x − 2) 2 • 5
y ′ = 15(5x − 2) 2
Math 400 3–4B Chain Rule
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©2013 Eitel
Example 2
The Chain Rule with the Power Rule
(
Find y ′ if y = 2 3x 2 + 5x
)
4
Think of the outside function as 2(u)4
take the derivative of the outside function 2• 4(u) 3
(
8 3x 2 + 5x
)
3
now multiply that by the derivative of the inside function
if u = 3x 2 + 5x then u′ = 6x + 5
n −1
u
u′4
6a•
7n8 647483 6
47
8
2
f ′(x) = 2• 4
3x + 5x • (6x + 5)
(
(
)
)
3
f ′(x) = 8 3x 2 + 5x • (6x + 5)
Example 3
The Chain Rule with the Power Rule
Find y ′ if y = 4(tan x )3
Think of the outside function as 4(u)3
take the derivative of the outside function 4 • 3(u)2
12(tan x) 2
now multiply that by the derivative of the inside function
if u = tan x then u′ = sec 2 x
n −1
u74
u ′4
47
8
6
4
8 6
2
2
f ′(x) = 12 (tan x ) • sec x
a}
•n
(
)
f ′(x) = 12tan2 x sec 2 x
Math 400 3–4B Chain Rule
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©2013 Eitel
Example 4
The Chain Rule with the Power Rule
(
) (
)
Find y ′ if y = 3 6x 2 − 1 = 6x 2 − 1
1/3
Think of the outside function as (u)1/3
take the derivative of the outside function
(
)
1
• (u) −2/3
3
−2/3
1
6x 2 − 1
3
now multiply that by the derivative of the inside function
if u = 6x 2 − 1 then u′ = 12x
n−1
n 64
4u744
8 67
u ′8
}
−2/3
1
f ′(x) =
6x 2 − 1
• (12x )
3
(
)
(
)
−2/3
1
• 12x • 6x 2 − 1
3
f ′(x) =
4x
f ′(x) =
(6x2 − 1)
2/3
Example 5
The Chain Rule with the y = Sin (u)
( )
Find f ′ if f (x) = sin 3x 4
if y = f (u) then y′ = f ′ (u)• u′
Think of the outside function as sin(u)
take the derivative of the outside function to get cos(u)
( )
cos 3x 4
now multiply that by the derivative of the inside function
if u = 3x 4 then u′ = 12x 3
f7
′ (u)
u′
6
4
4
8 }
f ′(x) = cos 3x 4 •12x 3
( )
( )
f ′(x) = 12x3 cos 3x 4
Math 400 3–4B Chain Rule
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©2013 Eitel
Example 6
The Chain Rule with y = e u
Find f ′(x) if f (x) = e x
2
if y = f (u) than y′ = f ′ (u)• u′
Think of the outside function as e u
take the derivative of the outside function to get e u
2
ex
now multiply that by the derivative of u
if u = x 2 then u′ = 2x
f}
′ (u)
f ′(x) = e
x2
u′
}
• 2x
f ′(x) = 2x e x
2
Example 7
1/2
Find f ′(x) if f (x) = e x = e x
if y = f (u) than y′ = f ′ (u)• u′
Think of the outside function as e u
take the derivative of the outside function to get e u
1/2
ex
now multiply that by the derivative of u
1
if u = x 1/2 then u′ = x −1/2
2
u′ 8
f ′ (u) 67
}
1/2
1
f ′(x) = e x • x −1/2
2
f ′(x) =
e x
2x −1/2
e x
f ′(x) =
2 x
Math 400 3–4B Chain Rule
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Example 8
The Chain Rule with y = ln( u)
(
Find y ′ if y = ln 3x 4 + 2x 3
)
if u = 3x 4 + 2x 3 then u′ = 12x 3 + 6x 2
if y = ln(u) then y′ =
y′ =
u′
u
12x 3 + 6x 2
3x 4 + 2x 3
6x 2 (2x + 1)
y′ = 3
x (3x + 2)
y′ =
6(2x + 1)
x( 3x + 2)
Example 9
The Chain Rule with y = ln( u)
Find y ′ if y = ln 3x 2 − 5x
1/2 1
y = ln 3x 2 − 5x
= • ln 3x 2 − 5x
2
(
)
(
)
if u = 3x 2 − 5x then u′ = 6x − 5
if y = ln(u) then y′ =
y′ =
y′ =
y′ =
Math 400 3–4B Chain Rule
u′
u
1
6x − 5
• 2
2 3x − 5x
(
6x − 5
2 3x 2 − 5x
)
6x − 5
2x ( 3x − 5)
Page 5 of 11
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Example 10
The Chain Rule with y = ln
⎛ u⎞
= ln u − lnv
⎝ v⎠
⎛ x2 −1 ⎞
Find y ′ if y = ln ⎜
⎟
⎝ sin(2x) ⎠
(
)
y = ln x 2 − 1 − ln (sin(2x) )
if y = ln(u) then y′ =
u′
u
der
of7
sin(2x)
64
48
der}
ofx 2
−
cos(2x) • 2
sin2x
2x
y′ = 2
−
x −1
2 cos(2x)
sin(2x)
y′ =
2x
x2 −1
Example 11
The Chain Rule with y = ln
⎛ u⎞
= ln u − lnv
⎝ v⎠
⎛ x • sin(2x)⎞
Find y ′ if y = ln ⎜
2
⎟
⎝
⎠
ex
2
y = ln ( x • sin(2x)) − ln ⎛⎝ e x ⎞⎠
y = ln(x) + ln(sin(2x)) − x 2 ln (e )
y = ln(x) + ln(sin(2x)) − x 2
u′
if y = ln(u) then y′ =
u
der
of7
sin(2x)
64
48
der }of x
y′ =
y′ =
Math 400 3–4B Chain Rule
1
x
+
cos(2x) • 2
−
sin(2x)
der}
of x 2
2x
1
2 cos(2x)
+
− 2x
x
sin(2x)
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©2013 Eitel
Repeated use of the Chain Rule
])
([
then y′ = f ′ ( g[ (h {x} ) ] ) • g′ [ (h{x } ) ] • (h {x } )
If y = f g (h {x} )
Repeated use of the Chain Rule
Example 12
([
Find f ′(x) if f (x) = sin cos ( 4x )
f ′(x) =
([
d
sin cos ( 4 x )
dx
([
])
([
])
f ′(x) = cos cos ( 4 x ) •
])
])
d
cos ( 4x )
dx
f ′(x) = cos cos ( 4 x ) • (− sin(4 x))
d
( 4x )
dx
f ′(x) = cos(cos( 4x )) • (− sin(4 x))( 4 )
f ′(x) = −4 • cos(cos( 4x )) • sin ( 4x )
an alternate notation for example 12
Find y ′ if y = sin (cos ( 4 x ))
if u = cos ( 4x ) then
y = sin (u) where u = cos ( 4x )
the4
der
of4
sin
u the6
der
of 4
cos
6
47
4
8
47
84x the der}of 4x
y′ = cos(cos ( 4x )) • − sin ( 4x ) •
4
y ′ = −4 • cos(cos ( 4x )) • sin ( 4 x )
Math 400 3–4B Chain Rule
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©2013 Eitel
Example 13
Repeated use of the Chain Rule
Find f ′(x) if f (x) = (cos ( 4 x ))
f ′(x) =
3
d
3
cos ( 4x ))
(
dx
f ′(x) = 3• (cos ( 4x )) •
2
d
cos ( 4x )
dx
f ′(x) = 3• (cos ( 4x )) • (− sin(4 x)) •
2
d
(4x )
dx
f ′(x) = 3• (cos ( 4x )) • (− sin(4 x)) • 4
2
f ′(x) = −12 • (cos ( 4x )) • sin( 4x )
2
an alternate notation for example 13
Find y ′ if y = (cos ( 4 x ))
3
if u = cos ( 4x ) then
y = u3 where u = cos ( 4x )
n
the
der7
of4
(u)
der
6
44
4
8 the6
4of
7cos(4
4
8 x) the der}of 4x
2
y ′ = 3• (cos ( 4x )) • − sin( 4x )
•
4
y ′ = −12 • (cos ( 4 x )) • sin(4 x )
2
Math 400 3–4B Chain Rule
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©2013 Eitel
Example 14
The Chain Rule with the Product Rule
(
)
Find f ′(x) if f (x) = 4 x • 3x 2 − 2 = 4x • 3x 2 − 2
the}
first
f ′(x) = 4 x
der
of 4
sec44
the
644
47
8
64
7sec
48 der of first
}
−1/2
1/2
1
3x 2 − 2
• 6x + 3x 2 − 2
•
4
2
(
)
(
)
f ′(x) = 12x2 3x 2 − 2
(
)
factor out a 3x 2 − 2
(
1/2
)
−1/2
(
)
+ 4 3x 2 − 2
)
1/2
−1/2
[(
−1/2 ⎛
f ′(x) = 3x 2 − 2
(
)]
2
2
⎞
⎝12x + 4 3x − 2 ⎠
12x 2 + 12x 2 − 8]
[
f ′(x) =
(3x2 − 2)
f ′(x) =
Math 400 3–4B Chain Rule
1/2
24 x 2 − 8
3x 2 − 2
Page 9 of 11
©2013 Eitel
Example 15
The Chain Rule with the Quotient Rule
2x
Find f ′(x) if f (x) =
3x 2 − 2
6the
4bottom
748
3x 2 − 2)
(
f ′(x) =
1/2
•
der }
of top
2
=
2x
( 3x 2 − 2)1/2
the
top
}
− 2x
of7
bottom
64der
44
444
8
−1/2
1
•
3x 2 − 2
• 6x
2
(
(13x422−423)
)
2
bottom squared
f ′(x) =
(
)1/2 − 6x 2 ( 3x2 − 2) −1/2
2
( 3x 2 − 2)
2 3x 2 − 2
(
)
factor out a 3x 2 − 2
3x 2 − 2)
(
f ′(x) =
−1/2
−1/2
[(
)
• 2 3x 2 − 2 − 6x 2
( 3x2 − 2)
2
]
6x 2 − 4 − 6x 2 ]
[
f ′(x) =
(3x2 − 2) 5/2
f ′(x) =
Math 400 3–4B Chain Rule
−4
(3x2 − 2)
5/2
Page 10 of 11
©2013 Eitel
Example 15
The Chain Rule with the Power Rule and the Quotient Rule
⎛ x 2 − 2⎞
Find f ′(x) if f (x) = ⎜
⎟
⎝ 2x + 3⎠
2
n
derof
the top
6the
4
4
74u4
8 the bottom der of top
6
4
74
8 der of }bottom
6
474
8
}
1
2
(2x + 3) • 2x − x − 2 •
2
⎛ x 2 − 2⎞
f ′(x) = 2⎜
•
⎟
(1
2x + 3) 2
⎝ 2x + 3⎠
424
3
(
)
bottom squared
⎛ x 2 − 2⎞ 4x 2 + 6x − 2x 2 + 4
f ′(x) = 2⎜
⎟•
(2x + 3)2
⎝ 2x + 3⎠
⎛ x 2 − 2⎞ 2x 2 + 6x + 4
f ′(x) = 2⎜
⎟•
(2x + 3) 2
⎝ 2x + 3⎠
f ′(x) =
Math 400 3–4B Chain Rule
4 x 2 + 12x + 8
(2x + 3)3
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©2013 Eitel