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CHAIN RULE: DAY 2WITH TRIG FUNCTIONS
Section 2.4A
Calculus AP/Dual, Revised ©2018
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 1
THE CHAIN RULE
A.𝒅
𝒅𝒙𝒇 𝒈 𝒙 = 𝒇′ 𝒈 𝒙 𝒈′ 𝒙
B. If 𝒇(𝒙) is a differentiable function and 𝒈(𝒙) is a differentiable
function, then 𝒚 = 𝒇(𝒈(𝒙)) is a differentiable function.
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 2
TRIG FUNCTION DERIVATIVES
C. Derivatives of Trig Functions (MEMORIZE THEM)
1.𝒅
𝒅𝒙𝐬𝐢𝐧 𝒙 = 𝐜𝐨𝐬𝒙
2.𝒅
𝒅𝒙𝐜𝐨𝐬𝒙 = −𝐬𝐢𝐧𝒙
3.𝒅
𝒅𝒙𝐭𝐚𝐧𝒙 = 𝐬𝐞𝐜𝟐 𝒙
4.𝒅
𝒅𝒙𝐜𝐬𝐜 𝒙 = −𝐜𝐬𝐜 𝒙 𝐜𝐨𝐭 𝒙
5.𝒅
𝒅𝒙𝐬𝐞𝐜 𝒙 = 𝐬𝐞𝐜𝒙 𝐭𝐚𝐧𝒙
6.𝒅
𝒅𝒙𝐜𝐨𝐭 𝒙 = −𝐜𝐬𝐜𝟐 𝒙
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 3
TRIG DERIVATIVES
So, when you have to ask your neighbor a question…
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 4
...PSSTPositive
SEC
SEC
TAN
sec sec tand
x x xdx
=
2tan secd
x xdx
=
TRIG DERIVATIVES
Think opposite of PSST… is to CHANGE
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 5
CHANGE
CSC
CSC
COT
csc csc cotd
x x xdx
= −
2cot cscd
x xdx
= −
Positive
SEC
SEC
TAN
→
( )Negative CHANGE
CSC
CSC
COT
REVIEW
Basic Functions Composite Functions
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 6
sin x
xe
ln x
( )sin 2 3x +
2xe
( )4ln 3x
COMPOSITE FUNCTIONS
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 7
( )sin x
( )sin 2 3x +
( )2 3 )x +
EXAMPLE 1
Solve 𝒚′ for 𝒚 = 𝐬𝐢𝐧 𝟐𝒙 + 𝟑
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 8
( )( ) ( )' ' 'y f g x g x=
( )' 2cos 2 3y x= +
( )' 2cos 2 3y x= +
Original
Derivative
( )sin 2 3x +
( )cos 2 3x +
2 3x +
2
EXAMPLE 2
Solve 𝒚′ for 𝒚 = 𝐜𝐨𝐬 𝟐𝒙
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 9
( )( ) ( )' ' 'y f g x g x=
( )( )sin 2 2x−
( )' 2 sin 2y x= −
( )cos 2x
( )sin 2x−
2x
2
EXAMPLE 3
Solve 𝒚′ for 𝒚 = 𝐜𝐨𝐬𝟐 𝒙
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 10
( )( ) ( )' ' 'y f g x g x=
( )( )( )' 2 cos sin 1y x x= −
( )( )' 2 cos siny x x= −
( )2
cos x
( )1
2 cos x
cos x
sin x−
x
1
EXAMPLE 4
Solve 𝒚′ for 𝒚 = 𝐜𝐬𝐜 𝒙𝟑
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 11
2 3 3' 3 csc coty x x x= −
EXAMPLE 5
Solve 𝒚′ for 𝒚 = 𝐜𝐨𝐬 (𝟑𝒙)𝟐
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 12
( )2' 18 sin 9y x x= −
YOUR TURN
Solve 𝒚′ for 𝒚 = 𝐭𝐚𝐧𝟐 𝒙
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 13
2' 2 tan secy x x=
WHEN YOU BUILD A CHAIN…
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 14
It can keep going… and going…
REVISIT ORDER
A. Wrapper
B. Shell
C. Gum
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 15
WITH TRIG CHAIN RULE FUNCTIONS
A. Make sure the exponent and negative are put in the appropriate spots
B. Use parenthesis
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 16
EXAMPLE 6
Solve 𝒅𝒚
𝒅𝒙for 𝒚 = 𝐬𝐢𝐧𝟑 𝟒𝒙
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 17
( ) ( ) ( )2' 12sin 4 cos 4f x x x=
( )( ) ( )' ' 'y f g x g x=
( ) ( )( )( )2' 3 sin cos 4 4y x x=
( )( )3
sin 4x
( )( )2
3 sin 4x
( )sin 4x
( )cos 4x
4x
4
EXAMPLE 7
Solve 𝒅𝒚
𝒅𝒙for 𝒇 𝒙 = 𝟑𝐜𝐨𝐬𝟑 𝒙𝟑 + 𝟏
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 18
( ) ( ) ( )2 2 3 3' 27 cos 1 sin 1f x x x x= − + +
EXAMPLE 8
Solve 𝒅𝒚
𝒅𝒙for 𝒇 𝒙 = 𝐜𝐨𝐭𝟑(𝒙𝟐 − 𝟓𝒙)
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 19
( ) ( ) ( ) ( )2 2 2 2' 3cot 5 csc 5 2 5f x x x x x x= − − − −
YOUR TURN
Solve 𝒅𝒚
𝒅𝒙for 𝒇 𝒙 = 𝟒𝐬𝐞𝐜𝟑 𝟐𝒙 + 𝟏
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 20
( ) ( ) ( )3' 24sec 2 1 tan 2 1f x x x= + +
EXAMPLE 9
Solve 𝒇′′ 𝒙 for 𝒇 𝒙 = 𝒙𝟒 𝐬𝐢𝐧 𝒙
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 21
( ) ( ) ( )3 2 3 44 cos sin 12 4 cos sinx x x x x x x x+ + + −
4 siny x x=
3 4' 4 sin cosy x x x x= +
'' ' 'y fg gf= +
( ) 2 3 4'' 12 sin 8 cos sinf x x x x x x x= + −
EXAMPLE 10
Solve 𝒇′ 𝒙 for 𝒇 𝒙 =𝟏−𝒙 𝟑
𝐜𝐨𝐬 𝟓𝒙𝟒
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 22
( ) ( ) ( ) ( )( )( ) ( )
( )
2 34 4 4
24
cos5 3 1 1 1 sin 5 5
cos5
d dx x x x x
dx dx
x
− − − − −
( )( ) ( ) ( )( ) ( )
( )
3 34 4
24
cos5 1 1 cos5
'cos5
d dx x x x
dx dxf x
x
− − −
=
EXAMPLE 10
Solve 𝒇′ 𝒙 for 𝒇 𝒙 =𝟏−𝒙 𝟑
𝐜𝐨𝐬 𝟓𝒙𝟒
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 23
( ) ( )( ) ( ) ( )2 34 3 4
2 4
cos5 3 1 20 1 sin 5
cos 5
x x x x x
x
− − + −
( ) ( ) ( )( )( )( )
( )
2 34 4 3
24
cos5 3 1 1 sin 5 20
cos5
x x x x x
x
− − − − −
( )( ) ( ) ( ) ( )
2 34 3 4
2 4
3 1 cos5 20 1 sin5'
cos 5
x x x x xf x
x
− − + −=
YOUR TURN
Solve 𝒇′ 𝒙 for 𝒇 𝒙 =−𝟐𝒙𝟐−𝟓
𝐜𝐨𝐬 𝟐𝒙𝟑
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 24
( )( )
( )
3 3 3 3
2 3
2 2cos2 6 sin 2 15 sin 2'
cos 2
x x x x x xf x
x
− + −=
EXAMPLE 11
Determine the point(s) in the interval of 𝟎, 𝟐𝝅 at which the graph of
𝒇 𝒙 =𝐜𝐨𝐬 𝒙
𝟐+𝐬𝐢𝐧 𝒙has a horizontal tangent.
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 25
( )( )
2
2sin 1'
2 sin
xf x
x
− −=
+
2sin 1 0x− − =
2sin 1x− =1
sin2
x = −
7 3 11 3, , ,
6 3 6 3
−
EXAMPLE 12
If 𝒚 = 𝐭𝐚𝐧𝒖, 𝒖 = 𝒗 −𝟏
𝒗, and 𝒗 = 𝐥𝐧 𝒙 what is the value of
𝒅𝒚
𝒅𝒙at 𝒙 = 𝒆?
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 26
dy dy du dv
dx du dv dx=
( ) ( )1
tan lndy dy du dv
u v xdx du dv v dx
= −
( )( )2
2
1 1sec 0 1
1
dy
dx e
= +
ln 1
11 1 1 0
1
x e
v e
u
=
= =
= − = − =
2
e
EXAMPLE 13
Determine the equation of the tangent line using the equation, 𝒚 =𝟏
𝒙+
𝐜𝐨𝐬 𝒙 at 𝝅, 𝟏 .
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 27
( )1/21 cosy x x−= +
( ) ( )1/22 1
' cos sin2
y x x x−−= − + −
2
1 sin'
2 cos
xy
x x= − −
( )
( )
( )2
sin1'
2 2 cosy
= − −
( ) 2
1 0'
2 1y
= − −
( ) 2
1'y
= −
( )2
11y x
− = − −
AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR)
If 𝒇 𝒙 = 𝐬𝐢𝐧 𝒙𝟐 − 𝝅 , then 𝒇′ 𝟑𝝅 =
(A) 𝟏
(B) −𝟐 𝟑𝝅
(C) 𝐜𝐨𝐬 𝟐 𝟑𝝅
(D) 𝟐 𝟑𝝅
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 28
If 𝒇 𝒙 = 𝐬𝐢𝐧 𝒙𝟐 − 𝝅 , then 𝒇′ 𝟑𝝅 =
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 29
AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR)
Vocabulary Connections and Process Answer and Justifications
DDerivative
Chain Rule
( ) ( )2sinf x x = −
( ) ( )( )2' cos 2f x x x= −
( ) ( )( ) ( )( )2
' cos 3 2 3f x = −Trigonometric
Derivative
( ) ( )( )' cos 3 2 3f x = −
( ) ( )( )' cos 2 2 3f x =
( ) ( )( )' 1 2 3f x =
AP MULTIPLE CHOICE PRACTICE QUESTION 2(NON-CALCULATOR)
Solve the derivative of 𝒇 𝜽 = 𝐬𝐢𝐧𝟐𝜽
(A) 𝐜𝐨𝐬 𝟐𝜽
𝐬𝐢𝐧 𝟐𝜽
(B) 𝐜𝐨𝐬 𝟐𝜽
𝟐 𝐬𝐢𝐧 𝟐𝜽
(C) 𝐜𝐨𝐬 𝟐𝜽
(D) 𝐬𝐞𝐜 𝟐𝜽
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 30
Solve the derivative of 𝒇 𝜽 = 𝐬𝐢𝐧𝟐𝜽
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 31
AP MULTIPLE CHOICE PRACTICE QUESTION 2 (NON-CALCULATOR)
Vocabulary Connections and Process Answer and Justifications
ADerivative
Chain Rule( )sin 2 cos 2 2
d
d
=
( ) ( ) ( )( )1/2 1/21
sin 2 sin 2 cos 2 22
d
d
−=
Trigonometric
Derivative
( )1/2 cos 2
sin 2sin 2
d
d
=
ASSIGNMENT
Page 136
43-71 odd, 84, 87, 93, 102
7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 32