CHAIN RULE: DAY 2 WITH TRIG FUNCTIONS€¦ · CHAIN RULE: DAY 2 WITH TRIG FUNCTIONS Section 2.4A...

CHAIN RULE: DAY 2WITH TRIG FUNCTIONS

Section 2.4A

Calculus AP/Dual, Revised ©2018

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7/30/2018 1:44 AM §2.4A: Chain Rule Day 2 1

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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.


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


TRIG DERIVATIVES

So, when you have to ask your neighbor a question…


...PSSTPositive

SEC

SEC

TAN

sec sec tand

x x xdx

=

2tan secd

x xdx

=

TRIG DERIVATIVES

Think opposite of PSST… is to CHANGE


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


sin x

xe

ln x

( )sin 2 3x +

2xe

( )4ln 3x

COMPOSITE FUNCTIONS


( )sin x

( )sin 2 3x +

( )2 3 )x +

EXAMPLE 1

Solve 𝒚′ for 𝒚 = 𝐬𝐢𝐧 𝟐𝒙 + 𝟑


( )( ) ( )' ' '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


( )( ) ( )' ' '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


( )( ) ( )' ' '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


2 3 3' 3 csc coty x x x= −

EXAMPLE 5

Solve 𝒚′ for 𝒚 = 𝐜𝐨𝐬 (𝟑𝒙)𝟐


( )2' 18 sin 9y x x= −

YOUR TURN

Solve 𝒚′ for 𝒚 = 𝐭𝐚𝐧𝟐 𝒙


2' 2 tan secy x x=

WHEN YOU BUILD A CHAIN…


It can keep going… and going…

REVISIT ORDER

A. Wrapper

B. Shell

C. Gum


WITH TRIG CHAIN RULE FUNCTIONS

A. Make sure the exponent and negative are put in the appropriate spots

B. Use parenthesis


EXAMPLE 6

Solve 𝒅𝒚

𝒅𝒙for 𝒚 = 𝐬𝐢𝐧𝟑 𝟒𝒙


( ) ( ) ( )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


( ) ( ) ( )2 2 3 3' 27 cos 1 sin 1f x x x x= − + +

EXAMPLE 8

Solve 𝒅𝒚

𝒅𝒙for 𝒇 𝒙 = 𝐜𝐨𝐭𝟑(𝒙𝟐 − 𝟓𝒙)


( ) ( ) ( ) ( )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


( ) ( ) ( )3' 24sec 2 1 tan 2 1f x x x= + +

EXAMPLE 9

Solve 𝒇′′ 𝒙 for 𝒇 𝒙 = 𝒙𝟒 𝐬𝐢𝐧 𝒙


( ) ( ) ( )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


( ) ( ) ( ) ( )( )( ) ( )

( )

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


( ) ( )( ) ( ) ( )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


( )( )

( )

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.


( )( )

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?


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 .


( )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


If 𝒇 𝒙 = 𝐬𝐢𝐧 𝒙𝟐 − 𝝅 , then 𝒇′ 𝟑𝝅 =



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


Solve the derivative of 𝒇 𝜽 = 𝐬𝐢𝐧𝟐𝜽



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


CHAIN RULE: DAY 2 WITH TRIG FUNCTIONS€¦ · CHAIN RULE: DAY 2 WITH TRIG FUNCTIONS Section 2.4A...

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