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Trigo 2:Chain Rule

Trigo 2:Chain Rule

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Mathematics

•

12th Grade

•

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Thavarajah Selvarajah

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14 Slides • 40 Questions

1

Trigo 2:
Chain Rule

by Thavarajah Selvarajah

2

​For the following, decide whether

chain rule ( PIT-STOP 2) is needed?

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3

4

​When

chain rule ( PIT-STOP 2) is needed?

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5

Multiple Choice

Given y=sin⁡(1−3x)y=\sin\left(1-3x\right)  ,then   ddx[sin⁡(1−3x)]\frac{d}{dx}\left[\sin\left(1-3x\right)\right]  =

1

Yes

2

No

6

Multiple Choice

Given y=sin⁡(3x)y=\sin\left(3x\right)  ,then   ddx(sin⁡3x)\frac{d}{dx}\left(\sin3x\right)  =

1

Yes

2

No

7

Multiple Choice

Given y=3sin⁡(x)y=3\sin\left(x\right)  ,then   ddx(3sin⁡x)\frac{d}{dx}\left(3\sin x\right)  =

1

Yes

2

No

8

Multiple Choice

Given y=3sin⁡23xy=3\sin\frac{2}{3}x  ,then   ddx(3sin⁡23x)\frac{d}{dx}\left(3\sin\frac{2}{3}x\right)  =

1

Yes

2

No

9

Multiple Choice

Given y=3sin⁡wy=3\sin w  ,then   ddw(3sin⁡w)\frac{d}{dw}\left(3\sin w\right)  =

1

Yes

2

No

10

​WENT WELL?

COULD YOU DISTINGUISH WHICH ONES NEEDS PIT-STOP 2

( CHAIN RULE) ?

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LEVEL 2 : THE PROCESS of Chain Rule :) ​
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12

Multiple Choice

Question image

What is  p?p?  

1

sin⁡ (1−3x)\sin\ \left(1-3x\right)  

2

−3x-3x  

3

1−3x1-3x  

4

3x3x  

13

Multiple Choice

Question image

What is  q?q?  

1

sin⁡ u\sin\ u  

2

cos⁡ u\cos\ u  

3

sin⁡ x\sin\ x  

4

cos⁡ x\cos\ x  

14

Multiple Choice

Question image

What is  r?r?  

1

cos⁡ (1−3x)\cos\ \left(1-3x\right)  

2

−3cos⁡ u-3\cos\ u  

3

−3cos⁡ (1−3x)-3\cos\ \left(1-3x\right)  

4

cos⁡ x\cos\ x  

15

Multiple Choice

Question image

Given is 1st step of Chain Rule. What is a?a?  

1

uu  

2

tan⁡ 2x\tan\ 2x  

3

2x2x  

16

Multiple Choice

Question image

Given is continuation step of previous Chain Rule. What is bb  and  c?c?  

1

b=2, c=tan⁡ ub=2,\ c=\tan\ u

2

b=−2, c=sec⁡2ub=-2,\ c=\sec^2u

3

b=2, c=sec⁡2ub=2,\ c=\sec^2u  

17

Multiple Choice

Question image

Given is first few steps of Chain Rule differentiation of the function. What is p?p?  

1

23cosec⁡ u\frac{2}{3}\operatorname{cosec}\ u  

2

cosec⁡ u\operatorname{cosec}\ u  

3

23cosec⁡ (1−2x)\frac{2}{3}\operatorname{cosec}\ \left(1-2x\right)

18

Multiple Choice

Question image

Given is the continuation steps of previous  Chain Rule differentiation of the function. What is q?q?  

1

−43cosec⁡u cot⁡ u-\frac{4}{3}\operatorname{cosec}u\ \cot\ u

2

−43cosec⁡(1−2x)cot⁡ (1−2x)-\frac{4}{3}\operatorname{cosec}\left(1-2x\right)\cot\ \left(1-2x\right)

3

23cosec⁡ (1−2x)\frac{2}{3}\operatorname{cosec}\ \left(1-2x\right)

4

43cosec⁡(1−2x)cot⁡ (1−2x)\frac{4}{3}\operatorname{cosec}\left(1-2x\right)\cot\ \left(1-2x\right)

19

​ARE YOU CONFIDENT DOING THE QUESTIONS INDEPENDENTLY?

20

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21

​LEVEL 3

​DOING PIT-STOP 2 (CHAIN RULE)

​

​INDEPENDENTLY

​ARE YOU SURE

​ YOU ARE READY?

22

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​LET'S DO THIS CHAMPS......

24

Multiple Choice

Given  y=sin⁡5xy=\sin5x  .Find dydx\frac{\text{d}y}{\text{d}x}  

1

cos⁡5x\cos5x  

2

−cos⁡5x-\cos5x  

3

5cos⁡5x5\cos5x  

4

−5cos⁡5x-5\cos5x  

25

Multiple Choice

Given  y=15sin⁡5xy=\frac{1}{5}\sin5x  .Find dydx\frac{\text{d}y}{\text{d}x}  

1

cos⁡5x\cos5x  

2

15cos⁡5x\frac{1}{5}\cos5x  

3

−15cos⁡5x-\frac{1}{5}\cos5x

4

−cos⁡5x-\cos5x  

26

Multiple Choice

Given  y=cos⁡2xy=\cos2x  .Find dydx\frac{\text{d}y}{\text{d}x}  

1

sin⁡2x\sin2x  

2

−sin⁡2x-\sin2x  

3

2sin⁡2x2\sin2x  

4

−2sin⁡2x-2\sin2x  

27

Multiple Choice

Given  y=cot⁡6xy=\cot6x  .Find dydx\frac{\text{d}y}{\text{d}x}  

1

cosec⁡26x\operatorname{cosec}^26x  

2

−cosec⁡26x-\operatorname{cosec}^26x  

3

−6cosec⁡26x-6\operatorname{cosec}^26x  

4

6cosec⁡26x6\operatorname{cosec}^26x  

28

Multiple Choice

Given  y=16cot⁡6xy=\frac{1}{6}\cot6x  .Find dydx\frac{\text{d}y}{\text{d}x}  

1

cosec⁡26x\operatorname{cosec}^26x  

2

−cosec⁡26x-\operatorname{cosec}^26x  

3

−6cosec⁡26x-6\operatorname{cosec}^26x  

4

6cosec⁡26x6\operatorname{cosec}^26x  

29

Multiple Choice

Given  y=16tan⁡(1−6x)y=\frac{1}{6}\tan\left(1-6x\right)  .Find dydx\frac{\text{d}y}{\text{d}x}  

1

sec⁡2(1−6x)\sec^2\left(1-6x\right)  

2

−sec⁡2(1−6x)-\sec^2\left(1-6x\right)  

3

16sec⁡2(1−6x)\frac{1}{6}\sec^2\left(1-6x\right)  

4

−16sec⁡2(1−6x)-\frac{1}{6}\sec^2\left(1-6x\right)

30

Multiple Choice

Given  y=−12cos⁡2xy=-\frac{1}{2}\cos2x  .Find dydx\frac{\text{d}y}{\text{d}x}  

1

sin⁡2x\sin2x  

2

−12sin⁡2x-\frac{1}{2}\sin2x  

3

12sin⁡2x\frac{1}{2}\sin2x  

4

−sin⁡2x-\sin2x  

31

Multiple Choice

Given  y=15sin⁡(1−5x)y=\frac{1}{5}\sin\left(1-5x\right)  .Find dydx\frac{\text{d}y}{\text{d}x}  

1

15cos⁡(1−5x)\frac{1}{5}\cos\left(1-5x\right)  

2

−15cos⁡(1−5x)-\frac{1}{5}\cos\left(1-5x\right)  

3

−cos⁡(1−5x)-\cos\left(1-5x\right)  

4

cos⁡(1−5x)\cos\left(1-5x\right)  

32

Multiple Choice

Given  y=25sin⁡(5x−1)y=\frac{2}{5}\sin\left(5x-1\right)  .Find  dydx\frac{\text{d}y}{\text{d}x}  

1

2cos⁡(5x−1)2\cos\left(5x-1\right)  

2

25cos⁡(5x−1)\frac{2}{5}\cos\left(5x-1\right)  

3

−25cos⁡(5x−1)-\frac{2}{5}\cos\left(5x-1\right)  

4

−2cos⁡(5x−1)-2\cos\left(5x-1\right)  

33

Multiple Choice

Given y=12cos⁡(1−2x)y=\frac{1}{2}\cos\left(1-2x\right)  , find dydx\frac{\text{d}y}{\text{d}x}  

1

12sin⁡(1−2x)\frac{1}{2}\sin\left(1-2x\right)  

2

−12sin⁡(1−2x)-\frac{1}{2}\sin\left(1-2x\right)  

3

sin⁡(1−2x)\sin\left(1-2x\right)  

4

−sin⁡(1−2x)-\sin\left(1-2x\right)  

34

Multiple Choice

Given y=−12cos⁡(2x−1)y=-\frac{1}{2}\cos\left(2x-1\right)  , find dydx\frac{\text{d}y}{\text{d}x}  

1

12sin⁡(2x−1)\frac{1}{2}\sin\left(2x-1\right)  

2

−12sin⁡(2x−1)-\frac{1}{2}\sin\left(2x-1\right)  

3

sin⁡(2x−1)\sin\left(2x-1\right)  

4

−sin⁡(2x−1)-\sin\left(2x-1\right)  

35

Multiple Choice

Given y=−16cot⁡6xy=-\frac{1}{6}\cot6x  , find dydx\frac{\text{d}y}{\text{d}x}  

1

cosec⁡26x\operatorname{cosec}^26x  

2

−16cosec⁡26x-\frac{1}{6}\operatorname{cosec}^26x  

3

16cosec⁡26x\frac{1}{6}\operatorname{cosec}^26x  

4

−cosec⁡26x-\operatorname{cosec}^26x  

36

Multiple Choice

Given  y=16cot⁡(1−6x)y=\frac{1}{6}\cot\left(1-6x\right)  

1

16cosec⁡2(1−6x)\frac{1}{6}\operatorname{cosec}^2\left(1-6x\right)  

2

−cosec⁡2(1−6x)-\operatorname{cosec}^2\left(1-6x\right)  

3

cosec⁡2(1−6x)\operatorname{cosec}^2\left(1-6x\right)  

4

−16cosec⁡2(1−6x)-\frac{1}{6}\operatorname{cosec}^2\left(1-6x\right)  

37

Multiple Choice

Given y=sec⁡7xy=\sec7x  ,find dydx\frac{\text{d}y}{\text{d}x}  

1

sec⁡7xtan⁡7x\sec7x\tan7x  

2

7sec⁡7xtan⁡7x7\sec7x\tan7x  

3

−7sec⁡7xtan⁡7x-7\sec7x\tan7x

4

17sec⁡7xtan⁡7x\frac{1}{7}\sec7x\tan7x

38

39

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​TIRED??

40

Multiple Choice

Given  y=17sec⁡7xy=\frac{1}{7}\sec7x  

1

sec⁡7xtan⁡7x\sec7x\tan7x  

2

7sec⁡7xtan⁡7x7\sec7x\tan7x  

3

−sec⁡7xtan⁡7x-\sec7x\tan7x  

4

−7sec⁡7xtan⁡7x-7\sec7x\tan7x  

41

Multiple Choice

Given y=−17sec⁡7xy=-\frac{1}{7}\sec7x  

1

−sec⁡7xtan⁡7x-\sec7x\tan7x  

2

sec⁡7xtan⁡7x\sec7x\tan7x  

3

17sec⁡7xtan⁡7x\frac{1}{7}\sec7x\tan7x  

4

−7sec⁡7xtan⁡7x-7\sec7x\tan7x  

42

Multiple Choice

Given  y=17sec⁡(1−7x)y=\frac{1}{7}\sec\left(1-7x\right)  ,find dydx\frac{dy}{dx}  

1

sec⁡(1−7x)tan⁡(1−7x)\sec\left(1-7x\right)\tan\left(1-7x\right)  

2

−sec⁡(1−7x)tan⁡(1−7x)-\sec\left(1-7x\right)\tan\left(1-7x\right)  

3

17sec⁡(1−7x)tan⁡(1−7x)\frac{1}{7}\sec\left(1-7x\right)\tan\left(1-7x\right)

4

−17sec⁡(1−7x)tan⁡(1−7x)-\frac{1}{7}\sec\left(1-7x\right)\tan\left(1-7x\right)

43

Multiple Choice

Given  y=−17sec⁡(1−7x)y=-\frac{1}{7}\sec\left(1-7x\right)  ,find dydx\frac{dy}{dx}  

1

sec⁡(1−7x)tan⁡(1−7x)\sec\left(1-7x\right)\tan\left(1-7x\right)  

2

−sec⁡(1−7x)tan⁡(1−7x)-\sec\left(1-7x\right)\tan\left(1-7x\right)  

3

17sec⁡(1−7x)tan⁡(1−7x)\frac{1}{7}\sec\left(1-7x\right)\tan\left(1-7x\right)

4

−17sec⁡(1−7x)tan⁡(1−7x)-\frac{1}{7}\sec\left(1-7x\right)\tan\left(1-7x\right)

44

Multiple Choice

Given  y=cosec⁡ 8xy=\operatorname{cosec}\ 8x  , find  dydx\frac{dy}{dx}  

1

−cosec⁡8xcot⁡8x-\operatorname{cosec}8x\cot8x  

2

8cosec⁡8xcot⁡8x8\operatorname{cosec}8x\cot8x  

3

−8cosec⁡8xcot⁡8x-8\operatorname{cosec}8x\cot8x  

4

cosec⁡8xcot⁡8x\operatorname{cosec}8x\cot8x  

45

Multiple Choice

Given  y=18cosec⁡ 8xy=\frac{1}{8}\operatorname{cosec}\ 8x  , find  dydx\frac{dy}{dx}  

1

−cosec⁡8xcot⁡8x-\operatorname{cosec}8x\cot8x  

2

8cosec⁡8xcot⁡8x8\operatorname{cosec}8x\cot8x  

3

−8cosec⁡8xcot⁡8x-8\operatorname{cosec}8x\cot8x  

4

cosec⁡8xcot⁡8x\operatorname{cosec}8x\cot8x  

46

Multiple Choice

Given  y=−18cosec⁡ 8xy=-\frac{1}{8}\operatorname{cosec}\ 8x  , find  dydx\frac{dy}{dx}  

1

−cosec⁡8xcot⁡8x-\operatorname{cosec}8x\cot8x  

2

8cosec⁡8xcot⁡8x8\operatorname{cosec}8x\cot8x  

3

−8cosec⁡8xcot⁡8x-8\operatorname{cosec}8x\cot8x  

4

cosec⁡8xcot⁡8x\operatorname{cosec}8x\cot8x  

47

Multiple Choice

Given  y=−18cosec⁡ (8x−1)y=-\frac{1}{8}\operatorname{cosec}\ \left(8x-1\right)  , find  dydx\frac{dy}{dx}  

1

−cosec⁡(8x−1)cot⁡(8x−1)-\operatorname{cosec}\left(8x-1\right)\cot\left(8x-1\right)  

2

8cosec⁡(8x−1)cot⁡(8x−1)8\operatorname{cosec}\left(8x-1\right)\cot\left(8x-1\right)  

3

−8cosec⁡(8x−1)cot⁡(8x−1)-8\operatorname{cosec}\left(8x-1\right)\cot\left(8x-1\right)  

4

cosec⁡(8x−1)cot⁡(8x−1)\operatorname{cosec}\left(8x-1\right)\cot\left(8x-1\right)  

48

Multiple Choice

Given  y=18cosec⁡ (1−8x)y=\frac{1}{8}\operatorname{cosec}\ \left(1-8x\right)  , find  dydx\frac{dy}{dx}  

1

−cosec⁡(1−8x)cot⁡(1−8x)-\operatorname{cosec}\left(1-8x\right)\cot\left(1-8x\right)  

2

8cosec⁡(1−8x)cot⁡(1−8x)8\operatorname{cosec}\left(1-8x\right)\cot\left(1-8x\right)  

3

−8cosec⁡(1−8x)cot⁡(1−8x)-8\operatorname{cosec}\left(1-8x\right)\cot\left(1-8x\right)  

4

cosec⁡(1−8x)cot⁡(1−8x)\operatorname{cosec}\left(1-8x\right)\cot\left(1-8x\right)  

49

Multiple Choice

Given y=tan⁡3xy=\tan3x  ,find dydx\frac{dy}{dx}  

1

sec⁡23x\sec^23x  

2

3sec⁡23x3\sec^23x  

3

sec⁡2x\sec^2x  

4

3sec⁡2x3\sec^2x  

50

Multiple Choice

Given y=13tan⁡3xy=\frac{1}{3}\tan3x  ,find dydx\frac{dy}{dx}  

1

sec⁡23x\sec^23x  

2

3sec⁡23x3\sec^23x  

3

sec⁡2x\sec^2x  

4

3sec⁡2x3\sec^2x  

51

Multiple Choice

Given y=−13tan⁡3xy=-\frac{1}{3}\tan3x  ,find dydx\frac{dy}{dx}  

1

sec⁡23x\sec^23x  

2

3sec⁡23x3\sec^23x  

3

−sec⁡23x-\sec^23x  

4

−3sec⁡23x-3\sec^23x  

52

Multiple Choice

Given y=−13tan⁡(1−3x)y=-\frac{1}{3}\tan\left(1-3x\right)  ,find dydx\frac{dy}{dx}  

1

sec⁡2(1−3x)\sec^2\left(1-3x\right)  

2

3sec⁡2(1−3x)3\sec^2\left(1-3x\right)  

3

−13sec⁡2(1−3x)-\frac{1}{3}\sec^2\left(1-3x\right)  

4

−3sec⁡2(1−3x)-3\sec^2\left(1-3x\right)  

53

Multiple Choice

Given y=−13tan⁡(3x−1)y=-\frac{1}{3}\tan\left(3x-1\right)  ,find dydx\frac{dy}{dx}  

1

sec⁡2(1−3x)\sec^2\left(1-3x\right)  

2

−sec⁡2(1−3x)-\sec^2\left(1-3x\right)  

3

−13sec⁡2(1−3x)-\frac{1}{3}\sec^2\left(1-3x\right)  

4

−3sec⁡2(1−3x)-3\sec^2\left(1-3x\right)  

54

​WHAT YOU SHOULD HAVE UNLOCKED HERE?

​

​1. able to identify which needs PIT-STOP 2 (Chain rule)

​2. How to carry out the process independently

​3. Please review the questions that you are unsure before trying the next subsection

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Trigo 2:
Chain Rule

by Thavarajah Selvarajah

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