Kurt Kareem I. Israel
Teacher Ferdinand Corpuz
Grade 11 – Lorenz
Date Answered: March 25, 2021
Basic Calculus
Semester 2 – Quarter 3 – Weeks 7 - 8
Module 3 – The Derivatives
LESSON 1:
The Chain Rule
Activity 1: What I Know (Page 3)
1. B.




𝑦 = √𝑡 3 + 1 when 𝑡 = 2
1
𝑑
(𝑡 3 + 1)
3
2√𝑡 +1 𝑑𝑥
𝑑
(𝑡 3 + 1) = 3𝑡 2
𝑑𝑡
1
(3𝑥 )2
𝑦′ =
2√𝑡 3 +1
𝑦 ′′ =

𝑦 ′′ = √𝑡 3

3𝑡 2
2√𝑡 3 +1
3𝑥2
2
3
(2√𝑡 +1)(6𝑡)−3𝑡 (
)
2√𝑥3 +1
2
(2√𝑥 3 +1)


=
12𝑡√𝑡 3 +1−
9𝑡4
√𝑡3+1
=
4𝑡 3 +4
=
12𝑡√𝑡 3 +1−9𝑡4
√𝑡 3 +1
12𝑡 4 −12𝑡−9𝑡 4
3𝑡 4 −12𝑡
=
(4𝑡 3 +4)√𝑡 3 +1
+1 (4𝑡 3 +4)
3(2)4−12(2)
72
(4(2)3 +4)√(2)3 +1
2
2
3
= 108
𝑢𝑛𝑖𝑡𝑠/sec
2. csc𝑜 𝜃 ′ = −4 cot 𝜃 csc2 𝜃

𝑑
𝑑𝜃
[csc2 𝜃 + cot 2 𝜃] =
𝑑
𝑑
𝑑𝜃
[csc 2 𝜃] +
𝑑
𝑑
𝑑𝜃
[cot 2 𝜃]

(2 csc 𝜃 )


2 csc 𝜃 csc 𝜃 (− cot 𝜃)) + 2 cot 𝜃)(− csc2 𝜃)
csc𝑜 𝜃 ′ = −4 cot 𝜃 csc2 𝜃
𝑑𝜃
[csc 𝜃 ] + (2 cot 𝜃 )
𝑑𝜃
[cot 𝜃]
3. D
3𝑥

𝑦 = 𝑥2 +1

(3)


𝑑
(
𝑥
)
𝑑𝑥 𝑥 2 +1
𝑥(𝑥 2 +1)−(𝑥 2 +1)𝑥
𝑦 ′ = (3)
(𝑥 2 +1)2
2 +1)
3(−𝑥
𝑦 ′ = (𝑥2 +1)2
= (3)
(𝑥 2 +1)−2𝑥𝑥
(𝑥 2 +1)2
=
3(−𝑥 2 +1)
(𝑥 2 +1)2
4. 𝑓 (𝑥 )′ = 8𝑥 3 + 2𝑥



𝑆𝑜𝑙𝑣𝑒: (𝑥 2 )(2𝑥 2 + 1) = (2𝑥 )(2𝑥 2 + 1) + (4𝑥)(𝑥 2 )
𝑓 (𝑥 )′ = 4𝑥 3 + 2𝑥 + 4𝑥 3
𝑓 (𝑥 )′ = 8𝑥 3 + 2𝑥
1
(4𝑡 3 +4)
5. A


𝑓 ′(𝑥 ) = (9 − 𝑥 2 ) = (9 − 32 ) = 0
𝑓 ′′ (𝑥 ) = 0 = 0
6. 2𝑥 sin(𝑥 ) + 𝑥 2 cos(𝑥)




𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
′
(𝑥 2 ) sin(𝑥 ) +
𝑑
𝑑𝑥
(sin(𝑥 ))𝑥 2
(𝑥 2 ) = 2𝑥
(sin(𝑥 )) = cos(𝑥 )
𝑦 = 2𝑥 sin(𝑥 ) + 𝑥 2 cos(𝑥)
7. 2 sec2 (𝑥 ) tan(𝑥 ) + 2𝑥 𝑠𝑒𝑐 2 (𝑥 2 )




𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
(sec2 (𝑥)) +
𝑑
𝑑𝑥
(tan(𝑥 2 ))
(sec2 (𝑥)) = 2𝑠𝑒𝑐 2 (𝑥)tan(𝑥)
(tan(𝑥 2 )) = sec2 (𝑥 2 )(2𝑥 )= 2 sec2 (𝑥 ) tan(𝑥 ) + sec2 (𝑥2)(2𝑥)
2 sec2 (𝑥 ) tan(𝑥 ) + 2𝑥 𝑠𝑒𝑐 2 (𝑥 2 )
8. D
𝑑





𝑑 (1+cos(𝑥))(1−cos(𝑥))−𝑑𝑥(1−cos(𝑥))(1+cos(𝑥))
(1−cos(𝑥))2
𝑑𝑥
𝑑
𝑑𝑥
𝑑
(1 + cos(𝑥 )) = − sin(𝑥 )
(1 − cos(𝑥 )) = sin(𝑥 )
𝑑𝑥
𝑑 (− sin(𝑥))(1−cos(𝑥))−sin(𝑥)(1+cos(𝑥))
(
(1−cos(𝑥))2
𝑑𝑥
2
sin(𝑥)
𝑦 ′ = − (1−cos(𝑥))2
9. B
𝑑





3
3
𝑑 (𝑥 +2)𝑥−𝑑𝑥(𝑥 +2)
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑥2
3
(𝑥 + 2) = 3𝑥 2
(𝑥 ) = 1
𝑑𝑥
𝑑 (3𝑥 2 −1)(𝑥 3 +2)
𝑑𝑥
′(
𝑥2
𝑓 𝑡) =
2𝑥 3 −2
𝑥2
=
=
2𝑥 3 −2
𝑥2
2(−2)3−2
(−2)2
7
= −2
10. C
35 2
𝑡
2

𝑠′ = −

−35(1) + 58 = 23
+ 58𝑡 + 91 = −35𝑡 + 58
11. 2𝑥𝑒 𝑥 2



𝑑
(𝑥 2 ) = 2𝑥
𝑑𝑥
′(
2
𝑓 𝑡) = 𝑒 𝑥 (2𝑥 )
𝑓 ′(𝑡) = 2𝑥𝑒 𝑥 2
2sin(𝑥)
= − (1−cos(𝑥))2
12. A



𝑑 1
(sin 2𝜃)2 (2 cos 2𝜃)
𝑑𝑥 2
2 cos 2𝜃
2√sin 2 𝜃
cos 2𝜃
√sin 2𝜃
13. A




𝑓 ′(𝑥 ) = 6𝑥 2 − 3
𝑓 ′′ (𝑥 ) = 12
𝑥 = 1, 𝑡ℎ𝑒𝑛 12(1)
12
14. A

𝑑
𝑑𝑥
(𝑦) = csc(𝑥 ) − cot(𝑥)
15. C


𝑑
1
𝑑
(2) 𝑑𝑥 ((2𝑥 + 𝑒 2𝑥 )−1 ) = − (2𝑥+𝑒2𝑥 ) 𝑑𝑥 (2𝑥 + 𝑒 2𝑥 )
𝑑
𝑑𝑥
(2𝑥 + 𝑒 2𝑥 ) = (2 + 𝑒 2𝑥 )(2)
2(2+𝑒 2𝑥 )
1

2 − (2𝑥+𝑒2𝑥 ) (2𝑥 + 𝑒 2𝑥 ) = − (2𝑥+𝑒2𝑥 )

𝑓 ′(𝑥 ) = − (2𝑥+𝑒2𝑥 )
4(1+𝑒 2𝑥 )
16. C



1
ℎ
ℎ
1
+ 2 = 1(ℎ)−1 + 2
1
1
1
−1ℎ−2 + 2 = ℎ2 + 2
ℎ2 −2
2ℎ 2
17. C


6𝑡 − 2 = 6(2) − 2
10𝑚/𝑠𝑒𝑐
18. A






(𝑡 2 − 1)3 . 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒.
𝑠 ′ = 3(𝑡 2 − 1)2 (2𝑡) = 6𝑡(𝑡 2 − 1)2 = 6𝑡(𝑡 4 − 212 + 1)
𝑠 ′ = 6𝑡 5 − 12𝑡 3 + 6𝑡
𝑑
𝑑
𝑑
(6𝑡 5 ) = 30𝑡 4 , (−12𝑡 3 ) = −36𝑡 2 , & (6𝑡) = 6
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑠 ′′ = 30𝑡 4 − 36𝑡 2 + 6 = 30(2)4 + 36(2)2 + 6
342 𝑢𝑛𝑖𝑡𝑠/𝑠𝑒𝑐 2
19. A



𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒: 6𝜋𝑟 2
𝑑
(6𝜋𝑟 2 ) = 2(6)𝜋𝑟 2−1
𝑑𝑥
𝑦 ′ = 12𝜋𝑟
20. 128𝜋𝑐𝑚3 /𝑠𝑒𝑐
4

𝑉 = 3 𝜋𝑟 3 & 𝑟 = 2𝑡

𝐹𝑖𝑟𝑠𝑡 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒: 𝑉 ′ = 3 𝜋(2𝑡)2 = 3(2𝑡)2 𝑑𝑡 (2𝑡)




𝑑
𝑑𝑡
4
3
4
𝑑
(2𝑡) = 2
𝜋(2𝑡)2 (2) = 32𝜋𝑡 2
𝑑
𝑆𝑒𝑐𝑜𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒: 𝑉 ′′ = 𝑑𝑡 (32𝜋𝑡 2 ) = 64𝜋𝑡
64𝜋(2) = 128𝜋𝑐𝑚3 /𝑠𝑒𝑐
Activity 2: What’s In (Page 5)
1. C. 12



𝑓 (𝑥 ) = 12𝑥 + 3
𝑃𝑜𝑤𝑒𝑟 𝑅𝑢𝑙𝑒: 𝑓 ′ (𝑥 ) = (1)121−1 + 0
𝑓 ′(𝑥 ) = 12
2. A. 0


The derivative of a constant is 0.
Therefore, 𝑓 ′ (𝑥 ) = 0.
3. A. 4x + 2




𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒: 𝑓 ′ (𝑥 ) = 2𝑥 2 & 𝑔′(𝑥 ) = 2𝑥
𝑑
2(2)𝑥 2−1 = 4𝑥
𝑑𝑥
𝑑
= 1(2)𝑥 1−1 = 2
𝑑𝑥
′(
𝑓 𝑥 ) = 4𝑥 + 2
4. A. 50𝑥 4 − 28𝑥 3 + 6𝑥 2 − 18𝑥



𝑓 (𝑥 ) = 10𝑥 5 − 7𝑥 4 + 2𝑥 3 − 9𝑥 2 + 192
𝑃𝑜𝑤𝑒𝑟 𝑅𝑢𝑙𝑒: 𝑓 ′(𝑥 ) = 5(10)𝑥 5−1 − 4(7)𝑥 4−1 + 3(2)𝑥 3−1 − 2(9)𝑥 2−1 + 0
𝑓 ′(𝑥) = 50𝑥 4 − 28𝑥 3 + 6𝑥 2 − 18𝑥
5. B. sec(𝑥 ) tan(𝑥)

𝑑
𝑑𝑥
𝑑
(tan 𝑥) = sec2 𝑥, 𝑓𝑖𝑛𝑑 𝑑𝑥 (sec 𝑥)
𝑑
1

𝐹𝑖𝑛𝑑 𝐶𝑜𝑚𝑚𝑜𝑛 𝐷𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒: 𝑑𝑥 [cos(𝑥)] = −

− cos2(𝑥) = cos2(𝑥)

𝑑
𝑠𝑖𝑛(𝑥)
𝑑𝑥
sin(𝑥)
(sec 𝑥) = sec(𝑥 ) tan(𝑥)
Activity 3: What’s New (Page 5)
1. 36𝑥 2 − 36𝑥 2 + 56𝑥 − 16


𝑓 (𝑥 ) = (3𝑥 2 − 2𝑥 + 4)2
𝑑
2(3𝑥 2 − 2𝑥 + 4) (3𝑥 2 − 2𝑥 + 4)

𝑑
𝑑𝑥
𝑑𝑥
2
(3𝑥 − 2𝑥 + 4) = (6𝑥 − 2)
𝑑
[cos(𝑥)]
𝑑𝑥
cos2(𝑥)


2(3𝑥 2 − 2𝑥 + 4)(6𝑥 − 2)
36𝑥 2 − 36𝑥 2 + 56𝑥 − 16
2. 2cos(2𝑥)




𝑑
𝑑𝑥
sin(2𝑥 )
𝑑
cos(2𝑥 )
𝑑
𝑑𝑥
𝑑𝑥
(2𝑥 )
(2𝑥 ) = 2 → cos(2𝑥 ) (2)
2cos(2𝑥)
Activity 4: What’s More (Page 8)
Activity 3. Find the derivative, dy/dx of the following using chain rule:
24
a. 𝑓 ′ (𝑥 ) = − (𝑥+1)4
3
2
2 𝑑𝑦

𝑑𝑦
2

𝑑𝑦

3 (𝑥+1) (− (𝑥+12 ) = − (𝑥+1)4

𝑓 ′(𝑥 ) = − (𝑥+1)4
( ) = 3 (𝑥+1)
𝑑𝑥 𝑥+1
(
2
2
( )
𝑑𝑥 𝑥+1
2
) = − (𝑥+1)2
𝑑𝑥 𝑥+1
2 2
2
24
24
972
d. 𝑓 ′(𝑥 ) = − (3𝑥+1)3
4

𝑑𝑦
3

4 (3𝑥+1)

𝑑𝑦

4(

𝑓 ′ (𝑥 ) = − (3𝑥+1)3
(
)
𝑑𝑥 3𝑥+1
3 𝑑𝑦
3
(
3
𝑑𝑥 3𝑥+1
)
(
3
3
3𝑥+1
3
)
𝑑𝑥 3𝑥+1
9
= − (3𝑥+1)2
3
972
) (− (3𝑥+1)2 ) = − (3𝑥+1)3
972
2𝑥 2 −2
f. 𝑓 ′ (𝑥 ) = √𝑥2





𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
−2
𝑥√𝑥 2 − 2
(𝑥 )√𝑥 2 − 2 +
(𝑥 ) = 1 𝑎𝑛𝑑
𝑑𝑥
𝑑𝑦
𝑑𝑥
(1)(√𝑥 2 − 2) +
2𝑥 2 −2
𝑓 ′(𝑥 ) = √𝑥2
𝑑𝑦
(√𝑥 2 − 2)(𝑥 )
(√𝑥 2 − 2) =
𝑥
√𝑥 2 −2
𝑥
√𝑥 2 −2
2𝑥 2 −2
𝑥 = √𝑥2
−2
−2
Activity 5: What I Have Learned (Page 9)
Activity 4.
1.

A. 𝑓 ′ (𝑥 ) = 4𝑥 3 − 24𝑥 2 + 36𝑥 − 8 𝑎𝑛𝑑 𝑓 ′(2) = 0
 𝑓 (𝑥 ) = (𝑥 2 − 4𝑥 + 1)2 ; 𝑓′(2)
𝑑
 𝐶ℎ𝑎𝑖𝑛 𝑅𝑢𝑙𝑒: 2(𝑥 2 − 4𝑥 + 1) 𝑑𝑥 (𝑥 2 − 4𝑥 + 1)





𝑑
𝑑𝑥
(𝑥 2 − 4𝑥 + 1) = 2𝑥 − 4
2(𝑥 2 − 4𝑥 + 1)(2𝑥 − 4)
𝑓 ′(𝑥 ) = 4𝑥 3 − 24𝑥 2 + 36𝑥 − 8
𝑓 ′(2) = 4(2)3 − 24(2)2 + 36(2) − 8
𝑓 ′(2) = 0
2.

A. 𝑓 ′′ (𝑥 ) = −4𝑥 2 cos 𝑥 2 − 2 sin 𝑥 2
 𝑦 = cos 𝑥 2
𝑑
𝑑
 𝑑𝑥 (cos(𝑥 2 )) = − sin 𝑥 2 𝑑𝑥 𝑥 2
𝑑
𝑥 2 = 2𝑥
 𝑓 𝑥 ) = − sin 𝑥 2 (2𝑥 ) = −2𝑥 sin(𝑥 2 )
𝑑
𝑑
 𝑑𝑥 (−2𝑥𝑠𝑖𝑛(𝑥 2 )) = −2 𝑑𝑥 (𝑥𝑠𝑖𝑛(𝑥 2 ))

𝑑𝑥
′(
𝑑
𝑑
 −2 (𝑑𝑥 (𝑥 ) sin(𝑥 2 ) + 𝑑𝑥 (sin(𝑥 2 ))(𝑥 ))
𝑑
𝑑
(sin(𝑥 2 )) = cos 𝑥 2 (2𝑥 )
 −2(1 sin 𝑥 2 + cos 𝑥 2 (2𝑥𝑥 ) = −2(sin(𝑥 2 ) + 2𝑥 2 cos(𝑥 2 ))
 𝑓 ′ (𝑥 ) = −4𝑥 2 cos 𝑥 2 − 2 sin 𝑥 2


𝑑𝑥
(𝑥 ) = 1 &
𝑑𝑥
B. 𝑓 ′′(𝑥 ) = −4𝑥 2 cos(𝑥 2 ) sin(𝑥 ) − 4𝑥 cos(𝑥 ) sin(𝑥 2 ) − sin(𝑥 ) + cos(𝑥 2 ) −
2 sin(𝑥 2 ) sin(x)
 𝑦 = sin 𝑥 cos 𝑥 2
𝑑
𝑑
 𝑑𝑥 (sin(𝑥 ) cos(𝑥 2 )) + 𝑑𝑥 (cos(𝑥 2 )) sin(𝑥 )

𝑑
𝑑
sin 𝑥 = cos 𝑥 & 𝑑𝑥 (cos(𝑥 2 )) = −sin(𝑥 2 )(2𝑥)
𝑑𝑥
′(
 𝑓 𝑥 ) = cos(𝑥 ) cos(𝑥 2 ) + (− sin(𝑥 2 ) (2𝑥 )) sin(𝑥 )
 𝑓 ′ (𝑥 ) = cos(𝑥 ) cos(𝑥 2 ) − 2𝑥 sin(𝑥 2 ) sin(𝑥)
𝑑
 𝑑𝑥 (cos(𝑥 ) cos(𝑥 2 )) − (2𝑥 sin(𝑥 2 ) sin(𝑥))

𝑑
(cos(𝑥 ) cos(𝑥 2 )) = − sin(𝑥 ) cos(𝑥 2 ) − 2𝑥 sin(𝑥 2 ) cos(𝑥 )
𝑥
𝑑
(2𝑥 sin(𝑥 2 )sin(𝑥)) = 2(sin(𝑥 2 ) sin(𝑥 ) + 𝑥(2𝑥 cos(𝑥 2 ) sin(𝑥 ) + cos(𝑥 ) sin(𝑥 2 )))
 𝑓 𝑥 ) = − sin(𝑥 ) cos(𝑥 2 ) − 2𝑥 sin(𝑥 2 ) cos(𝑥 ) − 2(sin(𝑥 2 ) sin(𝑥 ) +
𝑥(2𝑥 cos(𝑥 2 ) sin(𝑥 ) + cos(𝑥 )sin(𝑥 2 ))
 𝑓 ′′ (𝑥 ) = − sin(𝑥 ) cos(𝑥 2 ) − 2𝑥 sin(𝑥 2 ) cos(𝑥 ) − 2(sin(𝑥 2 ) sin(𝑥 ) +
𝑥(2𝑥 cos(𝑥 2 ) 𝑠𝑖𝑛
 𝑓 ′′ (𝑥 ) = −4𝑥 2 cos(𝑥 2 ) sin(𝑥 ) − 4𝑥 cos(𝑥 ) sin(𝑥 2 ) − sin(𝑥 ) + cos(𝑥 2 ) −
2 sin(𝑥 2 ) sin(𝑥)

𝑑𝑥
′′ (
LESSON 2:
Implicit Differentiation
Activity 1: What’s New (Page 11)
1. 𝑦 ′ (𝑥 ) = −2





2𝑥 + 𝑦 = 8
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠: 2𝑥 + 𝑦 = 8
𝑑
2 + 𝑑𝑥 (𝑦) = 0
𝑑
(𝑦) = −2
𝑑𝑥
′(
𝑦 𝑥 ) = −2
Activity 2: What’s More (Page 16)
𝑑2
a) 𝑑𝑥2 (𝑦) =





2𝑦 2 −𝑥 2
4𝑦 3
or
−𝑥 2 +2𝑦 2
4𝑦 3
4𝑦 2 + 5 = 2𝑥 2
𝑑
8𝑦 𝑑𝑥 (𝑦) = 4𝑥
𝑑
(𝑦 ) =
𝑑𝑥
𝑥
2𝑦
𝑑2
( )
2 𝑦 =
𝑑𝑥
𝑑2
𝑑𝑥 2
(𝑦 ) =
𝑑
(𝑦)
𝑑𝑥
2
2𝑦
𝑦−𝑥
2𝑦 2−𝑥 2
4𝑦 3
=
or
2𝑦 2 −𝑥 2
4𝑦 3
−𝑥 2 +2𝑦 2
4𝑦 3
LESSON 3:
Related Rates
Activity 1: What’s In (Page 19)
cos(𝑥)𝑦
1. 𝑦 ′ = 4 sin(4𝑦)−sin(𝑥)



𝑑𝑦
𝐹𝑖𝑛𝑑 𝑑𝑥 𝑖𝑓 cos 4𝑦 + 𝑦 sin 𝑥 = 3
𝑑𝑦
𝑑𝑥
𝑑
𝑑𝑥
[cos(4𝑦) + sin(𝑥 ) 𝑦] =
[cos(4𝑦)] +
𝑑
𝑑
𝑑𝑥
𝑑
𝑑𝑥
[3]
[sin(𝑥 ) 𝑦] = 0
𝑑

(− sin(4𝑦))

(−4)


−4𝑦 ′ sin(4𝑦) + cos(𝑥 ) 𝑦 + sin(𝑥 ) 𝑦 ′ = 0
−4𝑦 ′ sin(4𝑦) + sin(𝑥 ) 𝑦 ′ + cos(𝑥 ) 𝑦 = 0

𝑑𝑦
𝑑𝑥
𝑑
𝑑𝑥
𝑑𝑥
[4𝑦] +
𝑑𝑥
[sin(𝑥 )]𝑦 + (sin(𝑥 ))
𝑑
𝑑𝑥
[𝑦] = 0
(𝑦) sin(4𝑦) + cos(𝑥 ) 𝑦 + sin(𝑥 ) 𝑦 ′ = 0
cos(𝑥)𝑦
𝑦 ′ = 4 sin(4𝑦)−sin(𝑥)
2. 𝑦 = 0


𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
[𝑦 3 + 𝑥 3 = 𝑑𝑥 [8]
𝑑
[𝑦 3 ] +
𝑑
𝑑𝑥
[𝑥 3 ] = 0

(3𝑦 2 )


3𝑦 2 𝑦 ′ + 3𝑥 2 = 0
3(𝑦 2 𝑦 ′ + 𝑥 2 ) = 0

𝑦′ = −

𝐴𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 (0, 2): 𝑦 = − 22 = − 4 = 𝟎
𝑑𝑥
[𝑦] + 3𝑥 2 = 0
𝑥2
𝑦2
02
0
Activity 2: What’s More (Page 22)
Activity 3. Solve the following problems
1
1. 15𝜋 𝑚𝑖𝑙𝑒𝑠 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟



𝐴 = 𝜋𝑟 2
where r = 15
2
𝐴(𝑡) = 𝜋𝑟 (𝑡)
𝑑𝑥 ′
𝐴 (𝑡) = 2𝜋𝑟(𝑡)𝑟 ′(𝑡)


𝑟(𝑡1 ) = 15 𝑚𝑖𝑙𝑒𝑠
𝐴′(𝑡1 ) = 2𝜋𝑟(𝑡1 )𝑟 ′(𝑡1 )
𝑑𝑦
𝐴′ (𝑡1 )
2𝜋𝑟(𝑡1 )
1
2𝑚𝑖.2
𝑦

𝑟′(𝑡1 ) =

𝑟′(𝑡1 ) = 15𝜋

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑡ℎ𝑒 𝑐𝑖𝑡𝑦 ′𝑠 𝑟𝑎𝑑𝑖𝑢𝑠 𝑖𝑠 𝑔𝑟𝑜𝑤𝑖𝑛𝑔 𝑏𝑦
=
2𝜋𝑟(15 𝑚𝑖.)
1
15𝜋
𝑚𝑖𝑙𝑒𝑠 𝑜𝑟 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒𝑙𝑦 0.21 𝑚𝑖𝑙𝑒𝑠 𝑎 𝑦𝑒𝑎𝑟.
2. 𝑟 ′(𝑡) = 40.21 𝑚3 /𝑠
4

𝑦 = 3 𝜋𝑟 2

𝑦(𝑡) = 3 𝜋𝑟 2 (𝑡)

𝑦 ′(𝑡) = 3 𝜋(3𝑟 2 (𝑡)𝑟 ′(𝑡)) = 4𝜋𝑟 2 (𝑡)𝑟 ′(𝑡)




4
4
𝑟(𝑡1 ) = 4𝑚
𝑚
𝑟 ′(𝑡) = 4𝜋(4𝑚)2 (0.2 𝑠 ) = 12.8𝜋
′(
𝑚3
𝑠
= 40.21
𝑚3
𝑠
3
𝑟 𝑡) = 40.21 𝑚 /𝑠
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑡ℎ𝑒 𝑠𝑝ℎ𝑒𝑟𝑒 𝑖𝑠 𝑔𝑟𝑜𝑤𝑖𝑛𝑔 𝑏𝑦 40.21 𝑚3 /𝑠.
5. 1.73 𝑓𝑡/𝑠
𝑑𝑥
3𝑓𝑡

𝐺𝑖𝑣𝑒𝑛: 𝑑𝑡 = 𝑠𝑒𝑐𝑜𝑛𝑑 𝑎𝑛𝑑 𝑐 = 20𝑓𝑡

𝐹𝑖𝑛𝑑: 𝑑𝑡 : 𝑦 = 10


𝑑𝑥
𝑥 2 + 𝑦 2 = 202
𝑥 2 + 𝑦 2 = 400
𝑑𝑥
𝑑𝑥

2𝑥 ( 𝑑𝑡 ) + 2𝑦 ( 𝑑𝑡 ) = 0

𝑊ℎ𝑒𝑛 𝑦 = 10, 𝑥 = √400 − 102 = √300

√300 (2 𝑑𝑡 ) + (10 𝑑𝑥) = 0

(12)

𝑑𝑦

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟 𝑖𝑠 𝑠𝑙𝑖𝑑𝑖𝑛𝑔 𝑑𝑜𝑤𝑛 𝑡ℎ𝑒
𝑑𝑥
𝑑𝑡
=
𝑑𝑦
𝑑𝑦
𝑑𝑥
= (√300) 𝑑𝑡
𝑑𝑡
√300
10
= 1.73 𝑓𝑡/𝑠
𝑏𝑢𝑖𝑙𝑑𝑖𝑛𝑔 𝑎𝑡 𝑎 𝑟𝑎𝑡𝑒𝑜𝑓 1.73 𝑓𝑡 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑.
Activity 3: Assessment (Posttest) (Page 27)
1. B.




𝑦 = √𝑡 3 + 1 when 𝑡 = 2
1
𝑑
(𝑡 3 + 1)
3
2√𝑡 +1 𝑑𝑥
𝑑
(𝑡 3 + 1)
𝑑𝑡
1
′
𝑦 =
2√𝑡 3 +1
𝑦 ′′ =

𝑦 ′′ = √𝑡 3

3𝑡 2
(3𝑥 )2 =
2√𝑡 3 +1
(2√𝑡 3 +1)(6𝑡)−3𝑡2 (


= 3𝑡 2
3𝑥2
2√𝑥3 +1
2
(2√𝑥 3 +1)
12𝑡√𝑡 3 +1−
)
9𝑡4
√𝑡3+1
=
4𝑡 3 +4
=
12𝑡 4 −12𝑡−9𝑡 4
3𝑡 4 −12𝑡
=
(4𝑡 3 +4)√𝑡 3 +1
+1 (4𝑡 3 +4)
3(2)4−12(2)
72
(4(2)3 +4)√(2)3 +1
2
2
3
= 108
𝑢𝑛𝑖𝑡𝑠/sec
2. csc𝑜 𝜃 ′ = −4 cot 𝜃 csc2 𝜃
𝑑
𝑑
𝑑

[csc2 𝜃 + cot 2 𝜃] = 𝑑𝜃 [csc 2 𝜃] + 𝑑𝜃 [cot 2 𝜃]
𝑑𝜃

(2 csc 𝜃 )


2 csc 𝜃 csc 𝜃 (− cot 𝜃)) + 2 cot 𝜃)(− csc2 𝜃)
csc𝑜 𝜃 ′ = −4 cot 𝜃 csc2 𝜃
𝑑
𝑑𝜃
[csc 𝜃 ] + (2 cot 𝜃 )
𝑑
𝑑𝜃
[cot 𝜃]
12𝑡√𝑡 3 +1−9𝑡4
√𝑡 3 +1
1
(4𝑡 3 +4)
3. D
3𝑥

𝑦 = 𝑥2 +1

(3)

𝑦 ′ = (3)

𝑦′ =
𝑑
𝑥
(
)
𝑑𝑥 𝑥 2 +1
𝑥(𝑥 2 +1)−(𝑥 2 +1)𝑥
(𝑥 2 +1)2
= (3)
(𝑥 2 +1)−2𝑥𝑥
(𝑥 2 +1)2
=
3(−𝑥 2 +1)
(𝑥 2 +1)2
3(−𝑥 2 +1)
(𝑥 2 +1)2
4. 𝑓 (𝑥 )′ = 8𝑥 3 + 2𝑥



𝑆𝑜𝑙𝑣𝑒: (𝑥 2 )(2𝑥 2 + 1) = (2𝑥 )(2𝑥 2 + 1) + (4𝑥)(𝑥 2 )
𝑓 (𝑥 )′ = 4𝑥 3 + 2𝑥 + 4𝑥 3
𝑓 (𝑥 )′ = 8𝑥 3 + 2𝑥
5. A


𝑓 ′(𝑥 ) = (9 − 𝑥 2 ) = (9 − 32 ) = 0
𝑓 ′′ (𝑥 ) = 0 = 0
6. 2𝑥 sin(𝑥 ) + 𝑥 2 cos(𝑥)




𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
′
(𝑥 2 ) sin(𝑥 ) +
𝑑
𝑑𝑥
(sin(𝑥 ))𝑥 2
(𝑥 2 ) = 2𝑥
(sin(𝑥 )) = cos(𝑥 )
𝑦 = 2𝑥 sin(𝑥 ) + 𝑥 2 cos(𝑥)
7. 2 sec2 (𝑥 ) tan(𝑥 ) + 2𝑥 𝑠𝑒𝑐 2 (𝑥 2 )




𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
(sec2 (𝑥)) + 𝑑𝑥 (tan(𝑥 2 ))
(sec2 (𝑥)) = 2𝑠𝑒𝑐 2 (𝑥)tan(𝑥)
(tan(𝑥 2 )) = sec2 (𝑥 2 )(2𝑥 )= 2 sec2 (𝑥 ) tan(𝑥 ) + sec2 (𝑥2)(2𝑥)
2 sec2 (𝑥 ) tan(𝑥 ) + 2𝑥 𝑠𝑒𝑐 2 (𝑥 2 )
8. D
𝑑





𝑑 (1+cos(𝑥))(1−cos(𝑥))−𝑑𝑥(1−cos(𝑥))(1+cos(𝑥))
(1−cos(𝑥))2
𝑑𝑥
𝑑
𝑑𝑥
𝑑
(1 + cos(𝑥 )) = − sin(𝑥 )
(1 − cos(𝑥 )) = sin(𝑥 )
𝑑𝑥
𝑑 (− sin(𝑥))(1−cos(𝑥))−sin(𝑥)(1+cos(𝑥))
(
(1−cos(𝑥))2
𝑑𝑥
2
sin(𝑥)
𝑦 ′ = − (1−cos(𝑥))2
9. B
𝑑





3
3
𝑑 (𝑥 +2)𝑥−𝑑𝑥(𝑥 +2)
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑥2
3
(𝑥 + 2) = 3𝑥 2
(𝑥 ) = 1
𝑑𝑥
𝑑 (3𝑥 2 −1)(𝑥 3 +2)
𝑑𝑥
′(
𝑓 𝑡) =
𝑥2
2𝑥 3 −2
𝑥2
=
=
2𝑥 3 −2
𝑥2
2(−2)3−2
(−2)2
7
= −2
2sin(𝑥)
= − (1−cos(𝑥))2
10. C
35 2
𝑡
2

𝑠′ = −

−35(1) + 58 = 23
+ 58𝑡 + 91 = −35𝑡 + 58
11. 2𝑥𝑒 𝑥 2



𝑑
(𝑥 2 ) = 2𝑥
𝑑𝑥
′(
𝑓 𝑡) = 𝑒 𝑥 2 (2𝑥 )
𝑓 ′(𝑡) = 2𝑥𝑒 𝑥 2
12. A



𝑑 1
(sin 2𝜃)2 (2 cos 2𝜃)
𝑑𝑥 2
2 cos 2𝜃
2√sin 2 𝜃
cos 2𝜃
√sin 2𝜃
13. A




𝑓 ′(𝑥 ) = 6𝑥 2 − 3
𝑓 ′′ (𝑥 ) = 12
𝑥 = 1, 𝑡ℎ𝑒𝑛 12(1)
12
14. A

𝑑
𝑑𝑥
(𝑦) = csc(𝑥 ) − cot(𝑥)
15. C


𝑑
1
𝑑
(2) 𝑑𝑥 ((2𝑥 + 𝑒 2𝑥 )−1 ) = − (2𝑥+𝑒2𝑥 ) 𝑑𝑥 (2𝑥 + 𝑒 2𝑥 )
𝑑
𝑑𝑥
(2𝑥 + 𝑒 2𝑥 ) = (2 + 𝑒 2𝑥 )(2)
2(2+𝑒 2𝑥 )
1

2 − (2𝑥+𝑒2𝑥 ) (2𝑥 + 𝑒 2𝑥 ) = − (2𝑥+𝑒2𝑥 )

𝑓 ′(𝑥 ) = − (2𝑥+𝑒2𝑥 )
4(1+𝑒 2𝑥 )
16. C



1
ℎ
ℎ
+ = 1(ℎ)−1 +
2
1
1
1
2
1
−1ℎ−2 + 2 = ℎ2 + 2
ℎ2 −2
2ℎ 2
17. C



𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒: 3𝑡 2 − 2𝑡 + 5 = 6𝑡 − 2
6𝑡 − 2 = 6(2) − 2
10𝑚/𝑠𝑒𝑐
18. A






(𝑡 2 − 1)3 . 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒.
𝑠 ′ = 3(𝑡 2 − 1)2 (2𝑡) = 6𝑡(𝑡 2 − 1)2 = 6𝑡(𝑡 4 − 212 + 1)
𝑠 ′ = 6𝑡 5 − 12𝑡 3 + 6𝑡
𝑑
𝑑
𝑑
(6𝑡 5 ) = 30𝑡 4 , (−12𝑡 3 ) = −36𝑡 2 , & (6𝑡) = 6
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑠 ′′ = 30𝑡 4 − 36𝑡 2 + 6 = 30(2)4 + 36(2)2 + 6
342 𝑢𝑛𝑖𝑡𝑠/𝑠𝑒𝑐 2
19. A



𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒: 6𝜋𝑟 2
𝑑
(6𝜋𝑟 2 ) = 2(6)𝜋𝑟 2−1
𝑑𝑥
𝑦 ′ = 12𝜋𝑟
20. 128𝜋𝑐𝑚3 /𝑠𝑒𝑐
4

𝑉 = 3 𝜋𝑟 3 & 𝑟 = 2𝑡

𝐹𝑖𝑟𝑠𝑡 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒: 𝑉 ′ = 3 𝜋(2𝑡)2 = 3(2𝑡)2 𝑑𝑡 (2𝑡)




𝑑
𝑑𝑡
4
3
4
𝑑
(2𝑡) = 2
𝜋(2𝑡)2 (2) = 32𝜋𝑡 2
𝑑
𝑆𝑒𝑐𝑜𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒: 𝑉 ′′ = 𝑑𝑡 (32𝜋𝑡 2 ) = 64𝜋𝑡
64𝜋(2) = 128𝜋𝑐𝑚3 /𝑠𝑒𝑐