JC TUTORIALS
MAT2691
ASSIGNMENT 1
SEMESTER 1
2022
, QUESTION 1
QUESTION 1.1
Solution:
𝑦 = tan(𝑐𝑜𝑠 2 (𝑥))
𝑑𝑢
𝐿𝑒𝑡 𝑢 = 𝑐𝑜𝑠 2 (𝑥) , 𝑡ℎ𝑒𝑛 = −2𝑐𝑜𝑠(𝑥) sin(𝑥)
𝑑𝑥
𝑦 = tan(𝑢)
𝑑𝑦
= 𝑠𝑒𝑐 2 (𝑢)
𝑑𝑢
𝑑𝑦 𝑑𝑦 𝑑𝑢
= ×
𝑑𝑥 𝑑𝑢 𝑑𝑥
𝑑𝑦
= 𝑠𝑒𝑐 2 (𝑢) × (−2𝑐𝑜𝑠(𝑥) sin(𝑥))
𝑑𝑥
𝑑𝑦
= 𝑠𝑒𝑐 2 (𝑐𝑜𝑠 2 (𝑥)) × (−2𝑐𝑜𝑠(𝑥) sin(𝑥))
𝑑𝑥
𝑑𝑦
= −2𝑐𝑜𝑠(𝑥) sin(𝑥) 𝑠𝑒𝑐 2 (𝑐𝑜𝑠 2 (𝑥)) ∴ 2𝑐𝑜𝑠(𝑥) sin(𝑥) = sin(2𝑥)
𝑑𝑥
𝒅𝒚
= 𝐬𝐢𝐧(𝟐𝒙) 𝒔𝒆𝒄𝟐 (𝒄𝒐𝒔𝟐 (𝒙))
𝒅𝒙
QUESTION 1.2
, Solution:
𝑥 = 𝑐𝑜𝑠 −1 (√𝑡)
𝐿𝑒𝑡 𝑢 = √𝑡
=
� −1
� = 𝑐𝑜𝑠 (𝑢)
𝑥
1
𝑑𝑥 1
2 =−
𝑑𝑢 √1 − 𝑢2
�
�
�
𝑑𝑥 𝑑𝑥 𝑑𝑢
� = ×
𝑑𝑡 𝑑𝑢 𝑑𝑡
�
𝑑𝑥
� =− 1 1
×
𝑑𝑡
� √1 − 𝑢2 2√𝑡
�
𝑑𝑥 1 1
=− ×
𝑑𝑡 2 2√𝑡
1 √1 − (√𝑡)
2
𝑑𝑥
� 1 1
=− ×
�𝑑𝑡 √1 − 𝑡 2 2√𝑡
𝑑𝑥 1 1
=− ∴𝑡=
𝑑𝑡 2√𝑡(1 − 𝑡 2 ) 2
𝑑𝑥 1
=−
𝑑𝑡
1 1 2
2 (√( ) (1 − ( ) ))
2 2
𝑑𝑥 1
=−
𝑑𝑡 3
2 (√8)
𝒅𝒙 √𝟔
=
𝒅𝒕 𝟑
MAT2691
ASSIGNMENT 1
SEMESTER 1
2022
, QUESTION 1
QUESTION 1.1
Solution:
𝑦 = tan(𝑐𝑜𝑠 2 (𝑥))
𝑑𝑢
𝐿𝑒𝑡 𝑢 = 𝑐𝑜𝑠 2 (𝑥) , 𝑡ℎ𝑒𝑛 = −2𝑐𝑜𝑠(𝑥) sin(𝑥)
𝑑𝑥
𝑦 = tan(𝑢)
𝑑𝑦
= 𝑠𝑒𝑐 2 (𝑢)
𝑑𝑢
𝑑𝑦 𝑑𝑦 𝑑𝑢
= ×
𝑑𝑥 𝑑𝑢 𝑑𝑥
𝑑𝑦
= 𝑠𝑒𝑐 2 (𝑢) × (−2𝑐𝑜𝑠(𝑥) sin(𝑥))
𝑑𝑥
𝑑𝑦
= 𝑠𝑒𝑐 2 (𝑐𝑜𝑠 2 (𝑥)) × (−2𝑐𝑜𝑠(𝑥) sin(𝑥))
𝑑𝑥
𝑑𝑦
= −2𝑐𝑜𝑠(𝑥) sin(𝑥) 𝑠𝑒𝑐 2 (𝑐𝑜𝑠 2 (𝑥)) ∴ 2𝑐𝑜𝑠(𝑥) sin(𝑥) = sin(2𝑥)
𝑑𝑥
𝒅𝒚
= 𝐬𝐢𝐧(𝟐𝒙) 𝒔𝒆𝒄𝟐 (𝒄𝒐𝒔𝟐 (𝒙))
𝒅𝒙
QUESTION 1.2
, Solution:
𝑥 = 𝑐𝑜𝑠 −1 (√𝑡)
𝐿𝑒𝑡 𝑢 = √𝑡
=
� −1
� = 𝑐𝑜𝑠 (𝑢)
𝑥
1
𝑑𝑥 1
2 =−
𝑑𝑢 √1 − 𝑢2
�
�
�
𝑑𝑥 𝑑𝑥 𝑑𝑢
� = ×
𝑑𝑡 𝑑𝑢 𝑑𝑡
�
𝑑𝑥
� =− 1 1
×
𝑑𝑡
� √1 − 𝑢2 2√𝑡
�
𝑑𝑥 1 1
=− ×
𝑑𝑡 2 2√𝑡
1 √1 − (√𝑡)
2
𝑑𝑥
� 1 1
=− ×
�𝑑𝑡 √1 − 𝑡 2 2√𝑡
𝑑𝑥 1 1
=− ∴𝑡=
𝑑𝑡 2√𝑡(1 − 𝑡 2 ) 2
𝑑𝑥 1
=−
𝑑𝑡
1 1 2
2 (√( ) (1 − ( ) ))
2 2
𝑑𝑥 1
=−
𝑑𝑡 3
2 (√8)
𝒅𝒙 √𝟔
=
𝒅𝒕 𝟑