MAT2691 Semester 1 Assignment 1 2021

MAT2691 SEMESTER 1 ASSIGNMENT 1 2021



Question 1



1.1 𝑦 = cosh−1(𝑠𝑒𝑐𝑥)

𝐿𝑒𝑡 𝑓(𝑥) = 𝑠𝑒𝑐𝑥
𝑓 ′ (𝑥) = 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥

𝑦 = cosh−1(𝑓(𝑥))

𝑑𝑦 𝑓 ′ (𝑥)
=
𝑑𝑥 √[𝑓(𝑥)]2 − 1

𝑑𝑦 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
=
𝑑𝑥 √[𝑠𝑒𝑐𝑥]2 − 1

𝑑𝑦 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
=
𝑑𝑥 √𝑠𝑒𝑐 2 𝑥 − 1

𝑑𝑦 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
=
𝑑𝑥 √1 + tan2 𝑥 − 1

𝑑𝑦 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
=
𝑑𝑥 √tan2 𝑥

𝑑𝑦 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥
=
𝑑𝑥 𝑡𝑎𝑛𝑥

𝑑𝑦
= 𝑠𝑒𝑐𝑥
𝑑𝑥

1.2 𝑦 = 2𝑠𝑖𝑛𝑥

𝐿𝑒𝑡 𝑓(𝑥) = 𝑠𝑖𝑛𝑥
𝑓 ′ (𝑥) = 𝑐𝑜𝑠𝑥

𝑦 = 2 𝑓(𝑥)

2 𝑓(𝑥)
𝑦 = 𝑙𝑛2 [ ]
𝑙𝑛2

𝑑𝑦 𝑑 2 𝑓(𝑥)
= 𝑙𝑛2 [ ( )]
𝑑𝑥 𝑑𝑥 𝑙𝑛2

𝑑𝑦
= 𝑙𝑛2[𝑓 ′ (𝑥)2𝑓(𝑥) ]
𝑑𝑥

, 𝑑𝑦
= 𝑙𝑛2[𝑐𝑜𝑠𝑥2𝑠𝑖𝑛𝑥 ]
𝑑𝑥

𝑑𝑦
= 𝑙𝑛2(𝑐𝑜𝑠𝑥)2𝑠𝑖𝑛𝑥
𝑑𝑥
𝑥
1.3 𝑦 2 = 𝑦+1

𝑊𝑒 𝑐𝑎𝑛𝑛𝑜𝑡 𝑚𝑎𝑘𝑒 𝑦 𝑡ℎ𝑒 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎. 𝑇𝑜 𝑢𝑠𝑒 𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑖𝑜𝑛.
𝑑𝑦
𝑁𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 = 𝑦′
𝑑𝑥

𝑥
[𝑦]2 =
𝑦+1

𝐿𝑒𝑡 𝑓(𝑥) = 𝑥 𝑎𝑛𝑑 𝑔(𝑥) = 𝑦 + 1

𝑓 ′ (𝑥) = 1 𝑎𝑛𝑑 𝑔′ (𝑥) = 𝑦′ + 0

𝑓 ′ (𝑥) = 1 𝑎𝑛𝑑 𝑔′ (𝑥) = 𝑦 ′

𝑓
[𝑦]2 =
𝑔

𝑔𝑓 ′ − 𝑓𝑔′
2[𝑦]2−1 𝑦 ′ =
𝑔2

(𝑦 + 1)1 − 𝑥𝑦′
2[𝑦]1 𝑦 ′ =
(𝑦 + 1)2

𝑦 + 1 − 𝑥𝑦′
2𝑦𝑦 ′ =
(𝑦 + 1)2

𝑦 + 1 − 𝑥𝑦′
2𝑦𝑦 ′ × (𝑦 + 1)2 = × (𝑦 + 1)2
(𝑦 + 1)2

2𝑦𝑦 ′ (𝑦 + 1)2 = 𝑦 + 1 − 𝑥𝑦′

2𝑦𝑦 ′ (𝑦 + 1)2 + 𝑥𝑦′ = 𝑦 + 1

𝑦 ′ [2𝑦(𝑦 + 1)2 + 𝑥] = 𝑦 + 1

𝑦 ′ [2𝑦(𝑦 + 1)2 + 𝑥] 𝑦+1
2
=
[2𝑦(𝑦 + 1) + 𝑥] [2𝑦(𝑦 + 1)2 + 𝑥]

𝑦+1
𝑦′ =
[2𝑦(𝑦 + 1)2 + 𝑥]

𝑑𝑦 𝑦+1
=
𝑑𝑥 [2𝑦(𝑦 + 1)2 + 𝑥]