MAT1512 Assignment 6 2021

MAT1512 ASSIGNMENT 6 2021



Question 1



(𝑎) 𝑧 = ln(𝑥 + 𝑡 2 )


𝜕𝑧 𝜕
= (ln(𝑥 + 𝑡 2 ))
𝜕𝑥 𝜕𝑥
𝜕𝑧 1 𝜕
= 2
× (𝑥 + 𝑡 2 )
𝜕𝑥 𝑥 + 𝑡 𝜕𝑥
𝜕𝑧 1
= × (1 + 0)
𝜕𝑥 𝑥 + 𝑡 2
𝜕𝑧 1
=
𝜕𝑥 𝑥 + 𝑡 2


𝜕𝑧 𝜕
= (ln(𝑥 + 𝑡 2 ))
𝜕𝑡 𝜕𝑡
𝜕𝑧 1 𝜕
= 2
× (𝑥 + 𝑡 2 )
𝜕𝑡 𝑥 + 𝑡 𝜕𝑡
𝜕𝑧 1
= × (0 + 2𝑡)
𝜕𝑡 𝑥 + 𝑡 2
𝜕𝑧 2𝑡
=
𝜕𝑡 𝑥 + 𝑡 2


𝑥

(𝑏) 𝐹(𝑥, 𝑦) = ∫ cos(𝑒 𝑡 ) 𝑑𝑡
𝑦




𝑥
𝜕𝐹 𝜕
= (∫ cos(𝑒 𝑡 ) 𝑑𝑡)
𝜕𝑥 𝜕𝑥
𝑦

𝜕𝐹 𝜕 𝜕
= cos(𝑒 𝑥 ) × (𝑥) − cos(𝑒 𝑦 ) × (𝑦)
𝜕𝑥 𝜕𝑥 𝜕𝑥
𝜕𝐹
= cos(𝑒 𝑥 ) × 1 − cos(𝑒 𝑦 ) × 0
𝜕𝑥

, 𝜕𝐹
= cos(𝑒 𝑥 )
𝜕𝑥


𝑥
𝜕𝐹 𝜕
= (∫ cos(𝑒 𝑡 ) 𝑑𝑡)
𝜕𝑦 𝜕𝑦
𝑦

𝜕𝐹 𝜕 𝜕
= cos(𝑒 𝑥 ) × (𝑥) − cos(𝑒 𝑦 ) × (𝑦)
𝜕𝑦 𝜕𝑦 𝜕𝑦
𝜕𝐹
= cos(𝑒 𝑥 ) × 0 − cos(𝑒 𝑦 ) × 1
𝜕𝑦
𝜕𝐹
= − cos(𝑒 𝑦 )
𝜕𝑦



(𝑐) 𝑓(𝑥, 𝑦, 𝑧) = 𝑥𝑦 2 𝑒 −𝑥𝑧


𝜕𝑓 𝜕
= (𝑥𝑦 2 𝑒 −𝑥𝑧 ) (𝑂𝑛𝑙𝑦 𝑥 𝑖𝑠 𝑎 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒. 𝑦 𝑎𝑛𝑑 𝑧 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑡𝑟𝑒𝑎𝑡𝑒𝑑 𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠)
𝜕𝑥 𝜕𝑥
𝜕𝑓 𝜕
= 𝑦2 × (𝑥𝑒 −𝑥𝑧 ) (𝑇ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑟𝑢𝑙𝑒 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑢𝑠𝑒𝑑 𝑜𝑛 𝑥 × 𝑒 −𝑥𝑧 )
𝜕𝑥 𝜕𝑥
𝜕𝑓 𝜕 𝜕 −𝑥𝑧
= 𝑦 2 [ (𝑥) × 𝑒 −𝑥𝑧 + 𝑥 × (𝑒 )]
𝜕𝑥 𝜕𝑥 𝜕𝑥
𝜕𝑓 𝜕
= 𝑦 2 [1 × 𝑒 −𝑥𝑧 + 𝑥 × 𝑒 −𝑥𝑧 × (−𝑥𝑧)]
𝜕𝑥 𝜕𝑥
𝜕𝑓
= 𝑦 2 [𝑒 −𝑥𝑧 + 𝑥 × 𝑒 −𝑥𝑧 × (−1𝑧)]
𝜕𝑥
𝜕𝑓
= 𝑦 2 [𝑒 −𝑥𝑧 − 𝑥𝑧𝑒 −𝑥𝑧 ]
𝜕𝑥
𝜕𝑓
= 𝑦 2 𝑒 −𝑥𝑧 − 𝑥𝑦 2 𝑧𝑒 −𝑥𝑧
𝜕𝑥


𝜕𝑓 𝜕
= (𝑥𝑦 2 𝑒 −𝑥𝑧 ) (𝑂𝑛𝑙𝑦 𝑦 𝑖𝑠 𝑎 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒. 𝑥 𝑎𝑛𝑑 𝑧 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑡𝑟𝑒𝑎𝑡𝑒𝑑 𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠)
𝜕𝑦 𝜕𝑦
𝜕𝑓 𝜕 2
= 𝑥𝑒 −𝑥𝑧 × (𝑦 )
𝜕𝑦 𝜕𝑦
𝜕𝑓
= 𝑥𝑒 −𝑥𝑧 × 2𝑦
𝜕𝑦