Calculus A
MAT1512
Assignment 4
Question 1
2
(a) 𝑧 = 𝑙𝑛(𝑥 + 𝑡 )
∂𝑧 1
𝑧𝑥 = ∂𝑥
= 2
(𝑥+𝑡 )
∂𝑧 2𝑡
𝑧𝑡 = ∂𝑡
= 2
(𝑥+𝑡 )
𝑥
𝑡
(b) 𝐹(𝑥, 𝑦) = ∫ 𝑐𝑜𝑠(𝑒 )𝑑𝑡
𝑦
𝑥
∂𝐹 𝑡
∂𝑥
= ∫ 𝑐𝑜𝑠(𝑒 )𝑑𝑡
𝑦
∂𝐹 𝑥
∂𝑥
= 𝑐𝑜𝑠(𝑒 )
Now
𝑥
∂𝐹 𝑡
∂𝑦
= ∫ 𝑐𝑜𝑠(𝑒 )𝑑𝑡
𝑦
−𝑦
𝑡
= ∫ 𝑐𝑜𝑠(𝑒 )𝑑𝑡
𝑥
𝑦
=− 𝑐𝑜𝑠(𝑒 )
2 −𝑥𝑧
(c) 𝐹(𝑥, 𝑦, 𝑧) = 𝑥𝑦 𝑒
, ∂𝐹 2 −𝑥𝑧 2 −𝑥𝑧
∂𝑥
=𝑦 𝑒 − 𝑥𝑧𝑦 𝑒
2 −𝑥𝑧
=𝑦 𝑒 (1 − 𝑥𝑧)
∂𝐹 −𝑥𝑧
∂𝑦
= 2𝑥𝑦𝑒
∂𝐹 2 2 −𝑥𝑧
∂𝑧
=− 𝑥 𝑦 𝑒
Question 2
(a) 𝑈 = 𝑙𝑛(𝑥 + 2𝑦)
1
𝑈𝑥(𝑥, 𝑦) = 𝑥+2𝑦
2
𝑈𝑦(𝑥, 𝑦) = 𝑥+2𝑦
∂ 1 2
𝑈𝑥𝑦(𝑥, 𝑦) = ∂𝑦
( 𝑥+2𝑦 ) =− 2
(𝑥+2𝑦)
∂ 2 2
𝑈𝑦𝑥(𝑥, 𝑦) = ∂𝑥
( 𝑥+2𝑦 ) =− 2
(𝑥+2𝑦)
Therefore 𝑈𝑥𝑦 = 𝑈𝑦𝑥
𝑥𝑦
(b) 𝑈 = 𝑒 𝑠𝑖𝑛𝑦
MAT1512
Assignment 4
Question 1
2
(a) 𝑧 = 𝑙𝑛(𝑥 + 𝑡 )
∂𝑧 1
𝑧𝑥 = ∂𝑥
= 2
(𝑥+𝑡 )
∂𝑧 2𝑡
𝑧𝑡 = ∂𝑡
= 2
(𝑥+𝑡 )
𝑥
𝑡
(b) 𝐹(𝑥, 𝑦) = ∫ 𝑐𝑜𝑠(𝑒 )𝑑𝑡
𝑦
𝑥
∂𝐹 𝑡
∂𝑥
= ∫ 𝑐𝑜𝑠(𝑒 )𝑑𝑡
𝑦
∂𝐹 𝑥
∂𝑥
= 𝑐𝑜𝑠(𝑒 )
Now
𝑥
∂𝐹 𝑡
∂𝑦
= ∫ 𝑐𝑜𝑠(𝑒 )𝑑𝑡
𝑦
−𝑦
𝑡
= ∫ 𝑐𝑜𝑠(𝑒 )𝑑𝑡
𝑥
𝑦
=− 𝑐𝑜𝑠(𝑒 )
2 −𝑥𝑧
(c) 𝐹(𝑥, 𝑦, 𝑧) = 𝑥𝑦 𝑒
, ∂𝐹 2 −𝑥𝑧 2 −𝑥𝑧
∂𝑥
=𝑦 𝑒 − 𝑥𝑧𝑦 𝑒
2 −𝑥𝑧
=𝑦 𝑒 (1 − 𝑥𝑧)
∂𝐹 −𝑥𝑧
∂𝑦
= 2𝑥𝑦𝑒
∂𝐹 2 2 −𝑥𝑧
∂𝑧
=− 𝑥 𝑦 𝑒
Question 2
(a) 𝑈 = 𝑙𝑛(𝑥 + 2𝑦)
1
𝑈𝑥(𝑥, 𝑦) = 𝑥+2𝑦
2
𝑈𝑦(𝑥, 𝑦) = 𝑥+2𝑦
∂ 1 2
𝑈𝑥𝑦(𝑥, 𝑦) = ∂𝑦
( 𝑥+2𝑦 ) =− 2
(𝑥+2𝑦)
∂ 2 2
𝑈𝑦𝑥(𝑥, 𝑦) = ∂𝑥
( 𝑥+2𝑦 ) =− 2
(𝑥+2𝑦)
Therefore 𝑈𝑥𝑦 = 𝑈𝑦𝑥
𝑥𝑦
(b) 𝑈 = 𝑒 𝑠𝑖𝑛𝑦
