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MAT1512 Assignment 4 2021
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UNISA MAT1512 Calculus A Assignment FOUR solutions of 2021. Topics covered: Implicit differentiation. The chain rule. The product rule. The Mean Value Theorem. (BONUS: It is shown that each "c" value lies in the appropriate interval.) Equations of tangent lines. Equations of normal lines. Th...

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  • May 29, 2021
  • May 31, 2021
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  • 2021/2022
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MAT1512 ASSIGNMENT 4 2021


𝑑𝑦
𝑁𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 = 𝑦′
𝑑𝑥
Question 1



(a) 𝑥𝑠𝑖𝑛𝑦 + 𝑦𝑠𝑖𝑛𝑥 = 1


𝑑 𝑑
(𝑥𝑠𝑖𝑛𝑦 + 𝑦𝑠𝑖𝑛𝑥) = (1)
𝑑𝑥 𝑑𝑥


𝑑 𝑑
(𝑥𝑠𝑖𝑛𝑦) + (𝑦𝑠𝑖𝑛𝑥) = 0
𝑑𝑥 𝑑𝑥


𝑑 𝑑 𝑑 𝑑
(𝑥) × 𝑠𝑖𝑛𝑦 + 𝑥 × (𝑠𝑖𝑛𝑦) + (𝑦) × 𝑠𝑖𝑛𝑥 + 𝑦 × (𝑠𝑖𝑛𝑥) = 0
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥


𝑑𝑦 𝑑𝑦
1 × 𝑠𝑖𝑛𝑦 + 𝑥 × 𝑐𝑜𝑠𝑦 × + 𝑠𝑖𝑛𝑥 + 𝑦𝑐𝑜𝑠𝑥 = 0
𝑑𝑥 𝑑𝑥


𝑠𝑖𝑛𝑦 + 𝑥𝑐𝑜𝑠𝑦𝑦 ′ + 𝑦 ′ 𝑠𝑖𝑛𝑥 + 𝑦𝑐𝑜𝑠𝑥 = 0



𝑥𝑐𝑜𝑠𝑦𝑦 ′ + 𝑦 ′ 𝑠𝑖𝑛𝑥 = −𝑠𝑖𝑛𝑦 − 𝑦𝑐𝑜𝑠𝑥



𝑦 ′ [𝑥𝑐𝑜𝑠𝑦 + 𝑠𝑖𝑛𝑥] = −𝑠𝑖𝑛𝑦 − 𝑦𝑐𝑜𝑠𝑥


−𝑠𝑖𝑛𝑦 − 𝑦𝑐𝑜𝑠𝑥
𝑦′ =
[𝑥𝑐𝑜𝑠𝑦 + 𝑠𝑖𝑛𝑥]


𝑠𝑖𝑛𝑦 + 𝑦𝑐𝑜𝑠𝑥
𝑦′ = −
𝑥𝑐𝑜𝑠𝑦 + 𝑠𝑖𝑛𝑥

, 𝑦
(b) tan(𝑥 − 𝑦) =
1+𝑥 2


𝑑 𝑑 𝑦
(tan(𝑥 − 𝑦)) = ( )
𝑑𝑥 𝑑𝑥 1 + 𝑥 2

𝑑 𝑑
𝑑 (𝑦) × (1 + 𝑥 2 ) − 𝑦 × (1 + 𝑥 2 )
sec 2 (𝑥
− 𝑦) × (𝑥 − 𝑦) = 𝑑𝑥 𝑑𝑥
𝑑𝑥 (1 + 𝑥 2 )2

2 (𝑥
𝑦 ′ (1 + 𝑥 2 ) − 𝑦(0 + 2𝑥)
sec − 𝑦) × (1 − 𝑦 ′ ) =
(1 + 𝑥 2 )2

2 (𝑥 2 (𝑥 ′
𝑦 ′ (1 + 𝑥 2 ) − 𝑦(2𝑥)
sec − 𝑦) × 1 − sec − 𝑦) × 𝑦 =
(1 + 𝑥 2 )2

𝑦 ′ (1 + 𝑥 2 ) 2𝑥𝑦
sec 2(𝑥 − 𝑦) − sec 2(𝑥 − 𝑦) 𝑦 ′ = 2 2
−
(1 + 𝑥 ) (1 + 𝑥 2 )2

𝑦′ 2𝑥𝑦
sec 2(𝑥 − 𝑦) − sec 2(𝑥 − 𝑦) 𝑦 ′ = −
(1 + 𝑥 ) (1 + 𝑥 2 )2
2


𝑦′ 2𝑥𝑦
− 2
− sec 2 (𝑥 − 𝑦) 𝑦 ′ = − − sec 2 (𝑥 − 𝑦)
(1 + 𝑥 ) (1 + 𝑥 2 )2

1 2𝑥𝑦
𝑦 ′ [− − sec 2 (𝑥
− 𝑦)] = − − sec 2(𝑥 − 𝑦)
(1 + 𝑥 2 ) (1 + 𝑥 2 )2


1 (1 + 𝑥 2 ) sec 2(𝑥 − 𝑦) 2𝑥𝑦 (1 + 𝑥 2 )2 sec 2(𝑥 − 𝑦)
𝑦 ′ [− − ] = − −
(1 + 𝑥 2 ) (1 + 𝑥 2 ) (1 + 𝑥 2 )2 (1 + 𝑥 2 )2

−1 − (1 + 𝑥 2 ) sec 2(𝑥 − 𝑦) −2𝑥𝑦 − (1 + 𝑥 2 )2 sec 2(𝑥 − 𝑦)
𝑦′ [ ] =
(1 + 𝑥 2 ) (1 + 𝑥 2 )2

−2𝑥𝑦 − (1 + 𝑥 2 )2 sec 2(𝑥 − 𝑦) −1 − (1 + 𝑥 2 ) sec 2 (𝑥 − 𝑦)
𝑦′ = ÷ [ ]
(1 + 𝑥 2 )2 (1 + 𝑥 2 )

−2𝑥𝑦 − (1 + 𝑥 2 )2 sec 2(𝑥 − 𝑦) (1 + 𝑥 2 )
𝑦′ = × [ ]
(1 + 𝑥 2 )2 −1 − (1 + 𝑥 2 ) sec 2 (𝑥 − 𝑦)

−2𝑥𝑦 − (1 + 𝑥 2 )2 sec 2(𝑥 − 𝑦) 1
𝑦′ = 2
×[ ]
(1 + 𝑥 ) −1 − (1 + 𝑥 ) sec 2(𝑥 − 𝑦)
2

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