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MAT1512 Assignment 2 2021

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MAT1512 - Calculus A (MAT1512)
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UNISA Calculus A MAT1512 Assignment TWO of 2021 solutions. Where possible, alternative methods are used. The topics covered are: Differentiation. Derivatives. The power rule. Derivatives of polynomials. Derivatives of trigonometric ratios. Differentiating term by term. The chain rule. The product rule. The quotient rule. Simplifying algebraic expressions.

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MAT1512 ASSIGNMENT 2 2021



Question 1



(a) 𝑓(𝑥) = 3𝑥 2 − 4𝑥 + 1

𝑓(𝑥 + ℎ) = 3(𝑥 + ℎ)2 − 4(𝑥 + ℎ) + 1

𝑓(𝑥 + ℎ) = 3(𝑥 2 + 2𝑥ℎ + ℎ2 ) − 4𝑥 − 4ℎ + 1

𝑓(𝑥 + ℎ) = 3𝑥 2 + 6𝑥ℎ + 3ℎ2 − 4𝑥 − 4ℎ + 1


𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 3𝑥 2 + 6𝑥ℎ + 3ℎ2 − 4𝑥 − 4ℎ + 1 − (3𝑥 2 − 4𝑥 + 1)

𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 3𝑥 2 + 6𝑥ℎ + 3ℎ2 − 4𝑥 − 4ℎ + 1 − 3𝑥 2 + 4𝑥 − 1

𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 6𝑥ℎ + 3ℎ2 − 4ℎ


𝑓(𝑥 + ℎ) − 𝑓(𝑥)
𝑓 ′ (𝑥) = lim
ℎ→0 ℎ

6𝑥ℎ + 3ℎ2 − 4ℎ
𝑓 ′ (𝑥) = lim
ℎ→0 ℎ

ℎ(6𝑥 + 3ℎ − 4)
𝑓 ′ (𝑥) = lim
ℎ→0 ℎ

1(6𝑥 + 3ℎ − 4)
𝑓 ′ (𝑥) = lim
ℎ→0 1

𝑓 ′ (𝑥) = lim 6𝑥 + 3ℎ − 4
ℎ→0


𝑓 ′ (𝑥) = 6𝑥 + 3(0) − 4

𝑓 ′ (𝑥) = 6𝑥 + 0 − 4

𝑓 ′ (𝑥) = 6𝑥 − 4

, 2𝑥+1
(b) 𝑓(𝑥) = 𝑥+3



2(𝑥 + ℎ) + 1
𝑓(𝑥 + ℎ) =
(𝑥 + ℎ) + 3

2𝑥 + 2ℎ + 1
𝑓(𝑥 + ℎ) =
𝑥+ℎ+3

2𝑥 + 2ℎ + 1 2𝑥 + 1
𝑓(𝑥 + ℎ) − 𝑓(𝑥) = −
𝑥+ℎ+3 𝑥+3

(2𝑥 + 2ℎ + 1) (2𝑥 + 1)
𝑓(𝑥 + ℎ) − 𝑓(𝑥) = −
(𝑥 + ℎ + 3) (𝑥 + 3)

(2𝑥 + 2ℎ + 1) × (𝑥 + 3) (2𝑥 + 1) × (𝑥 + ℎ + 3)
𝑓(𝑥 + ℎ) − 𝑓(𝑥) = −
(𝑥 + ℎ + 3) × (𝑥 + 3) (𝑥 + 3) × (𝑥 + ℎ + 3)

(2𝑥 + 2ℎ + 1)(𝑥 + 3) − (2𝑥 + 1)(𝑥 + ℎ + 3)
𝑓(𝑥 + ℎ) − 𝑓(𝑥) =
(𝑥 + ℎ + 3)(𝑥 + 3)

𝑥(2𝑥 + 2ℎ + 1) + 3(2𝑥 + 2ℎ + 1) − [2𝑥(𝑥 + ℎ + 3) + 1(𝑥 + ℎ + 3)]
𝑓(𝑥 + ℎ) − 𝑓(𝑥) =
(𝑥 + ℎ + 3)(𝑥 + 3)

2𝑥 2 + 2𝑥ℎ + 𝑥 + 6𝑥 + 6ℎ + 3 − [2𝑥 2 + 2𝑥ℎ + 6𝑥 + 𝑥 + ℎ + 3]
𝑓(𝑥 + ℎ) − 𝑓(𝑥) =
(𝑥 + ℎ + 3)(𝑥 + 3)

2𝑥 2 + 2𝑥ℎ + 7𝑥 + 6ℎ + 3 − [2𝑥 2 + 2𝑥ℎ + 7𝑥 + ℎ + 3]
𝑓(𝑥 + ℎ) − 𝑓(𝑥) =
(𝑥 + ℎ + 3)(𝑥 + 3)

2𝑥 2 + 2𝑥ℎ + 7𝑥 + 6ℎ + 3 − 2𝑥 2 − 2𝑥ℎ − 7𝑥 − ℎ − 3
𝑓(𝑥 + ℎ) − 𝑓(𝑥) =
(𝑥 + ℎ + 3)(𝑥 + 3)

6ℎ − ℎ
𝑓(𝑥 + ℎ) − 𝑓(𝑥) =
(𝑥 + ℎ + 3)(𝑥 + 3)

5ℎ
𝑓(𝑥 + ℎ) − 𝑓(𝑥) =
(𝑥 + ℎ + 3)(𝑥 + 3)

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