University of South Africa DSC 1520 Assignment 3 Solutions| All Correct Answers

DSC1520 ASSIGNMENT 3
POSSIBLE SOLUTIONS

Question 1
Find the derivative of the function: 𝐺(𝗑) = 𝗑(𝗑2 − 4√𝗑 + 4)
1
Replace √𝑥 with 𝑥2, expand the brackets and simplify before differentiating
1
𝐺(𝑥) = 𝑥(𝑥2 − 4𝑥2 + 4)
3 3
𝐺(𝑥) = 𝑥 − 4𝑥2 + 4𝑥

Apply the “Power Rule” of differentiation.

If 𝐺(𝗑) = 𝗑𝑛 then 𝐺′(𝗑) = 𝑛𝗑𝑛−1

Also, if 𝐺(𝗑) = 𝑎𝗑𝑛 then 𝐺′(𝗑) = 𝑎𝑛𝗑𝑛−1

Note: The derivative of 𝑎𝗑 is 𝑎 and the derivative of a constant term 𝑐 is 0.

3 3
𝐺′(𝑥) = 3𝑥3−1 − 4 ( ) 𝑥2−1 + 4
2
1
′ 2
𝐺 (𝑥) = 3𝑥 − 6𝑥2 + 4
1
Replace 𝑥2 with √𝑥

𝐺′(𝑥) = 3𝑥2 − 6√𝑥 + 4




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Question 2
Differentiate the function
𝗑2 + 6
𝑓(𝗑) =
2𝗑 + 5
Apply the “Quotient rule” of differentiation.

Let 𝑓(𝗑) = 𝑢
𝑣

𝑢 = 𝑥2 + 6; 𝑑𝑢 = 2𝑥 and 𝑣 = 2𝑥 + 5; 𝑑𝑣 = 2
𝑣𝑑𝑢 − 𝑢𝑑𝑣
𝑓′(𝗑) =
𝑣2
(2𝑥 + 5)2𝑥 − (𝑥2 + 6)2
ƒ′(𝑥) =
(2𝑥 + 5)2

4𝑥2 + 10𝑥 − 2𝑥2 − 12
ƒ′(𝑥) =
(2𝑥 + 5)2

4𝑥2 − 2𝑥2 + 10𝑥 − 12
ƒ′(𝑥) =
(2𝑥 + 5)2

2𝑥2 + 10𝑥 − 12
ƒ′(𝑥) =
(2𝑥 + 5)2
Factor out 2, the common factor on the numerator

2(𝑥2 + 5𝑥 − 6)
ƒ′(𝑥) =
(2𝑥 + 5)2
Now, factorize the bracket on the numerator
2(𝑥 + 6)(𝑥 − 1)
ƒ′(𝑥) =
(2𝑥 + 5)2


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Question 3
Find the derivative of the function
𝗑+1
𝑃(𝗑) = 𝗑5𝑒3𝗑 +
𝗑
Apply the “Product rule” on 𝗑5𝑒3𝗑 and “Quotient rule” on 𝗑+1
𝗑

Let 𝑃(𝑥) = ƒ(𝑥) + 𝑔(𝑥) where

ƒ(𝑥) = 𝑥5𝑒3𝑥 and 𝑔(𝑥) = 𝑥+1
𝑥

𝑃′(𝑥) = ƒ′(𝑥) + 𝑔′(𝑥)

ƒ(𝑥) = 𝑢𝑣 where

𝑢 = 𝑥5; 𝑑𝑢 = 5𝑥4 and 𝑣 = 𝑒3𝑥; 𝑑𝑣 = 3𝑒3𝑥

𝑓′(𝗑) = 𝑢𝑑𝑣 + 𝑣𝑑𝑢 [Product rule]

ƒ′(𝑥) = 𝑥5. 3𝑒3𝑥 + 𝑒3𝑥. 5𝑥4

ƒ′(𝑥) = 3𝑥5𝑒3𝑥 + 5𝑥4𝑒3𝑥 = 𝑒3𝑥(3𝑥5 + 5𝑥4)

𝑔(𝑥) = 𝑢 where
𝑣

𝑢 = 𝑥 + 1; 𝑑𝑢 = 1 and 𝑣 = 𝑥; 𝑑𝑣 = 1
𝑣𝑑𝑢 − 𝑢𝑑𝑣
𝑔′(𝑥) =
𝑣2
𝑥(1) − (𝑥 + 1)1 𝑥−𝑥−1 1
𝑔′(𝑥) = = =−
𝑥2 𝑥2 𝑥2
1
∴ 𝑃′(𝑥) = 𝑒3𝑥(3𝑥5 + 5𝑥4) −
𝑥2



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Directors
Mr. Kingston Mangwadu – B. Sc. (Hons) Civil Engineering III (University of Zimbabwe) Contact: +27 83 427 5621
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