Memo (COMPLETE
ANSWERS) Due 5 August
2025
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, Calculus Assignment Solutions
Question 1: Limits (6 Marks)
(1.1) lim(x→−1) (x³ − 2)/(x⁴ + 4x − 3)
= (-1³ − 2)/(-1⁴ + 4(-1) − 3)
= (-3)/(-6)
= 1/2
(1.2) lim(x→3) (x³ − 8)/(x² − 4)
= (27 − 8)/(9 − 4)
= 19/5
Question 2: Differentiation (22 Marks)
(2.1) y = 1 / (x − √(cos x))
Let u = x − √(cos x), then y = u⁻¹
dy/dx = -1/u² * du/dx
du/dx = 1 + sin x / (2√(cos x))
dy/dx = -[1 + sin x / (2√(cos x))] / (x − √(cos x))²
(2.2) y = (x³ + 2)(x² − 2x + 1)
dy/dx = 3x²(x² − 2x + 1) + (x³ + 2)(2x − 2)
= 3x⁴ − 6x³ + 3x² + 2x⁴ − 2x³ + 4x − 4
= 5x⁴ − 8x³ + 3x² + 4x − 4
(2.3) y = √(1 + tan²x) * sec²x
= √(sec²x) * sec²x = sec³x
dy/dx = 3sec²x * secx * tanx = 3sec³x * tanx
(2.4) y = ln(e^x * [(x−2)/(x+3)]^34)
= x + 34 * ln((x−2)/(x+3))
dy/dx = 1 + 34 * [ (x + 3)(1) − (x − 2)(1) ] / (x + 3)²
= 1 + 170 / (x + 3)²