(i) 2x - 
(ii) (5x3 + 3x – 1)(x – 1)
(iii) x-3(5 + 3x)
(v) x-4(3 – 4x-5)
(vi) 
f
(ii) (5x3 + 3x – 1)(x – 1)
(iii) x-3(5 + 3x)
(vi)
f
(i) Let f(x) = 2x -
f’(x) = 
= 
= 
= 2
(ii) Let f(x) = (5x3 + 3x – 1)(x – 1)
f’(x) = (x – 1) 
+ (5x3 + 3x – 1)
= (x – 1) 
= (x – 1) 
= (x – 1) [5(3x3-1) – 3] + (5x3 + 3x – 1)
= (x – 1) [15x2 – 3] + (5x3 + 3x – 1)
= 15x3 – 3x – 15x2 + 3 + 5x2 + 3x – 1
= 20x3 – 15x2 + 2
(iii) Let f(x) = x-3(5 + 3x)
f’(x) = 
= x-3
= x-3
= x-3
= 3x-3 - (15 – 9x) x-4
= 
= 
= 
(iv) Let f(x) = x5(3 – 6x-9) = 3x5 – 6x-4
f’(x) = 
= 
= 
= (3)(5) x4 – (6) (-4) x-5
= 15x4 + 24x-5
(v) Let f(x) = x-4(3 – 4x-5)
f’(x) = x-4
= x-4 
= x-4[0 – (4)(-5)x6] + (3 - 4x-5)(-4)x-5
= x-4[20 x6] – x-5(12 - 16x5)(-4)x-5
= 20 x-10 – 12x-5 + 16x-10 = 36 x-10 – 12 x-5
(vi) Let f(x) = 
f’(x) = 
= 
- 
= 
- 
= 
= 
- 
= 
- 
= 
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for some constant a.