(i) Let f (x) = 2x −

3

4

... (i)
Differentiating (1) with respect to x, we get
f ′(x) = 2·1 – 0 ⇒ f ′(x) = 2.
(ii) Let f(x) = (5x3 + 3x – 1) (x – 1) ... (1)
Differentiating (1) with respect to x, we get
f ′(x)=(5x3 +3x−1)′ (x−1)+(5x3 +3x−1)(x−1)′
⇒ f ′(x) = (5·3x2 + 3 – 0) (x – 1) + (5x3 + 3x – 1) (1 – 0)
= (15x2 + 3) (x – 1) + (5x3 + 3x – 1) (1) = 15x3 + 3x – 15x2 – 3 + 5x3 + 3x – 1
∴ f ′(x) = 20x3–15x2 + 6x – 4.
(iii) Let f(x) = x–3 (5 + 3x) ... (1)
Differentiating (1) with respect to x, we get
f ′(x) = (x–3)′ (5 + 3x) + (x–3) (5 + 3x)′
⇒ f ′(x) = (–3) x–3–1 (5 + 3x) + (x–3)(0 + 3)
= –3x–4 (5 + 3x) + x–3 ·(3) = –15x–4–9x–3+3x–3
= - 15x−4−6x−3 =

−15

x4

−

6

x3

∴ f' (x) = −

3

x2

(5 + 2x).
(iv) Let f(x) = x5(3–6x–9) ... (1)
Differentiating (1) with respect to x, we get
f ′(x) = (x5)′(3–6x–9) + x5(3–6x–9)′
= 5x4(3–6x–9) + x5(0+6·9x–10) = 15x4–30x–5+54x–5
∴ f' (x) = 15x4+24x−5 = 15x4+

24

x5

.
(v) Let f(x) = x−4(3−4x−5) ... (1)
Differentiating (1) with respect to x, we get
f′(x) = (x–4)′(3–4x–5) + x–4(3–4x–5)′
⇒ f ′(x) = –4x–5(3–4x–5) + x–4(0+20x–6) = –12x–5 + 16x–10+20x–10
= –12x–5+36x–10
∴ f' (x) =

−12

x5

+

36

x10

.
(vi) Let f (x) =

2

x+1

−

x2

3x−1

... (1)
Differentiating (1) with respect to x, we get
f' (x) = [

(x+1)(2)′−(2).(x+1)′

(x+1)2

] - [

(3x−1)(x2)′−x2(3x−1)′

(3x−1)2

]
=

−2

(x+1)2

- [

(3x−1)(2x)−x2(3)

(3x−1)2

]

−2

(x+1)2

+[

6x2−2x−3x2

(3x−1)2

]
=

−2

(x+1)2

−[

3x2−2x

(3x−1)2

]
∴ f' (x) =

−2

(x+1)2

−

x(3x−2)

(3x−1)2

.