(i) Let f (x) = 2x − $\frac{3}{4}$ ... (i) Differentiating (1) with respect to x, we get f ′(x) = 2·1 – 0 ⇒ f ′(x) = 2. (ii) Let f(x) = ($5x^3$ + 3x – 1) (x – 1) ... (1) Differentiating (1) with respect to x, we get f ′(x)=($5x^3$ +3x−1)′ (x−1)+($5x^3$ +3x−1)(x−1)′ ⇒ f ′(x) = (5·$3x^2$ + 3 – 0) (x – 1) + ($5x^3$ + 3x – 1) (1 – 0) = ($15x^2$ + 3) (x – 1) + ($5x^3$ + 3x – 1) (1) = $15x^3$ + 3x – $15x^2$ – 3 + $5x^3$ + 3x – 1 ∴ f ′(x) = $20x^3 - 15x^2$ + 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) = $-15x^{-4} - 9x^{-3} + 3x^{-3}$ = - $15x^{-4} - 6x^{-3}$ = $\frac{-15}{x^4} - \frac{6}{x^3}$ ∴ f' (x) = $-\frac{3}{x^2}$ (5 + 2x). (iv) Let f(x) = $x^5 (3 - 6x^{-9})$ ... (1) Differentiating (1) with respect to x, we get f ′(x) = $(x5)' (3 - 6x^{-9})$ + $x^5 (3 - 6x^{-9})'$ = $5x^4 (3 - 6x^{-9})$ + $x^5 (0 + 6 \cdot 9 x^{-10})$ = $15x^4 - 30x^{-5} + 54x^{-5}$ ∴ f' (x) = $15x^4 + 24x^{-5}$ = $15x^4 + \frac{24}{x^5}$. (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) = $\frac{-12}{x^5} + \frac{36}{x^{10}}$. (vi) Let f (x) = $\frac{2}{x+1} - \frac{x^2}{3x-1}$ ... (1) Differentiating (1) with respect to x, we get f' (x) = $\left[ \frac{(x+1)(2)' - (2)(x+1)'}{(x+1)^2} \right]$ - $\left[ \frac{(3x-1)(x^2)' - x^2(3x-1)'}{(3x-1)^2} \right]$ = $\frac{-2}{(x+1)^2}$ - $\left[ \frac{(3x-1)(2x) - x^2(3)}{(3x-1)^2} \right]$ $\frac{-2}{(x+1)^2} + \left[ \frac{6x^2 - 2x - 3x^2}{(3x-1)^2} \right]$ = $\frac{-2}{(x+1)^2} - \left[ \frac{3x^2 - 2x}{(3x-1)^2} \right]$ ∴ f' (x) = $\frac{-2}{(x+1)^2} - \frac{x(3x-2)}{(3x-1)^2}$.