(i) Let f(x) = (x – a) (x – b) ... (1) Differentiating (1) with respect to x, we get f ′(x) = (x – a) (x – b)′ + (x – a)′ (x – b) ⇒ f ′(x) = (x – a) + (x – b) = 2x – a – b (ii) Let f(x) = $(ax^2 + b)^2$ ... (1) Differentiating (1) with respect to x, we get f ′(x) = $2(ax^2 + b)$ × 2ax ⇒ f ′(x) = 4ax $(ax^2 + b)$. (iii) Let f (x) = $\frac{x-a}{x-b}$ ... (1) Differentiating (1) with respect to x, we get f' (x) = $\frac{(x-b)(x-a)'-(x-a)(x-b)'}{(x-b)^2}$ ⇒ f' (x) = $\frac{(x-b)-(x-a)}{(x-b)^2}$ = $\frac{x-b-x+a}{(x-b)^2}$ = $\frac{a-b}{(x-b)^2}$.