Q.For some constants $a$ and $b$, find the derivative of

Concept: Quotient Rule Differentiation (for part iii) and standard polynomial/power rules.

(i) Expand: $(x - a)(x - b) = x^2 - (a+b)x + ab$.

Differentiate term by term:

$\frac{d}{dx}[x^2] = 2x$, $\frac{d}{dx}[-(a+b)x] = -(a+b)$, $\frac{d}{dx}[ab] = 0$.

Result: $2x - (a+b)$.

(ii) Let $u = ax^2 + b$. Then by the chain rule:

$\frac{d}{dx}(u^2) = 2u \cdot \frac{du}{dx} = 2(ax^2 + b) \cdot (2ax) = 4ax(ax^2 + b)$. …

The Quotient Rule is the natural tool here because we have one function divided by another. It says: if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$. The key insight is that the numerator of the derivative is "bottom times derivative of top, minus top times derivative of bottom" — the order matters, and the minus sign is the most common source of error.

Let’s work through each part, building up to the quotient.

1. Part (i): $y = (x - a)(x - b)$

   This is a product, so we could use the Product Rule, but it’s simpler to expand first.

   Expand: $(x - a)(x - b) = x^2 - (a + b)x + ab$.

   Differentiate term by term:

   $\frac{dy}{dx} = 2x - (a + b)$.

   That’s the derivative — clean and direct.

2. Part (ii): $y = (ax^2 + b)^2$

   This is a composition: outer function $(\cdot)^2$, inner function $ax^2 + b$. Use the Chain Rule.

   Let $u = ax^2 + b$, then $y = u^2$.

   $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2u \cdot (2ax) = 2(ax^2 + b) \cdot 2ax = 4ax(ax^2 + b)$.

   Alternatively, expand: $(ax^2 + b)^2 = a^2 x^4 + 2ab x^2 + b^2$, then differentiate to get $4a^2 x^3 + 4ab x = 4ax(ax^2 + b)$ — same result.

3. Part (iii): $y = \dfrac{x - a}{x - b}$

   Here we apply the Quotient Rule. Identify:

   $u = x - a$, so $\frac{du}{dx} = 1$.

   $v = x - b$, so $\frac{dv}{dx} = 1$.

   Plug into the formula:

   $\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} = \frac{(x - b)(1) - (x - a)(1)}{(x - b)^2}$. …