IGCSE Add Maths Differentiation: 5 Common Mistakes (and How to Avoid Them)

Differentiation is one of the highest-scoring topics in IGCSE Additional Mathematics, but it’s also where careless slips cost the most marks. These are the five mistakes I see most often, and the habit that fixes each one.

1. Forgetting the inner derivative (chain rule)

When you differentiate a bracket raised to a power, you must multiply by the derivative of what’s inside the bracket.

The common wrong answer is $5(3x + 1)^4$, missing the $\times 3$.

Habit: every time you bring a power down, ask “what’s inside, and what’s its derivative?”

2. Using only half of the product rule

For $y = uv$, the rule is $\frac{dy}{dx} = u'v + uv'$. Both terms are needed, and each part may need the chain rule too.

Example: $y = x^2(2x - 1)^3$

• $u = x^2$, so $u' = 2x$
• $v = (2x - 1)^3$, so $v' = 3(2x - 1)^2 \times 2 = 6(2x - 1)^2$

Habit: write out $u$, $u'$, $v$ and $v'$ on separate lines before combining them. Factorising at the end is often required to find stationary points.

3. Getting the quotient rule upside down

For $y = \dfrac{u}{v}$:

The order on top matters because of the minus sign. Example: $y = \dfrac{x}{x + 1}$

Swap the order and you get $\dfrac{-1}{(x + 1)^2}$, the wrong sign.

Habit: remember it as “bottom times derivative of top, minus top times derivative of bottom, all over bottom squared”.

4. Differentiating fractions directly

You can’t differentiate $\dfrac{4}{x^2}$ by differentiating the top and bottom separately. Rewrite it as a negative power first:

The same goes for roots: $\sqrt{x} = x^{\frac{1}{2}}$.

Habit: before differentiating, rewrite every term in the form $ax^n$.

5. Substituting into the wrong equation (stationary points)

To find stationary points, solve $\frac{dy}{dx} = 0$ to get the $x$-values, then substitute them into the original equation to get $y$, not into $\frac{dy}{dx}$ (which would always give 0).

Example: $y = x^3 - 3x^2 + 2$

• $\frac{dy}{dx} = 3x^2 - 6x = 3x(x - 2) = 0$, so $x = 0$ or $x = 2$
• $y(0) = 2$ and $y(2) = 8 - 12 + 2 = -2$
• $\frac{d^2y}{dx^2} = 6x - 6$: at $x = 0$ it’s $-6$ (negative, so maximum at $(0, 2)$); at $x = 2$ it’s 6 (positive, so minimum at $(2, -2)$)

Habit: finish every stationary-point question by stating both coordinates and the nature (maximum or minimum).

Quick self-check

Differentiate each one, then check your answers:

1. $y = (2 - 5x)^4$
2. $y = \dfrac{6}{x^3}$
3. $y = x(x + 3)^2$