Introducing Second Derivatives Revision Notes | GCSE AQA Higher Further Maths

1. Why Differentiate Twice?

You already know that differentiating $y$ gives $\frac{dy}{dx}$, the gradient of the curve. But here is the question that opens up a whole new layer: what if you differentiate the gradient itself?

Think of a car journey. The speedometer shows your speed — that is like $\frac{dy}{dx}$, the rate of change of position. The rate at which your speed is changing is acceleration. Differentiating once gives speed; differentiating again gives acceleration. The second derivative works exactly the same way, but for the gradient of a curve.

When the gradient of a curve is itself changing — steepening or flattening — the second derivative captures that change. This is what $\frac{d^2y}{dx^2}$ measures: the rate of change of the gradient function.

2. The Notation $\frac{d^2y}{dx^2}$

The notation looks more complex than it is. The $d^2$ in the numerator and the $dx^2$ in the denominator are not squares in the arithmetic sense — they are a compact way of recording that the operation of differentiation has been performed twice. Read it aloud as "dee-two-y by dee-x-squared".

The process is straightforward: differentiate once to get $\frac{dy}{dx}$, then treat that result as a new function and differentiate it again. Each term follows the same power rule you already know — multiply by the power, then reduce the power by one.

3. Finding the Second Derivative — Worked Examples

The method is always the same two-step process. Work through the example below carefully, covering each step before you reveal it.

Worked example: Finding $\frac{d^2y}{dx^2}$ for a polynomial

Given $y = x^3 - 5x^2 + 4x - 7$, find $\frac{d^2y}{dx^2}$.

1. $y = x^3 - 5x^2 + 4x - 7$ — Identify every term before differentiating. The constant $-7$ will vanish after the first differentiation.
2. $\frac{dy}{dx} = 3x^2 - 10x + 4$ — $x^3 \to 3x^2$, $-5x^2 \to -10x$, $4x \to 4$, constant $-7 \to 0$. This is the gradient function. [M1]
3. $\frac{d^2y}{dx^2} = 6x - 10$ — $3x^2 \to 6x$, $-10x \to -10$, constant $4 \to 0$. This is the second derivative. [A1]
4. $\frac{d^2y}{dx^2} = 6x - 10$ — Write the answer using the correct notation. Never write $\frac{d^2y}{dx^2}$ as $\frac{dy}{dx}$ — they are different quantities.

Answer: $\frac{d^2y}{dx^2} = 6x - 10$

Notice that the second derivative is a linear expression here. That means the rate of change of the gradient varies with $x$ — the curve is bending at different rates at different points. This is exactly the kind of information $\frac{d^2y}{dx^2}$ is designed to reveal.

Worked example: Evaluating the second derivative at a specific point

Given $y = x^3 - 5x^2 + 4x - 7$, find the value of $\frac{d^2y}{dx^2}$ when $x = 2$.

1. $\frac{d^2y}{dx^2} = 6x - 10$ — From the previous example. No need to re-derive from scratch.
2. $\frac{d^2y}{dx^2} = 6(2) - 10$ — Replace every $x$ with the given value. [M1]
3. $= 12 - 10 = 2$ — $6 \times 2 = 12$, then $12 - 10 = 2$. [A1]
4. $\frac{d^2y}{dx^2} = 2 > 0$ at $x = 2$ — A positive value means the gradient is increasing at this point — the curve is bending upwards here.

Answer: $\frac{d^2y}{dx^2} = 2$ when $x = 2$

4. What the Second Derivative Actually Tells You

The first derivative $\frac{dy}{dx}$ tells you the gradient at a point. The second derivative $\frac{d^2y}{dx^2}$ tells you how that gradient is changing — whether the curve is becoming steeper, flatter, or switching direction.

Picture driving along a curved road. If the road is bending to the left (curving upwards like a bowl), the gradient is increasing — $\frac{d^2y}{dx^2} > 0$. If the road bends to the right (curving downwards like a hill), the gradient is decreasing — $\frac{d^2y}{dx^2} < 0$. When $\frac{d^2y}{dx^2} = 0$, the rate of change of gradient is momentarily zero.

5. Practice and Formula Reference

Before moving to exam technique, consolidate the method with a short practice set. For each question, write out both derivatives in full before evaluating or interpreting.

6. Exam Technique and Common Mistakes

Now that the method is clear, here is where marks are most commonly dropped. The errors below appear regularly in exam scripts — recognise them before the paper, not during it.

One final point on exam language: if a question asks you to 'interpret' the value of $\frac{d^2y}{dx^2}$ at a point, write a sentence that connects the number to the gradient. For example: "Since $\frac{d^2y}{dx^2} = 2 > 0$ at $x = 2$, the gradient of the curve is increasing at this point." That phrasing directly addresses what the second derivative measures and is the kind of answer that earns interpretation marks.

The second derivative $\frac{d^2y}{dx^2}$ is a compact, precise tool: it records how the gradient function is changing, and its correct notation and two-step method are both directly assessed at GCSE level.