The first derivative tells us how fast something is changing. The second derivative tells us how fast that change is changing. In physics, it is acceleration. In geometry, it reveals whether a curve is bending upwards like a smile, or downwards like a frown. Welcome to the second derivative.
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Imagine you are in a car. Your speedometer shows $60$ km/h and is steadily increasing — you are speeding up. A moment later, the speedometer still reads $60$ km/h, but now it is decreasing — you are slowing down. In both cases your speed is the same, but something about your motion feels very different. What mathematical concept could describe this difference?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: A stationary point is always a maximum or minimum.
Right: Stationary points can also be horizontal points of inflection where the concavity changes.
📚 Core Content
If the first derivative $f'(x)$ tells us the gradient of a curve at any point, then the second derivative $f''(x)$ tells us how that gradient is changing. We find it by simply differentiating twice:
$$f''(x) = \frac{d}{dx}\bigl(f'(x)\bigr)$$
Alternative notations include:
The sign of $f''(x)$ tells us about the concavity of the curve:
In the study of motion, if $s(t)$ represents the displacement of an object at time $t$, then:
| Quantity | Notation | Meaning |
| Displacement | $s(t)$ | Position |
| Velocity | $v(t) = s'(t)$ | Rate of change of displacement |
| Acceleration | $a(t) = s''(t)$ | Rate of change of velocity |
So acceleration is literally the second derivative of displacement. Positive acceleration means velocity is increasing; negative acceleration (deceleration) means velocity is decreasing.
🧮 Worked Examples
🧪 Activities
1 $f(x) = x^4 - 2x^3 + x$
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2 $y = 3x^2 - 5x + 2$
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3 Determine the concavity of $y = x^3 - 6x^2$ at $x = 2$.
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4 A particle has displacement $s(t) = 2t^3 - 9t^2 + 12t$. Find its acceleration at $t = 3$.
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1 If $f''(x) = 6x - 2$, explain what $f''(1) = 4$ and $f''(0) = -2$ tell you about the shape of the curve at those points.
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2 A car's velocity is constant at $60$ km/h. What is its acceleration? Explain why the second derivative of displacement is zero in this case.
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Earlier you thought about a car moving at $60$ km/h while speeding up versus slowing down.
The difference is acceleration — the second derivative of displacement. When speeding up, velocity is increasing, so acceleration is positive. When slowing down, velocity is decreasing, so acceleration is negative. The speedometer alone (velocity) cannot tell you this; only the rate of change of velocity (acceleration) captures the full picture of the motion.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. (a) Find $f''(x)$ for $f(x) = x^4 - 4x^3 + 2$. 2 MARKS
(b) Determine the values of $x$ for which the curve $y = f(x)$ is concave up. 2 MARKS
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7. A particle moves along a straight line with displacement $s(t) = t^3 - 9t^2 + 24t$, where $t$ is in seconds and $s$ is in metres. 4 MARKS
(a) Find expressions for the velocity and acceleration of the particle.
(b) Find the time(s) when the particle is instantaneously at rest.
(c) Determine the acceleration of the particle at the instant(s) when it is at rest.
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8. The graph of $y = f(x)$ is shown below (description: it is an increasing curve that bends upwards for $x < 1$, has an inflection point at $x = 1$, and bends downwards for $x > 1$). Without calculating, sketch the general shape of $y = f'(x)$ and $y = f''(x)$, and explain your reasoning. 3 MARKS
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1. $f'(x) = 4x^3 - 6x^2 + 1$, so $f''(x) = 12x^2 - 12x$
2. $\frac{dy}{dx} = 6x - 5$, so $\frac{d^2y}{dx^2} = 6$
3. $\frac{dy}{dx} = 3x^2 - 12x$, so $\frac{d^2y}{dx^2} = 6x - 12$. At $x = 2$, $y'' = 0$. This is an inflection point — neither concave up nor concave down at exactly $x = 2$.
4. $v(t) = 6t^2 - 18t + 12$, $a(t) = 12t - 18$. At $t = 3$, $a(3) = 18$ m/s$^2$.
1. $f''(1) = 4 > 0$ means the curve is concave up at $x = 1$ (bending upwards). $f''(0) = -2 < 0$ means the curve is concave down at $x = 0$ (bending downwards).
2. The acceleration is $0$. Since velocity is constant, its rate of change is zero. Therefore $s''(t) = 0$.
1. B — $f'(x) = 6x - 2$, so $f''(x) = 6$.
2. B — $y'' = 6x - 6 < 0$ when $x < 1$, so the curve is concave down for $x < 1$.
3. B — $a(t) = v'(t) = s''(t)$. Acceleration is the second derivative of displacement.
4. C — $v(t) = 6t - 4$, $a(t) = 6$. The acceleration is constant at $6$ m/s$^2$ for all $t$.
5. B — $y' = 3x^2$, $y'' = 6x$. Setting $6x = 0$ gives $x = 0$.
6. B — Constant velocity means the rate of change of velocity is zero, so acceleration is $0$.
7. A — $12x^2 \\ge 0$ for all real $x$, so the curve is concave up everywhere.
Q6 (4 marks): (a) $f'(x) = 4x^3 - 12x^2$ [1], so $f''(x) = 12x^2 - 24x$ [1]. (b) Concave up when $12x^2 - 24x > 0$, i.e. $12x(x - 2) > 0$ [1], giving $x < 0$ or $x > 2$ [1].
Q7 (4 marks): $v(t) = 3t^2 - 18t + 24$ [1] and $a(t) = 6t - 18$ [1]. Instantaneously at rest when $3t^2 - 18t + 24 = 0 \Rightarrow t^2 - 6t + 8 = 0 \Rightarrow (t-2)(t-4)=0$, so $t = 2,\,4$ [1]. $a(2) = -6$ m/s$^2$ and $a(4) = 6$ m/s$^2$ [1].
Q8 (3 marks): Since $f$ is increasing, $f'(x) > 0$ everywhere (sketch above the $x$-axis) [1]. $f$ is concave up for $x < 1$ and concave down for $x > 1$, so $f''(x) > 0$ for $x < 1$, $f''(x) < 0$ for $x > 1$, and $f''(1) = 0$ (sketch crossing the axis at $x=1$) [1]. Because $f$ changes from concave up to concave down at $x = 1$, the derivative $f'(x)$ has a maximum at $x = 1$ [1].
The Second Derivative
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