Second Derivatives

This lesson covers the second derivative d²y/dx², its use in determining the nature of stationary points, concavity, and points of inflection, as required by the Edexcel A-Level Mathematics specification (9MA0).

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What Is the Second Derivative?

The second derivative is the derivative of the derivative:

d²y/dx² = d/dx(dy/dx)

If y = f(x), then:

• f'(x) = dy/dx = the first derivative (gradient / rate of change)
• f''(x) = d²y/dx² = the second derivative (rate of change of the gradient)

Worked Example 1

Find d²y/dx² for y = 3x⁴ - 2x³ + 5x.

Solution: dy/dx = 12x³ - 6x² + 5

d²y/dx² = 36x² - 12x = 12x(3x - 1)

Worked Example 2

Find f''(x) for f(x) = eˣ + ln x.

Solution: f'(x) = eˣ + 1/x f''(x) = eˣ - 1/x² = eˣ - x⁻²

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Determining the Nature of Stationary Points

A stationary point occurs where dy/dx = 0. The second derivative test tells you the nature:

Second Derivative at Stationary Point	Nature
f''(x) > 0	Minimum (curve is concave up — shaped like a "U")
f''(x) < 0	Maximum (curve is concave down — shaped like an "n")
f''(x) = 0	Inconclusive — could be minimum, maximum, or point of inflection. Use another method.

Worked Example 3

Find the stationary points of y = x³ - 6x² + 9x + 1 and determine their nature.

Solution: dy/dx = 3x² - 12x + 9 = 3(x² - 4x + 3) = 3(x - 1)(x - 3)

Stationary points where dy/dx = 0: x = 1 or x = 3.

d²y/dx² = 6x - 12

At x = 1: d²y/dx² = 6(1) - 12 = -6 < 0 → Maximum y = 1 - 6 + 9 + 1 = 5. Maximum at (1, 5).

At x = 3: d²y/dx² = 6(3) - 12 = 6 > 0 → Minimum y = 27 - 54 + 27 + 1 = 1. Minimum at (3, 1).

Worked Example 4

Find and classify the stationary points of y = xe⁻ˣ.

Solution: Using the product rule: dy/dx = e⁻ˣ - xe⁻ˣ = e⁻ˣ(1 - x)

Set dy/dx = 0: e⁻ˣ(1 - x) = 0 → x = 1 (since e⁻ˣ > 0 always)

Now find d²y/dx² using the product rule on dy/dx = e⁻ˣ(1 - x): d²y/dx² = -e⁻ˣ(1 - x) + e⁻ˣ(-1) = e⁻ˣ(x - 1 - 1) = e⁻ˣ(x - 2)

At x = 1: d²y/dx² = e⁻¹(1 - 2) = -e⁻¹ < 0 → Maximum

Maximum at (1, e⁻¹) = (1, 1/e).

Exam Tip: The second derivative test is quick and usually conclusive. If f''(x) = 0 at the stationary point, fall back to examining the sign of f'(x) either side of the point.

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Concavity

The second derivative tells you about the concavity (curvature) of the graph:

Condition	Meaning
f''(x) > 0	Curve is concave up (bending upwards, "holds water")
f''(x) < 0	Curve is concave down (bending downwards, "spills water")
f''(x) = 0	Possible change in concavity (potential point of inflection)

Think of It This Way

• Concave up (f'' > 0): the gradient is increasing. If you're walking along the curve, it gets steeper (or less negative).
• Concave down (f'' < 0): the gradient is decreasing. The curve is flattening or turning downward.

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Points of Inflection

A point of inflection is where the concavity changes — the curve switches from concave up to concave down, or vice versa.

Conditions

A point of inflection occurs at x = a if:

1. f''(a) = 0 (necessary but not sufficient)
2. f'' changes sign at x = a (the crucial test)

Worked Example 5

Find the point of inflection for y = x³ - 3x² + 2.

Solution: dy/dx = 3x² - 6x d²y/dx² = 6x - 6 = 6(x - 1)

Set d²y/dx² = 0: x = 1

Check sign change:

• For x < 1 (e.g., x = 0): d²y/dx² = -6 < 0 (concave down)
• For x > 1 (e.g., x = 2): d²y/dx² = 6 > 0 (concave up)

Sign changes ✓ → Point of inflection at x = 1.

y = 1 - 3 + 2 = 0

Point of inflection at (1, 0).

Worked Example 6

Does y = x⁴ have a point of inflection at x = 0?

Solution: dy/dx = 4x³ d²y/dx² = 12x²

At x = 0: d²y/dx² = 0 ✓

But check sign change:

• For x < 0: d²y/dx² = 12x² > 0 (positive)
• For x > 0: d²y/dx² = 12x² > 0 (positive)

No sign change ✗ → Not a point of inflection. In fact, (0, 0) is a minimum.

Exam Tip: f''(x) = 0 alone does NOT guarantee a point of inflection. You must check that f'' changes sign.

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Stationary Points of Inflection

A stationary point of inflection is a point where:

• dy/dx = 0 (it's a stationary point)
• d²y/dx² = 0 (inconclusive second derivative test)
• d²y/dx² changes sign (it's a point of inflection, not a max or min)

Worked Example 7

Classify the stationary point of y = x³ at the origin.

Solution: dy/dx = 3x². At x = 0: dy/dx = 0. ✓ Stationary point. d²y/dx² = 6x. At x = 0: d²y/dx² = 0. Inconclusive.

Check the sign of dy/dx either side:

• For x < 0: dy/dx = 3x² > 0 (positive)
• For x > 0: dy/dx = 3x² > 0 (positive)

The gradient does not change sign → this is a stationary point of inflection, not a max or min.

Also check d²y/dx²: it changes from negative (x < 0) to positive (x > 0), confirming inflection.

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The Second Derivative in Physical Contexts

First Derivative	Second Derivative	Physical Meaning
ds/dt = velocity	d²s/dt² = acceleration	Rate of change of velocity
dP/dt = rate of growth	d²P/dt²	Acceleration of growth
dT/dt = rate of cooling	d²T/dt²	Whether cooling is speeding up or slowing down

Worked Example 8

A particle moves along a line with displacement s = t³ - 6t² + 9t metres at time t seconds.

(a) Find the velocity and acceleration. (b) Find when the particle is momentarily at rest. (c) Find when the acceleration is zero.

Solution: (a) v = ds/dt = 3t² - 12t + 9 a = dv/dt = d²s/dt² = 6t - 12

(b) At rest: v = 0 → 3t² - 12t + 9 = 0 → t² - 4t + 3 = 0 → (t-1)(t-3) = 0 t = 1 s or t = 3 s

(c) a = 0 → 6t - 12 = 0 → t = 2 s

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Common Mistakes

Mistake	Correction
Assuming f''(x) = 0 means a point of inflection	f'' must also change sign for a point of inflection
Assuming f''(x) = 0 means it's not a max or min	x⁴ has f'' = 0 at x = 0 but it's still a minimum
Confusing concave up and concave down	Concave up = f'' > 0 (U-shape). Concave down = f'' < 0 (n-shape)
Not showing the sign change of f''	In exam "show" questions, you must demonstrate the sign change

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Summary

• The second derivative d²y/dx² is the derivative of the first derivative.
• At a stationary point: f'' > 0 → minimum, f'' < 0 → maximum, f'' = 0 → inconclusive.
• Concave up: f'' > 0. Concave down: f'' < 0.
• A point of inflection occurs where f'' = 0 AND f'' changes sign.
• A stationary point of inflection has f' = 0 and f'' = 0, with f'' changing sign.
• In mechanics: the second derivative of displacement gives acceleration.

A-Level Deep Dive: Second Derivatives

Spec mapping

Edexcel 9MA0-01 specification section 9 — Differentiation covers the nature of stationary points using the second derivative; understand and apply the relationship between concavity and the sign of $f''(x)$f′′(x) (refer to the official specification document for exact wording). The second derivative $\dfrac{d^2 y}{dx^2}$dx2d2y​ is defined as $\dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right)$dxd​(dxdy​) and is examined principally in 9MA0-01 (Pure) but reappears in 9MA0-03 (Mechanics) where acceleration is the second derivative of displacement with respect to time. It also informs section 8 (Integration, where reversing differentiation twice recovers original functions up to two constants) and section 7 (Differential equations, where second-order behaviour determines stability of solutions). The Edexcel formula booklet does not list classification rules — the sign-test routine must be memorised.

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Worked example with full mark scheme

Question (8 marks): Given $y = x^3 - 3x^2 - 9x + 5$y=x3−3x2−9x+5:

(a) Find $\dfrac{dy}{dx}$dxdy​ and the coordinates of the stationary points of the curve. (4)

(b) Use the second derivative to determine the nature of each stationary point. (4)

Solution with mark scheme:

(a) Step 1 — differentiate.

$\dfrac{dy}{dx} = 3x^2 - 6x - 9$dxdy​=3x2−6x−9

M1 — applying the power rule term-by-term. A frequent slip is forgetting that the derivative of the constant $+5$+5 is $0$0, with candidates carrying $+5$+5 into the derivative.

Step 2 — set the derivative equal to zero.

$3x^2 - 6x - 9 = 0 \implies x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0$3x2−6x−9=0⟹x2−2x−3=0⟹(x−3)(x+1)=0

So $x = 3$x=3 or $x = -1$x=−1.

M1 — equating $\dfrac{dy}{dx} = 0$dxdy​=0 and solving correctly. Dividing through by $3$3 first is good practice; running the quadratic formula on $3x^2 - 6x - 9 = 0$3x2−6x−9=0 also works but invites arithmetic error.

Step 3 — find the corresponding $y$y-values.

At $x = 3$x=3: $y = 27 - 27 - 27 + 5 = -22$y=27−27−27+5=−22. At $x = -1$x=−1: $y = -1 - 3 + 9 + 5 = 10$y=−1−3+9+5=10.

A1 — both $y$y-coordinates correct. Stationary points are $(3, -22)$(3,−22) and $(-1, 10)$(−1,10).

A1 — coordinates written as ordered pairs (the Edexcel convention) rather than listed separately as $x$x- and $y$y-values.

(b) Step 1 — find the second derivative.

$\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}(3x^2 - 6x - 9) = 6x - 6$dx2d2y​=dxd​(3x2−6x−9)=6x−6

M1 — differentiating $\dfrac{dy}{dx}$dxdy​ correctly. Common error: differentiating $y$y a second time from scratch and dropping a sign on the $-9x$−9x term.

Step 2 — evaluate at each stationary point.

At $x = 3$x=3: $\dfrac{d^2 y}{dx^2} = 6(3) - 6 = 12 > 0$dx2d2y​=6(3)−6=12>0, so $(3, -22)$(3,−22) is a minimum.

At $x = -1$x=−1: $\dfrac{d^2 y}{dx^2} = 6(-1) - 6 = -12 < 0$dx2d2y​=6(−1)−6=−12<0, so $(-1, 10)$(−1,10) is a maximum.

M1 — substituting both $x$x-values into the second derivative.

A1 — correct sign and classification at $(3, -22)$(3,−22).

A1 — correct sign and classification at $(-1, 10)$(−1,10).

Total: 8 marks (M4 A4).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): A curve has equation $y = 2x^3 + 3x^2 - 12x + 1$y=2x3+3x2−12x+1.

(a) Show that the curve has stationary points at $x = 1$x=1 and $x = -2$x=−2. (2)

(b) Use the second derivative to classify each stationary point. (3)

(c) State the value of $x$x at which the curve changes from concave down to concave up. (1)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — differentiating to obtain $\dfrac{dy}{dx} = 6x^2 + 6x - 12$dxdy​=6x2+6x−12 and equating to zero.
• A1 (AO1.1b) — factorising as $6(x - 1)(x + 2) = 0$6(x−1)(x+2)=0 and stating $x = 1$x=1 and $x = -2$x=−2.

(b)

• M1 (AO1.1b) — second derivative $\dfrac{d^2 y}{dx^2} = 12x + 6$dx2d2y​=12x+6.
• A1 (AO2.1) — at $x = 1$x=1, $\dfrac{d^2 y}{dx^2} = 18 > 0$dx2d2y​=18>0, so minimum.
• A1 (AO2.1) — at $x = -2$x=−2, $\dfrac{d^2 y}{dx^2} = -18 < 0$dx2d2y​=−18<0, so maximum.

(c)

• B1 (AO2.2a) — concavity changes when $12x + 6 = 0$12x+6=0, i.e. $x = -\tfrac{1}{2}$x=−21​.

Total: 6 marks split AO1 = 3, AO2 = 3. This question pattern is canonical Paper 1 — AO1 marks for procedure, AO2 marks for the interpretive sign analysis. The (c) part rewards candidates who recognise that an inflection occurs where the second derivative changes sign.

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Synoptic links

Connects to:

1. Stationary points classification (section 9): the first derivative $\dfrac{dy}{dx} = 0$dxdy​=0 locates stationary points, but only the second derivative $\dfrac{d^2 y}{dx^2}$dx2d2y​ classifies them as maxima or minima. Without the second-derivative test, candidates must fall back on the slower sign-change-of-$f'$f′ method.

2. Concavity and inflection (section 9): $\dfrac{d^2 y}{dx^2} > 0$dx2d2y​>0 corresponds to concave up (the curve "holds water"); $\dfrac{d^2 y}{dx^2} < 0$dx2d2y​<0 corresponds to concave down. Points where concavity changes — when $f''$f′′ changes sign — are points of inflection.

3. Mechanics (9MA0-03), acceleration: in kinematics, displacement $s(t)$s(t), velocity $v = \dfrac{ds}{dt}$v=dtds​, and acceleration $a = \dfrac{dv}{dt} = \dfrac{d^2 s}{dt^2}$a=dtdv​=dt2d2s​. Newton's second law $F = ma$F=ma is therefore a second-order statement about displacement. Constant acceleration corresponds to $\dfrac{d^2 s}{dt^2} = \text{const}$dt2d2s​=const.

4. Taylor series (further A-Level / undergraduate): $f(x) \approx f(a) + f'(a)(x - a) + \tfrac{1}{2}f''(a)(x - a)^2 + \ldots$f(x)≈f(a)+f′(a)(x−a)+21​f′′(a)(x−a)2+… — the second derivative controls the leading curvature term in any local approximation. This is why $f''(a)$f′′(a) "is" the curvature.