Differentiation Applications (Edexcel IAL P2)

Edexcel International A-Level Mathematics· 2018 Issue 3 WMA12 Section 7.1· 25 min read

1. Increasing and Decreasing Functions★★☆☆☆⏱ 5 min

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A function is increasing on an interval if its first derivative is non-negative ( $f^{'} (x) \geq 0$ ) for all $x$ in that interval, and strictly increasing if $f^{'} (x) > 0$ . A function is decreasing if $f^{'} (x) \leq 0$ , and strictly decreasing if $f^{'} (x) < 0$ . This is a common 3-4 mark proof question in P2 exams.

📘 Definition

Increasing/Decreasing Function

For $f (x)$ defined on interval $[a, b]$ : 1. Increasing: $f^{'} (x) \geq 0$ for all $x \in (a, b)$ 2. Decreasing: $f^{'} (x) \leq 0$ for all $x \in (a, b)$

Example:

$f (x) = x^{3} + 2 x$ is strictly increasing for all real $x$ , as $f^{'} (x) = 3 x^{2} + 2 > 0$ for all $x$ .

📐 Worked Example

Prove that $f (x) = 2 x^{3} + 5 x - 1$ is strictly increasing for all real values of $x$ .

1
Calculate the first derivative using the P1 power rule:
$f'(x) = 6x^2 + 5$
2
Analyse the sign of $f^{'} (x)$ : $x^{2} \geq 0$ for all real $x$ , so $6 x^{2} \geq 0$ , meaning $6 x^{2} + 5 \geq 5 > 0$ for all $x$ .
3
State conclusion: since $f^{'} (x) > 0$ for all real $x$ , $f (x)$ is strictly increasing for all real $x$ .

Exam tip:

You must explicitly justify the sign of the first derivative for all values in the interval to get full marks for proof questions, not just state it is positive/negative.

2. Stationary Points and Classification★★★☆☆⏱ 7 min

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Stationary points occur where $f^{'} (x) = 0$ , meaning the tangent to the curve is horizontal. There are two common types in P2: local maxima (peaks) and local minima (troughs). You can classify them using either the first derivative test (checking sign changes of $f^{'} (x)$ around the point) or the faster second derivative test.

📘 Definition

Stationary Point

$f^{'} (c) = 0$ at $x = c$

A point on the curve $y = f (x)$ where the gradient is zero, with a horizontal tangent line.

Example:

$f (x) = x^{2}$ has a stationary minimum point at $x = 0$ , where $f^{'} (0) = 0$ .

📐 Worked Example

Find and classify the stationary points of $f (x) = x^{3} - 6 x^{2} + 9 x + 2$ .

1
Calculate the first derivative and set equal to zero to find stationary point x-values:
$f'(x) = 3x^2 - 12x + 9 = 0 \\ 3(x^2 - 4x + 3) = 0 \\ 3(x-1)(x-3) = 0 \\ x = 1 \text{ or } x=3$
2
Calculate the second derivative for classification:
$f''(x) = 6x - 12$
3
Evaluate $f^{''} (x)$ at each stationary point: For $x = 1$ , $f^{''} (1) = 6 (1) - 12 = - 6 < 0$ , so this is a local maximum. For $x = 3$ , $f^{''} (3) = 6 (3) - 12 = 6 > 0$ , so this is a local minimum.
4
Calculate corresponding y-values: Maximum at $(1, 6)$ , minimum at $(3, 2)$ .

3. Curve Sketching Using Differentiation★★★☆☆⏱ 6 min

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To sketch a polynomial curve, combine differentiation data (stationary points and classification) with basic coordinate geometry (intercepts, end behaviour). This is a common 4-6 mark question in P2 exams, and does not require perfect scale drawing, only correct shape and labelled key points.

Find the y-intercept by setting $x = 0$
Find x-intercepts by setting $y = 0$ and solving the polynomial equation
Find and classify all stationary points using differentiation
Determine end behaviour: for highest power $a x^{n}$ , if $a > 0$ , as $x \to + \infty$ , $y \to + \infty$ ; if $n$ even, $x \to - \infty$ , $y \to + \infty$ , if $n$ odd, $y \to - \infty$ . Reverse for $a < 0$ .
Plot all key points and draw a smooth curve through them.

📐 Worked Example

Sketch the curve $y = x^{3} - 3 x^{2} - 4 x$ , marking all intercepts and stationary points.

1
Find y-intercept: set $x = 0$ , so $y = 0$ , intercept at $(0, 0)$ .
2
Find x-intercepts: set $y = 0$ and factorise:
$x^3 - 3x^2 - 4x = 0 \\ x(x-4)(x+1) = 0 \\ x = -1, 0, 4$
3
Find stationary points: calculate first derivative and solve for $f^{'} (x) = 0$ :
$f'(x) = 3x^2 -6x -4 = 0 \\ x = \frac{6 \pm \sqrt{84}}{6} \approx 2.53 \text{ and } -0.527$
4
Classify with second derivative: $f^{''} (x) = 6 x - 6$ . $x = - 0.527$ is a maximum (y≈1.13), $x = 2.53$ is a minimum (y≈-13.13).
5
End behaviour: $x^{3}$ term positive, so as $x \to + \infty$ , $y \to + \infty$ , as $x \to - \infty$ , $y \to - \infty$ . Plot all labelled points and draw smooth curve.

Exam tip:

You will lose marks if you do not label all intercepts and stationary point coordinates on your sketch, even if the shape is correct.

4. Practical Optimisation Problems★★★★☆⏱ 7 min

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Optimisation problems ask you to find the maximum or minimum value of a real-world quantity (e.g. volume of a box, cost of materials) subject to constraints. You will first need to derive a function for the quantity to optimise in terms of a single variable, before using stationary point techniques to find the required value.

📘 Definition

Optimisation

The process of finding the maximum or minimum value of a function subject to given context constraints, one of the highest-weight P2 differentiation question types.

Example:

Finding the maximum volume of an open-top box made from a fixed-size sheet of card.

📐 Worked Example

An open-topped rectangular box is made from a 20cm x 20cm square sheet of card, by cutting out a square of side $x$ cm from each corner and folding up the sides. Find the maximum possible volume of the box, and the value of $x$ that gives this volume.

1
Derive the volume function: after cutting out squares of side $x$ , base dimensions are $(20 - 2 x)$ cm, height is $x$ cm. Volume $V$ is:
$V(x) = x(20 - 2x)^2 = 4x^3 - 80x^2 + 400x$
2
Define valid domain: $x > 0$ and $20 - 2 x > 0$ , so $0 < x < 10$ .
3
Differentiate $V (x)$ and set to zero to find stationary points:
$V'(x) = 12x^2 - 160x + 400 = 0 \\ 3x^2 - 40x + 100 = 0 \\ (3x - 10)(x - 10) = 0 \\ x = \frac{10}{3} \text{ or } x=10$
4
Discard $x = 10$ (outside valid domain). Classify remaining point with second derivative:
$V''(x) = 24x - 160 \\ V''(10/3) = 80 - 160 = -80 < 0 \\ \text{So this is a maximum point}.$
5
Calculate maximum volume: $V (10/3) = 16000/27 \approx 592.6$ cm³. Conclusion: maximum volume is 16000/27 cm³ when $x = 10/3$ cm.

Exam tip:

Always state the domain of your variable based on context, and explicitly classify the stationary point to confirm you have found a maximum/minimum, not just any stationary point.

5. Common Pitfalls

Wrong move:

Failing to justify the sign of $f^{'} (x)$ when proving a function is increasing/decreasing

Why:

Examiners require explicit proof the derivative has no sign changes over the interval, not just a statement of its sign.

Correct move:

Show all terms in the derivative are non-negative/non-positive, with no sign changes, to justify your conclusion.

Wrong move:

Assuming $f^{''} (c) = 0$ means the stationary point is an inflection point

Why:

The second derivative test is inconclusive when $f^{''} (c) = 0$ ; the point could still be a maximum or minimum.

Correct move:

Use the first derivative test to check for sign changes of $f^{'} (x)$ immediately either side of $x = c$ to classify the point.

Wrong move:

Failing to discard invalid stationary points in optimisation problems

Why:

Context constraints (e.g. length cannot be negative) may make some solutions to $f^{'} (x) = 0$ impossible.

Correct move:

Define the valid domain of your variable first, and reject any stationary points outside this domain before classifying.

Wrong move:

Forgetting to calculate the actual optimised quantity (e.g. volume) in context questions

Why:

The question asks for the maximum/minimum value, not just the x-coordinate of the stationary point.

Correct move:

Substitute the stationary point x-value back into your original function, and include units in your final answer.

Wrong move:

Skipping classification of stationary points for curve sketching

Why:

You will lose marks if you plot a maximum as a minimum or vice versa, even if you found the correct x-values.

Correct move:

Classify each stationary point using either first or second derivative test before sketching the curve.

6. Quick Reference Cheatsheet

Task	Method	Key Check
Identify increasing/decreasing function	Calculate $f^{'} (x)$ , check sign over interval	$f^{'} (x) \geq 0$ = increasing; $f^{'} (x) \leq 0$ = decreasing
Find stationary points	Solve $f^{'} (x) = 0$ for $x$	Check solutions are valid for the given context
Classify stationary points (fast)	Use second derivative test: $f^{''} (c) < 0$ = max; $f^{''} (c) > 0$ = min	If $f^{''} (c) = 0$ , use first derivative sign change test
Sketch polynomial curve	Find intercepts, stationary points, end behaviour	Label all key points clearly on your sketch
Solve optimisation problems	Derive function in 1 variable, find stationary point, classify	State final answer with units and context

7. Frequently Asked

Can I use the first derivative test instead of the second derivative test to classify stationary points?

Yes, both methods are accepted in Edexcel IAL P2 exams. The second derivative test is faster for polynomials, but if $f^{''} (c) = 0$ you must use the first derivative test to check for sign changes of $f^{'} (x)$ around $x = c$ .

Do I need to show all steps for optimisation problems?

Yes, you must show derivation of the function to optimise, differentiation steps, classification of the stationary point, and a concluding statement with units to earn full marks.

Going deeper

formula_bookEdexcel IAL Maths Formula BookletNo specific formula entry for this topic, uses basic differentiation rules
prerequisite_guideP1 Basic Differentiation

What's Next

Now that you have mastered P2 differentiation applications, you can progress to advanced differentiation techniques in Pure Mathematics 3, including the chain, product and quotient rules, and differentiation of trigonometric, exponential and logarithmic functions. You should also practice these skills with P2 past papers to familiarise yourself with exam phrasing and mark scheme requirements, as optimisation and curve sketching questions are consistently high-weight. Make sure you are fully confident with P1 basic differentiation before moving to P3 content, as all advanced differentiation techniques build on this foundational skill. These applications are also frequently combined with P1 coordinate geometry in multi-mark exam questions, so revising that topic will help you tackle more complex items efficiently.