# Applications of Differentiation

> IB Mathematics AA HL · IB AA HL
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u5-applications-of-differentiation/

This module covers core practical applications of differentiation for IB AA HL, including tangents/normals, classifying stationary points, increasing/decreasing intervals, and common real-world optimisation problems.

**Prerequisites:** [Basic differentiation rules](https://www.owlsprep.com/study/ib-math-aa-hl-u4-differentiation-rules/); [Chain, product and quotient rules](https://www.owlsprep.com/study/ib-math-aa-hl-u4-differentiation-techniques/)

## Learning objectives

- Find equations of tangents and normals to curves at given points
- Determine intervals of increase/decrease for functions
- Find and classify stationary points using first/second derivative tests
- Solve real-world optimisation problems in context

## Tangents and Normals to Curves

**Tangent and Normal Lines** — A tangent to a curve at a point touches the curve locally at that point, with gradient equal to the derivative at the point. A normal is perpendicular to the tangent at the same point, with gradient equal to the negative reciprocal of the tangent gradient.

*Notation:* Gradient of tangent at $x=a$: $f'(a)$

*Example:* If tangent gradient is 2, normal gradient is $-\frac{1}{2}$

**Worked example:** Find the equation of the tangent and normal to the curve $y = x^3 - 2x + 1$ at the point $(1, 0)$

1. Compute the first derivative (gradient function):
2. $$\frac{dy}{dx} = 3x^2 - 2$$
3. Evaluate the gradient at $x=1$:
4. $$\frac{dy}{dx}\bigg|_{x=1} = 3(1)^2 - 2 = 1$$
5. Use point-slope form $y - y_1 = m(x - x_1)$ for the tangent:
6. $$y - 0 = 1(x - 1) \implies y = x - 1$$
7. Calculate the normal gradient (negative reciprocal of tangent gradient):
8. $$m_n = -\frac{1}{1} = -1$$
9. Find the normal equation using point-slope form:
10. $$y - 0 = -1(x - 1) \implies y = -x + 1$$

> **Exam tip:** Always confirm if the question asks for tangent or normal, and double-check the negative sign for the normal gradient.

*Calculator:* allowed

## Increasing/Decreasing Functions & Stationary Points

**Stationary Point** — A stationary point of $f(x)$ is any point where $f'(x) = 0$, so the gradient of the curve is zero. Stationary points are classified as local maximum, local minimum, or stationary inflection point.

*Example:* $f(x) = x^3$ has a stationary inflection at $x=0$

We use the sign of the first derivative to determine if a function is increasing or decreasing on an interval: if $f'(x) > 0$, the function is increasing; if $f'(x) < 0$, it is decreasing.

**Worked example:** Find the intervals where $f(x) = x^3 - 3x^2 + 2$ is increasing/decreasing, and find all stationary points.

1. Compute the first derivative:
2. $$f'(x) = 3x^2 - 6x = 3x(x - 2)$$
3. Find stationary points by solving $f'(x) = 0$:
4. $$3x(x - 2) = 0 \implies x=0, \quad x=2$$
5. Test the sign of $f'(x)$ across intervals:
6. 1. $x < 0$: $f'(x) = 3(-)(-) = + > 0 \implies$ increasing
7. 2. $0 < x < 2$: $f'(x) = 3(+)(-) = - < 0 \implies$ decreasing
8. 3. $x > 2$: $f'(x) = 3(+)(+) = + > 0 \implies$ increasing

> **Exam tip:** Always test the sign of $f'(x)$ around critical points, do not assume the order of turning points.

*Calculator:* allowed

## Classifying Stationary Points

**Comparing methods**

Two methods are accepted in IB exams for classification, both are valid unless the question specifies a method:

- **First Derivative Test** — Check the sign change of $f'(x)$ around the stationary point $x=a$. A sign change from + to - is maximum, - to + is minimum, no change is inflection.
  - Pros: Works for all stationary points, including inflections
  - Cons: Requires testing two points, more computation

- **Second Derivative Test** — Evaluate $f''(a)$ at stationary point $x=a$: $f''(a) < 0$ = maximum, $f''(a) > 0$ = minimum, $f''(a)=0$ = inconclusive.
  - Pros: Fast for simple functions like polynomials
  - Cons: Inconclusive when $f''(a)=0$, cannot always identify inflections

**Worked example:** Classify the stationary points of $f(x) = x^3 - 3x^2 + 2$ using the second derivative test.

1. We already found stationary points at $x=0$ and $x=2$.
2. Compute the second derivative:
3. $$f''(x) = 6x - 6$$
4. Evaluate at $x=0$:
5. $$f''(0) = -6 < 0 \implies \text{local maximum at } x=0$$
6. Evaluate at $x=2$:
7. $$f''(2) = 12 - 6 = 6 > 0 \implies \text{local minimum at } x=2$$

**Check your understanding**

Test your understanding of the second derivative test:

1. If $f'(a) = 0$ and $f''(a) = 0$, what is the correct next step?

   - Definitely a stationary inflection point
   - The second derivative test is inconclusive, use the first derivative test
   - Definitely a local maximum
   - Definitely a local minimum

   *Why:* When $f''(a)=0$, we cannot draw a conclusion from the second derivative test, so we must check the sign change of $f'(x)$ around $x=a$.

*Calculator:* allowed

## Optimisation Problems

Optimisation problems ask you to find the maximum or minimum value of a quantity (volume, area, cost, etc.) in a real-world context. The standard method is: (1) express the quantity as a function of one variable, (2) find stationary points, (3) classify to confirm it is the required maximum/minimum.

**Worked example:** A rectangular box with a square base has total surface area 600 cm². Find the maximum possible volume of the box.

1. Let base side = $x$ cm, height = $h$ cm. Total surface area:
2. $$2x^2 + 4xh = 600$$
3. Solve for $h$ in terms of $x$:
4. $$h = \frac{300 - x^2}{2x}$$
5. Write volume as a function of $x$:
6. $$V = x^2 h = 150x - \frac{1}{2}x^3$$
7. Differentiate and solve for stationary points:
8. $$\frac{dV}{dx} = 150 - \frac{3}{2}x^2 = 0 \implies x^2 = 100 \implies x = 10$$
9. Classify with second derivative test:
10. $$\frac{d^2V}{dx^2} = -3x \implies \frac{d^2V}{dx^2}\bigg|_{x=10} = -30 < 0$$
11. Calculate maximum volume:
12. $$V = 150(10) - \frac{1}{2}(10)^3 = 1000 \text{ cm}^3$$

> **Exam tip:** Always confirm your solution makes physical sense (e.g. lengths are positive) and explicitly classify the stationary point to earn full marks.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Writing normal gradient as $1/f'(a)$ (forgetting the negative sign)
  - Why it fails: Perpendicular lines have gradients that multiply to $-1$, not 1
  - Correct: Always use $m_n = -\frac{1}{m_t}$ for the normal gradient, including the negative sign
- **Wrong:** Automatically classifying a point where $f'(a)=0, f''(a)=0$ as an inflection
  - Why it fails: $f''(a)=0$ does not guarantee an inflection, we need a sign change of $f''(x)$
  - Correct: When $f''(a)=0$, use the first derivative test to classify the stationary point
- **Wrong:** Stopping after finding the stationary point in optimisation, no classification
  - Why it fails: IB exams award marks for confirming you have found the required extrema
  - Correct: Always classify the stationary point to confirm it is the maximum/minimum needed
- **Wrong:** Writing increasing/decreasing intervals as closed intervals including stationary points
  - Why it fails: At stationary points $f'(x)=0$, so the function is neither increasing nor decreasing
  - Correct: Use open intervals for increasing/decreasing, e.g. $(0,2)$ not $[0,2]$

## Cheatsheet

| Concept | Key Result |
| --- | --- |
| Tangent gradient at $x=a$ | $m_t = f'(a)$ |
| Normal gradient at $x=a$ | $m_n = -1/f'(a)$ |
| Increasing interval | $f'(x) > 0$ |
| Decreasing interval | $f'(x) < 0$ |
| Stationary point | $f'(x) = 0$ |
| Local maximum (2nd test) | $f'(a)=0, f''(a) < 0$ |
| Local minimum (2nd test) | $f'(a)=0, f''(a) > 0$ |
| Optimisation step 1 | Write quantity as single-variable function |
| Optimisation step 2 | Find and classify stationary points |

## What's next

Applications of differentiation is one of the most heavily tested core topics in IB Mathematics AA HL, forming the foundation for more advanced calculus topics including implicit differentiation, parametric differentiation, and integration applications. Optimisation and stationary point classification are frequent long-answer questions in both Paper 1 and Paper 2, so mastering the methods here is critical for achieving a high overall score. This topic also introduces key reasoning skills for extrema problems that are reused in later topics like integration and differential equations. Next, you can build on this knowledge by exploring applications to implicitly and parametrically defined curves, then move on to related rates problems, another common exam application of differentiation.

- [Indefinite and definite integration](https://www.owlsprep.com/study/ib-math-aa-hl-u5-indefinite-and-definite-integration/)
- [Integration techniques: substitution, parts, partial fractions](https://www.owlsprep.com/study/ib-math-aa-hl-u5-integration-techniques-substitution-parts-partial/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ib-math-aa-hl-u5-applications-of-differentiation/