# Differentiation

> CIE IGCSE Mathematics · 0580 (2025-2027)
> Source: https://www.owlsprep.com/study/cie-0580-u2-differentiation/

This guide covers all differentiation content required for CIE IGCSE Maths 0580 Extended, including differentiating positive-integer power polynomials, using the gradient function, and finding/classifying stationary points.

**Prerequisites:** [Gradients of straight lines](https://www.owlsprep.com/study/cie-0580-u2-straight-line-graphs/); [Polynomial simplification](https://www.owlsprep.com/study/cie-0580-u2-polynomials/)

## Learning objectives

- Differentiate terms of the form $ax^n$ (n positive integer or 0) and sums of up to 3 such terms
- Interpret the derivative as the gradient function of a curve at any point
- Calculate the coordinates of stationary points of polynomial functions
- Classify stationary points as maxima or minima using first or second derivative tests

## Basic Differentiation of $ax^n$ Terms

**Power Rule for Differentiation** — For a term $f(x) = ax^n$ where $a$ is a constant and $n$ is a positive integer or 0, the derivative is $f'(x) = nax^{n-1}$. The derivative of any constant (where $n=0$, so $x^0 = 1$) is 0.

When differentiating a sum of up to 3 $ax^n$ terms, differentiate each term individually and add the results together. No additional rules (chain, product, quotient) are required for this syllabus.

**Worked example:** Differentiate $f(x) = 3x^2 + 5x + 7$

1. Differentiate the first term $3x^2$:

   $$2 \times 3x^{2-1} = 6x$$
2. Differentiate the second term $5x$:

   $$1 \times 5x^{1-1} = 5 \times 1 = 5$$
3. Differentiate the constant term 7: derivative = 0
4. Combine results:

   $$f'(x) = 6x + 5$$

> **Exam tip:** Always double check that you have reduced the power by 1 after multiplying by the original exponent, this is the most common mark-losing mistake on basic differentiation questions.

*Calculator:* forbidden

## Derivative as Gradient Function

The derivative of a function $f(x)$ is called the gradient function. If you substitute a value of $x$ into $f'(x)$, you get the exact gradient of the curve $y=f(x)$ at that $x$-coordinate, which can be used to find equations of tangents or normals.

**Worked example:** Find the gradient of the curve $y = 2x^3 - 4x + 1$ at the point where $x=2$

1. Differentiate the function to get the gradient function:

   $$\frac{dy}{dx} = 6x^2 - 4$$
2. Substitute $x=2$ into the derivative:

   $$6(2)^2 - 4 = 6 \times 4 - 4 = 20$$
3. The gradient of the curve at $x=2$ is 20

> **tip**
>
> If asked for the equation of a tangent to the curve, use the calculated gradient and substitute into the straight line formula $y - y_1 = m(x - x_1)$ after finding the corresponding $y$ value from the original function.

*Calculator:* allowed

## Finding Stationary Points

**Stationary Point** — A point on a curve where the gradient is zero ($f'(x) = 0$). The tangent to the curve at this point is horizontal, so it is either a local maximum or minimum value of the function.

To find stationary points, first calculate the derivative of the function, set it equal to zero and solve for $x$. Substitute the resulting $x$ value back into the original function to find the corresponding $y$ coordinate of the point.

**Worked example:** Find the coordinates of the stationary point of $y = x^2 - 4x + 3$

1. Calculate the first derivative:

   $$\frac{dy}{dx} = 2x - 4$$
2. Set derivative equal to zero and solve for $x$:

   $$2x - 4 = 0 \implies x = 2$$
3. Substitute $x=2$ back into the original function to find $y$:

   $$y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$$
4. The stationary point is at $(2, -1)$

*Calculator:* allowed

## Classifying Stationary Points

You can classify stationary points as maxima or minima using one of two accepted methods for your syllabus: the gradient sign test, or the second derivative test.

**Comparing methods**

Choose the fastest method for the question given:

- **Gradient Sign Test** — Calculate $f'(x)$ for values just less than and just greater than the $x$-coordinate of the stationary point. If gradient changes from positive to negative: maximum. If negative to positive: minimum.
  - Pros: No extra differentiation required
  - Cons: Requires multiple calculations

- **Second Derivative Test** — Differentiate $f'(x)$ to get the second derivative $f''(x)$. Substitute the stationary point $x$ value: if $f''(x) < 0$ it is a maximum, if $f''(x) > 0$ it is a minimum.
  - Pros: Fast, only one calculation after finding second derivative
  - Cons: Requires differentiating twice

**Worked example:** Classify the stationary point $(2, -1)$ of the function $y = x^2 - 4x + 3$

1. First derivative is $\frac{dy}{dx} = 2x - 4$, calculate the second derivative:

   $$\frac{d^2y}{dx^2} = 2$$
2. Substitute $x=2$ into the second derivative: $f''(2) = 2$, which is positive
3. Conclusion: The stationary point is a minimum

> **Exam tip:** The second derivative test is almost always the fastest method for polynomial functions in your exam, so prioritize learning this method to save time.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to reduce the power by 1 when differentiating, e.g. writing derivative of $3x^2$ as $6x^2$
  - Why it fails: You misremember the order of steps in the power rule
  - Correct: Always apply the rule $\frac{d}{dx}(ax^n) = nax^{n-1}$, so subtract 1 from the exponent after multiplying by the original exponent
- **Wrong:** Differentiating a constant term as the constant itself, e.g. derivative of 7 is 7
  - Why it fails: You misapply the power rule to $x^0$ terms
  - Correct: Remember the derivative of any constant is always 0, as $0 \times ax^{-1} = 0$
- **Wrong:** Substituting $x$ into the original function instead of the derivative when calculating gradient at a point
  - Why it fails: You confuse the original function (gives $y$ coordinate) with the gradient function (gives slope)
  - Correct: Use $f'(x)$ for gradient calculations, only use $f(x)$ to find the $y$ coordinate of a point
- **Wrong:** Mixing up second derivative test results, e.g. thinking positive $f''(x)$ is a maximum
  - Why it fails: You mix up sign interpretation
  - Correct: Use the mnemonic: Positive = Happy U-shaped curve (minimum), Negative = Sad ∩-shaped curve (maximum)
- **Wrong:** Trying to use chain/product/quotient rules for combined terms
  - Why it fails: You extend beyond syllabus scope unnecessarily
  - Correct: All exam functions are simple sums of $ax^n$ terms, so differentiate each term individually

## Cheatsheet

| Task | Key Steps | Rule/Formula |
| --- | --- | --- |
| Differentiate $ax^n$ | Multiply by $n$, subtract 1 from exponent | $\frac{d}{dx}(ax^n) = nax^{n-1}$ |
| Find gradient at $x=k$ | 1. Differentiate $f(x)$ to get $f'(x)$ 2. Substitute $x=k$ | $m = f'(k)$ |
| Find stationary point | 1. Set $f'(x)=0$, solve for $x$ 2. Substitute $x$ into $f(x)$ for $y$ | $f'(x) = 0$ |
| Classify stationary point (second derivative) | 1. Differentiate $f'(x)$ to get $f''(x)$ 2. Substitute stationary point $x$ | $f''(x) >0$ = minimum, $f''(x) <0$ = maximum |

## What's next

Now that you have mastered differentiation for CIE IGCSE 0580 Extended, you can apply these skills to solve optimisation problems, where you use maxima and minima to find the best possible value of a quantity (e.g. maximum area of a shape with fixed perimeter). You will also encounter differentiation questions combined with coordinate geometry, where you are asked to find equations of tangents and normals to curves. Make sure you practice past paper questions specifically for this topic to familiarise yourself with common exam phrasing and avoid the common pitfalls listed above.

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