# Differentiation — Standard Derivatives and the Chain Rule

> Additional Mathematics · CIE IGCSE 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u14-differentiation-standard-derivatives-and-the/

This guide covers all standard derivatives and the chain rule as required for CIE IGCSE Additional Mathematics 0606 (2025–2027 syllabus), including correct notation and exam-appropriate problem-solving steps for both Papers 1 and 2.

**Prerequisites:** Knowledge of composite functions, radians, and basic exponential/logarithmic functions; Familiarity with gradient of a curve at a point

## Learning objectives

- Recall standard derivatives for power, trigonometric, exponential and logarithmic functions
- Apply the chain rule to differentiate composite functions of all allowed types
- Use correct differentiation notation including $f'(x)$, $\frac{dy}{dx}$ and second derivatives
- Solve exam-style differentiation questions for both calculator and non-calculator papers

## Standard Derivatives & Notation

**First Derivative** — The function that gives the gradient of the original function $f(x)$ at any valid point $x$.

*Notation:* $f'(x)$, $\frac{dy}{dx}$

All standard derivatives below must be memorized, as they are not provided in the 0606 exam formula booklet. All trigonometric derivatives apply only when angles are measured in radians.

| Function $f(x)$ | Derivative $f'(x)$ |
| --- | --- |
| $x^n$, $n$ rational | $n x^{n-1}$ |
| $\text{sin }x$ | $\text{cos }x$ |
| $\text{cos }x$ | $-\text{sin }x$ |
| $\text{tan }x$ | $\text{sec}^2 x$ |
| $e^x$ | $e^x$ |
| $\text{ln }x$ | $\frac{1}{x}$ |

**Worked example:** Find the first derivative of $y = 4x^3 + 2\text{ln }x - 3\text{cos }x$

1. Differentiate each term separately using the standard derivatives table.
2. $$\frac{d}{dx}(4x^3) = 4 \times 3x^{2} = 12x^2$$
3. $$\frac{d}{dx}(2\text{ln }x) = 2 \times \frac{1}{x} = \frac{2}{x}$$
4. $$\frac{d}{dx}(-3\text{cos }x) = -3 \times (-\text{sin }x) = 3\text{sin }x$$
5. Combine all terms to get the final derivative.
6. $$\frac{dy}{dx} = 12x^2 + \frac{2}{x} + 3\text{sin }x$$

> **Exam tip:** Always check you have included the negative sign for the derivative of $\boldsymbol{\frac{d}{dx}(\text{cos }x) = -\text{sin }x}$ — this is one of the most common mark-losing mistakes in Paper 1.

*Calculator:* forbidden

## Introduction to the Chain Rule

**Chain Rule** — Rule used to differentiate composite functions, by multiplying the derivative of the outer function (keeping the inner function unchanged) by the derivative of the inner function.

*Notation:* $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ where $u = g(x)$ is the inner function of $y = f(g(x))$

> **mnemonic**
>
> OUTER FIRST, INNER NEXT: Differentiate outer function (leave inner as is) → multiply by derivative of inner function.

**Worked example:** Find $\frac{dy}{dx}$ for $y = (2x + 5)^7$ using the chain rule.

1. Identify inner and outer functions: let $u = 2x + 5$, so $y = u^7$.
2. $$\frac{dy}{du} = 7u^6$$
3. $$\frac{du}{dx} = 2$$
4. Apply the chain rule formula, then substitute back $u = 2x +5$.
5. $$\frac{dy}{dx} = 7u^6 \times 2 = 14(2x +5)^6$$

**Check your understanding**

1. What is the derivative of $(3x - 1)^4$?

   - $4(3x -1)^3$
   - $12(3x -1)^3$
   - $12(3x -1)^4$
   - $3(3x -1)^3$

   *Answer:* $12(3x -1)^3$

   *Why:* Correct: Derivative of outer function is $4(3x-1)^3$, derivative of inner function is 3, multiplied gives $12(3x-1)^3$.

*Calculator:* forbidden

## Chain Rule for Composite Standard Functions

Combine the chain rule with all standard derivatives (trigonometric, exponential, logarithmic) to differentiate any composite function allowed in the 0606 syllabus.

**Worked example:** Differentiate $y = e^{\text{sin }x}$ with respect to $x$.

1. Let inner function $u = \text{sin }x$, so $y = e^u$.
2. $$\frac{dy}{du} = e^u = e^{\text{sin }x}, \frac{du}{dx} = \text{cos }x$$
3. $$\frac{dy}{dx} = e^{\text{sin }x} \times \text{cos }x = \text{cos }x e^{\text{sin }x}$$

**Worked example:** Find $f'(x)$ if $f(x) = \text{ln}(\text{tan }x)$ where $0 < x < \frac{\pi}{2}$.

1. Let inner function $u = \text{tan }x$, so $f(u) = \text{ln }u$.
2. $$f'(u) = \frac{1}{u} = \frac{1}{\text{tan }x}, u' = \text{sec}^2 x$$
3. $$f'(x) = \frac{1}{\text{tan }x} \times \text{sec}^2 x = \frac{1}{\text{sin }x \text{ cos }x}$$

> **Exam tip:** You do not need to simplify trigonometric derivatives further unless explicitly asked to in the question, though simplifying may help with follow-up problem parts.

*Calculator:* allowed

## Second Derivatives of Composite Functions

**Second Derivative** — The derivative of the first derivative of a function, used to find the nature of stationary points and rates of change of gradients.

*Notation:* $f''(x)$, $\frac{d^2y}{dx^2}$

**Worked example:** Find $\frac{d^2y}{dx^2}$ for $y = \text{cos}(3x)$.

1. First calculate the first derivative using the chain rule.
2. $$\frac{dy}{dx} = -\text{sin}(3x) \times 3 = -3\text{sin}(3x)$$
3. Differentiate the first derivative again to get the second derivative, applying the chain rule a second time.
4. $$\frac{d^2y}{dx^2} = -3 \times \text{cos}(3x) \times 3 = -9\text{cos}(3x)$$

> **note**
>
> Second derivatives are frequently used in follow-up questions to distinguish between maximum and minimum points, covered in the applications of differentiation unit.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting the negative sign when differentiating $\text{cos }x$
  - Why it fails: The derivative of $\text{cos }x$ is $-\text{sin }x$, not $\text{sin }x$, so missing the sign costs 1 mark per occurrence.
  - Correct: Memorize trigonometric derivative pairs explicitly: $\text{sin }x \to +\text{cos }x$, $\text{cos }x \to -\text{sin }x$, $\text{tan }x \to +\text{sec}^2 x$.
- **Wrong:** Forgetting to multiply by the derivative of the inner function when applying the chain rule
  - Why it fails: The chain rule requires two factors (outer derivative × inner derivative), omitting the inner derivative gives an incorrect result.
  - Correct: Use the 'outer first, inner next' mnemonic to ensure you include both factors for every composite function differentiation.
- **Wrong:** Using degrees instead of radians for trigonometric differentiation
  - Why it fails: Standard trigonometric derivatives only apply when angles are measured in radians, the default for all 0606 calculus questions.
  - Correct: Always convert any angle given in degrees to radians before differentiating trigonometric functions.
- **Wrong:** Differentiating $\text{ln }x$ as $\text{ln}(\frac{1}{x})$ or $x$ instead of $\frac{1}{x}$
  - Why it fails: The derivative of $\text{ln }x$ is a reciprocal function, not another logarithmic or linear function.
  - Correct: Add the pair $\text{ln }x \to \frac{1}{x}$ to your memorized standard derivatives and test yourself regularly.
- **Wrong:** Expanding high-power composite functions (e.g. $(2x+7)^9$) before differentiating
  - Why it fails: Expanding these functions is time-consuming and prone to arithmetic errors, wasting valuable exam time.
  - Correct: Always use the chain rule for composite power functions, no matter how high the exponent.

## Cheatsheet

| Rule/Function | Derivative Result |
| --- | --- |
| $x^n$ (n rational) | $n x^{n-1}$ |
| $\text{sin }x$ | $\text{cos }x$ |
| $\text{cos }x$ | $-\text{sin }x$ |
| $\text{tan }x$ | $\text{sec}^2 x$ |
| $e^x$ | $e^x$ |
| $\text{ln }x$ | $\frac{1}{x}$ |
| Chain Rule ($y=f(g(x))$) | $f'(g(x)) \times g'(x)$ |
| Second Derivative | Derivative of first derivative |

## What's next

Now that you have mastered standard derivatives and the chain rule, you are ready to learn the product and quotient rules for differentiating products and fractions of functions, followed by applications of differentiation including finding stationary points, rates of change, and optimization problems for CIE IGCSE Additional Mathematics 0606. These skills are heavily tested across both Papers 1 and 2, so practice with a wide range of composite function questions to build speed and accuracy before your exam.

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