# Differentiation rules

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

This module covers all core differentiation rules required for IB Math AA HL, from basic constant and power rules to product, quotient and chain rules for composite functions. You will learn application strategies and avoid common exam mistakes.

**Prerequisites:** [Limit definition of the derivative](https://www.owlsprep.com/study/ib-math-aa-hl-u5-introduction-to-derivatives/)

## Learning objectives

- Recall and apply all core differentiation rules for common function types
- Correctly apply product, quotient and chain rules to combinations of functions
- Identify which rule to use for different differentiation problems
- Avoid common sign and coefficient errors in exam questions

## Basic Differentiation Rules

**Basic Differentiation Rules** — A set of shortcut rules that allow you to compute derivatives directly without using the limit definition, drastically speeding up calculations

*Notation:* \frac{d}{dx}[f(x)]

*Example:* The derivative of $x^n$ is $nx^{n-1}$ by the power rule

All basic rules follow from the limit definition of the derivative, but memorizing them is essential for solving complex problems in the limited exam time. Core basic rules for common functions are:

- Constant rule: $\frac{d}{dx}[c] = 0$ for any constant $c$
- Power rule: $\frac{d}{dx}[x^n] = nx^{n-1}$ for any real $n$
- Exponential rule: $\frac{d}{dx}[e^x] = e^x$, $\frac{d}{dx}[a^x] = a^x \ln a$
- Natural logarithm rule: $\frac{d}{dx}[\ln x] = \frac{1}{x}$
- Trigonometric rules: $\frac{d}{dx}[\sin x] = \cos x$, $\frac{d}{dx}[\cos x] = -\sin x$

**Worked example:** Find the derivative of $f(x) = 3x^4 - 2\sqrt{x} + \frac{5}{x^2} + 7$

1. Rewrite all terms with power exponents to apply the power rule:
2. $$3x^4 - 2x^{1/2} + 5x^{-2} + 7x^0$$
3. Apply the power rule term by term, using the constant multiple and sum rules:
4. $$f'(x) = 3(4x^3) - 2\left(\frac{1}{2}x^{-1/2}\right) + 5(-2x^{-3}) + 0$$
5. Simplify the final result:
6. $$f'(x) = 12x^3 - \frac{1}{\sqrt{x}} - \frac{10}{x^3}$$

> **Exam tip:** Always rewrite roots and reciprocals as power terms before differentiating to avoid coefficient errors.

## Product and Quotient Rules

**Product & Quotient Rules** — Rules for differentiating products and ratios of two distinct differentiable functions, respectively.

*Notation:* (fg)' = f'g + fg', \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}

The product rule applies when two functions are multiplied together, while the quotient rule applies when one function is divided by another. A common mnemonic helps remember the quotient rule order:

> **mnemonic**
>
> Low d high minus high d low, over the square of what's below

**Worked example:** Find the derivative of $g(x) = \frac{2x + 1}{x - 3}$

1. Identify numerator $f(x) = 2x+1$ and denominator $g(x) = x-3$, then find their derivatives: $f'(x) = 2$, $g'(x) = 1$
2. Apply the quotient rule:
3. $$g'(x) = \frac{(2)(x - 3) - (2x + 1)(1)}{(x - 3)^2}$$
4. Simplify the numerator: $2x - 6 - 2x - 1 = -7$
5. Final simplified result:
6. $$g'(x) = \frac{-7}{(x - 3)^2}$$

> **Exam tip:** Swapping the order of terms in the quotient numerator flips the sign of your answer, always double-check the order.

## The Chain Rule for Composite Functions

**Chain Rule** — A rule for differentiating composite functions, where one function is applied to the output of another.

*Notation:* \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}, (f(g(x)))' = f'(g(x)) \cdot g'(x)

The chain rule is the most frequently used differentiation rule in IB exams. The strategy is simple: differentiate the outer function first, leave the inner function unchanged, then multiply by the derivative of the inner function.

**Worked example:** Differentiate $y = (3x^2 - 5x + 2)^4$

1. Define inner function $u = 3x^2 - 5x + 2$ and outer function $y = u^4$
2. Differentiate each function separately:
3. $$\frac{dy}{du} = 4u^3, \quad \frac{du}{dx} = 6x - 5$$
4. Apply the chain rule:
5. $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 4u^3(6x - 5)$$
6. Substitute back $u = 3x^2 - 5x + 2$ to get the final result:
7. $$\frac{dy}{dx} = (24x - 20)(3x^2 - 5x + 2)^3$$

**Check your understanding**

Test your understanding of when to use the chain rule:

1. Do you need to apply the chain rule to find the derivative of $y = x^2 \sin x$?

   - Yes, because it is a product of two functions
   - No, chain rule is only for composite functions, product rule is needed here
   - Yes, because $x^2$ is a power function

   *Answer:* No, chain rule is only for composite functions, product rule is needed here

   *Why:* Correct! $y = x^2 \sin x$ is a product of two functions, not a composite function, so only the product rule is required.

## Combined Differentiation Problems

Most IB exam questions require combining multiple differentiation rules, almost always the chain rule with product or quotient rules. A reliable strategy is to start with the outermost operation, then work inwards, applying the required rule at each step.

**Worked example:** Find the derivative of $f(x) = e^{-x} \cos(2x)$

1. The outermost operation is a product, so start with the product rule:
2. $$f'(x) = \frac{d}{dx}[e^{-x}] \cos(2x) + e^{-x} \frac{d}{dx}[\cos(2x)]$$
3. Differentiate each term using the chain rule:
4. $$\frac{d}{dx}[e^{-x}] = e^{-x}(-1) = -e^{-x}, \quad \frac{d}{dx}[\cos(2x)] = -\sin(2x)(2) = -2\sin(2x)$$
5. Substitute back into the product rule expression:
6. $$f'(x) = (-e^{-x})\cos(2x) + e^{-x}(-2\sin(2x))$$
7. Factor out common terms to simplify:
8. $$f'(x) = -e^{-x}\left(\cos(2x) + 2\sin(2x)\right)$$

> **Exam tip:** Label each function layer and write out every step to avoid forgetting a chain rule derivative term, a very common exam error.

## Common pitfalls

- **Wrong:** Forgetting to multiply by the derivative of the inner function when applying the chain rule
  - Why it fails: Only the outer function is differentiated, leaving out a required factor that changes the entire result
  - Correct: Always remember to multiply by the derivative of the inner function after differentiating the outer function
- **Wrong:** Swapping the order of terms in the numerator of the quotient rule
  - Why it fails: This flips the sign of the entire derivative, leading to an incorrect answer that is commonly marked wrong
  - Correct: Use the mnemonic 'low d high minus high d low' to remember the correct order of terms
- **Wrong:** Differentiating products by multiplying derivatives: $(fg)' = f'g'$
  - Why it fails: Confusing the product rule with the sum rule, where the derivative of a sum is the sum of derivatives
  - Correct: Always use the full product rule: $(fg)' = f'g + fg'$
- **Wrong:** Writing $\frac{d}{dx}[\ln x] = \frac{1}{\ln x}$
  - Why it fails: Confusing the derivative of the logarithmic function with the power rule for $x^n$
  - Correct: Memorize that the derivative of $\ln x$ with respect to $x$ is always $\frac{1}{x}$
- **Wrong:** Forgetting $\ln a$ when differentiating $a^x$, writing $\frac{d}{dx}[a^x] = a^x$
  - Why it fails: Confusing general exponential functions with $e^x$, which is its own derivative
  - Correct: Always multiply by $\ln a$ when differentiating exponential functions with base $a \neq e$

## Cheatsheet

| Rule Name | Formula |
| --- | --- |
| Constant Rule | $\frac{d}{dx}[c] = 0$ |
| Power Rule | $\frac{d}{dx}[x^n] = nx^{n-1}$ |
| Product Rule | $(fg)' = f'g + fg'$ |
| Quotient Rule | $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ |
| Chain Rule | $(f(g(x)))' = f'(g(x))g'(x)$ |
| Exponential | $\frac{d}{dx}[e^x] = e^x, \frac{d}{dx}[a^x] = a^x \ln a$ |
| Logarithm | $\frac{d}{dx}[\ln x] = \frac{1}{x}$ |
| Sine | $\frac{d}{dx}[\sin x] = \cos x$ |
| Cosine | $\frac{d}{dx}[\cos x] = -\sin x$ |

## What's next

Differentiation rules are the foundation for almost all further calculus topics in IB Math AA HL. Once you master these core rules, you can move on to more advanced differentiation techniques including implicit differentiation, parametric differentiation, and finding higher order derivatives. You will also use these rules constantly when finding critical points, sketching curves, solving optimization problems, and reversing the process for integration by substitution in later units. Consistent practice applying combined rules is key to speed and accuracy in the exam.

- [Higher Order Derivatives](https://www.owlsprep.com/study/ib-math-aa-hl-u5-higher-order-derivatives/)
- [Applications of Differentiation](https://www.owlsprep.com/study/ib-math-aa-hl-u5-applications-of-differentiation/)
- [Implicit, parametric and logarithmic differentiation](https://www.owlsprep.com/study/ib-math-aa-hl-u5-implicit-parametric-and-logarithmic-differentiation/)

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