# Differentiation rules: power, product, quotient, chain

> IB Mathematics Applications & Interpretation HL · Unit 4: Calculus
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u4-differentiation-rules-power-product-quotient/

This module covers the core rules for differentiating common functions, including power, product, quotient and chain rules. You will learn to apply these rules individually and combine them to differentiate complex functions for optimisation and rate of change problems.

**Prerequisites:** [Introduction to derivatives and first principles](https://www.owlsprep.com/study/ib-math-ai-hl-u4-introduction-to-derivatives/)

## Learning objectives

- Apply the power rule to differentiate polynomial and power functions
- Correctly use product and quotient rules for products/quotients of functions
- Apply the chain rule to differentiate composite functions
- Combine multiple rules to differentiate complex functions

## The Power Rule

**Power Rule** — The derivative of a power function $x^n$ is $n x^{n-1}$ for any real constant $n$. For scalar multiples and sums, we have $\frac{d}{dx}(ax^n + bx^m) = a n x^{n-1} + b m x^{m-1}$.

*Notation:* $f(x) = x^n$

*Example:* For $f(x) = x^3$, $f'(x) = 3x^2$

**Worked example:** Differentiate $f(x) = 4x^5 - \frac{2}{x^2} + 7\sqrt{x}$ with respect to $x$.

1. Rewrite all terms in the standard power form $ax^n$:
2. $$f(x) = 4x^5 - 2x^{-2} + 7x^{1/2}$$
3. Apply the power rule to each term individually:
4. $$f'(x) = 4 \cdot 5 x^{5-1} - 2 \cdot (-2) x^{-2-1} + 7 \cdot \frac{1}{2} x^{(1/2)-1}$$
5. Simplify the result and rewrite in original notation:
6. $$f'(x) = 20x^4 + 4x^{-3} + \frac{7}{2}x^{-1/2} = 20x^4 + \frac{4}{x^3} + \frac{7}{2\sqrt{x}}$$

> **Exam tip:** Always rewrite roots and reciprocals as power terms before applying the rule to avoid sign and exponent errors.

## The Product Rule

**Product Rule** — If a function is the product of two differentiable functions $u(x)$ and $v(x)$, its derivative is given by $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$.

*Notation:* $y = u(x) v(x)$

*Example:* For $y = x^2 \sin x$, $y' = x^2 \cos x + 2x \sin x$

> **mnemonic**
>
> First times derivative of second plus second times derivative of first

**Worked example:** Find the derivative of $y = (3x^2 + 2x)(x^3 - 1)$.

1. Assign the two factors to $u$ and $v$:
2. $$u = 3x^2 + 2x, \quad v = x^3 - 1$$
3. Calculate derivatives of each individual function:
4. $$\frac{du}{dx} = 6x + 2, \quad \frac{dv}{dx} = 3x^2$$
5. Substitute into the product rule formula:
6. $$\frac{dy}{dx} = (3x^2 + 2x)(3x^2) + (x^3 - 1)(6x + 2)$$
7. Expand and simplify the result:
8. $$\frac{dy}{dx} = 9x^4 + 6x^3 + 6x^4 + 2x^3 - 6x - 2 = 15x^4 + 8x^3 - 6x - 2$$

## The Quotient Rule

**Quotient Rule** — For a quotient of two differentiable functions where $v(x) \neq 0$, the derivative is given by $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$, where $u$ is the numerator and $v$ is the denominator.

*Notation:* $y = \frac{u(x)}{v(x)}$

*Example:* For $y = \frac{\sin x}{x}$, $y' = \frac{x \cos x - \sin x}{x^2}$

> **mnemonic**
>
> Low d high minus high d low, over the square of what's below. (Low = denominator, high = numerator)

**Worked example:** Differentiate $y = \frac{2x + 1}{x^2 - 3}$ with respect to $x$.

1. Assign numerator to $u$ and denominator to $v$:
2. $$u = 2x + 1, \quad v = x^2 - 3$$
3. Calculate derivatives of $u$ and $v$:
4. $$\frac{du}{dx} = 2, \quad \frac{dv}{dx} = 2x$$
5. Substitute into the quotient rule formula:
6. $$\frac{dy}{dx} = \frac{(x^2 - 3)(2) - (2x + 1)(2x)}{(x^2 - 3)^2}$$
7. Expand, factor and simplify the numerator:
8. $$\frac{dy}{dx} = \frac{2x^2 - 6 - 4x^2 - 2x}{(x^2 - 3)^2} = \frac{-2(x^2 + x + 3)}{(x^2 - 3)^2}$$

> **Exam tip:** Always double-check the sign of the numerator, this is the most frequently marked error in IB exams for quotient rule questions.

## Chain Rule and Combining Multiple Rules

**Chain Rule** — For a composite function $y = f(g(x))$, let $u = g(x)$ (inner function). The derivative is $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$: differentiate outer function, multiply by derivative of inner function.

*Notation:* $y = f(g(x))$

*Example:* For $y = (x^2 + 1)^3$, $y' = 3(x^2 + 1)^2 \times 2x = 6x(x^2 + 1)^2$

**Worked example:** Find the derivative of $y = \frac{x \sqrt{2x + 1}}{(x^2 - 4)^2}$. Simplify your answer.

1. Rewrite the function to make differentiation easier:
2. $$y = x (2x+1)^{1/2} (x^2 - 4)^{-2}$$
3. First differentiate the inner composite terms using chain rule:
4. $$\frac{d}{dx}\left((2x+1)^{1/2}\right) = \frac{1}{2}(2x+1)^{-1/2} \cdot 2 = (2x+1)^{-1/2}$$
5. $$\frac{d}{dx}\left((x^2-4)^{-2}\right) = -2(x^2-4)^{-3} \cdot 2x = -4x(x^2-4)^{-3}$$
6. Apply the product rule to the three terms, then simplify:
7. $$\frac{dy}{dx} = \frac{-5x^3 - 3x^2 - 12x - 4}{(x^2 - 4)^3 \sqrt{2x+1}}$$

## Common pitfalls

- **Wrong:** Forgetting to multiply by the derivative of the inner function in chain rule
  - Why it fails: Only the outer function is differentiated, leading to an incorrect scaling factor for the derivative
  - Correct: Always identify the inner function first, differentiate it, and multiply its derivative to the derivative of the outer function
- **Wrong:** Swapping the order of terms in the quotient rule numerator
  - Why it fails: Reversing the order gives all terms the wrong sign, leading to zero marks for the question
  - Correct: Use the mnemonic 'low d high minus high d low' to confirm the correct order before simplifying
- **Wrong:** Differentiating a product by multiplying the derivatives of each factor
  - Why it fails: Confusing the derivative of a product with the derivative of a sum, leading to an incorrect result
  - Correct: Always use the full product rule: derivative of product is first d second plus second d first
- **Wrong:** Incorrect negative exponent for reciprocal terms when applying power rule
  - Why it fails: Forgetting that $\frac{1}{x^n} = x^{-n}$ leads to wrong exponent and wrong sign in the derivative
  - Correct: Rewrite all roots and reciprocals as explicit power terms before applying the power rule
- **Wrong:** Forgetting to apply chain rule to inner composite functions when combining rules
  - Why it fails: Focusing on the outer product/quotient rule leads to missing the inner chain derivative
  - Correct: Work from the outside in: apply the outermost rule first, and apply chain rule to every composite term as you go

## Cheatsheet

| Rule Name | Formula | Key Notes |
| --- | --- | --- |
| Power Rule | $\frac{d}{dx}(ax^n) = n a x^{n-1}$ | Works for all real $n$ |
| Product Rule | $\frac{d}{dx}(uv) = u v' + v u'$ | $u, v$ are functions of $x$ |
| Quotient Rule | $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}$ | $u$ = numerator, $v$ = denominator |
| Chain Rule | $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ | For composite $y = f(g(x))$, $u = g(x)$ |

## What's next

Now you have mastered the core differentiation rules, you can apply these to solve a wide range of applied calculus problems common in IB AI HL. These rules are the foundation for almost all remaining calculus topics in this unit, including finding stationary points, solving optimisation problems, and differentiating exponential, logarithmic and trigonometric functions. You will also rely on these rules for related rates problems, where the chain rule links quantities that change together over time, which appear regularly in Paper 2. Consistent practice to automate these rules now will make all future calculus topics far easier, as you will not need to waste time recalling basic rules when solving complex multi-step problems.

- [Related Rates](https://www.owlsprep.com/study/ib-math-ai-hl-u4-related-rates/)
- [Derivatives of standard functions](https://www.owlsprep.com/study/ib-math-ai-hl-u4-derivatives-of-standard-functions/)
- [Tangents, normals, critical points, inflection points](https://www.owlsprep.com/study/ib-math-ai-hl-u4-tangents-normals-critical-points-inflection/)

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