Differentiation rules: power, product, quotient, chain

IB Mathematics Applications & Interpretation HL· Topic 5: Calculus, 5.4 Differentiation· 6 min read

1. The Power Rule★☆☆☆☆⏱ 10 min

📘 Definition

Power Rule

$f (x) = x^{n}$

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 $\frac{d}{d x} (a x^{n} + b x^{m}) = an x^{n - 1} + bm x^{m - 1}$ .

Example:

For $f (x) = x^{3}$ , $f^{'} (x) = 3 x^{2}$

📐 Worked Example

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

1
Rewrite all terms in the standard power form $a x^{n}$ :
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.

2. The Product Rule★★☆☆☆⏱ 12 min

📘 Definition

Product Rule

$y = u (x) v (x)$

If a function is the product of two differentiable functions $u (x)$ and $v (x)$ , its derivative is given by $\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$ .

Example:

For $y = x^{2} sin x$ , $y^{'} = x^{2} cos x + 2 x sin x$

📐 Worked Example

Find the derivative of $y = (3 x^{2} + 2 x) (x^{3} - 1)$ .

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
$\frac{dy}{dx} = 9x^4 + 6x^3 + 6x^4 + 2x^3 - 6x - 2 = 15x^4 + 8x^3 - 6x - 2$

3. The Quotient Rule★★★☆☆⏱ 15 min

📘 Definition

Quotient Rule

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

For a quotient of two differentiable functions where $v (x) \neq = 0$ , the derivative is given by $\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$ , where $u$ is the numerator and $v$ is the denominator.

Example:

For $y = \frac{s i n x}{x}$ , $y^{'} = \frac{x c o s x - s i n x}{x ^{2}}$

📐 Worked Example

Differentiate $y = \frac{2 x + 1}{x ^{2} - 3}$ 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.

4. Chain Rule and Combining Multiple Rules★★★★☆⏱ 20 min

📘 Definition

Chain Rule

$y = f (g (x))$

For a composite function $y = f (g (x))$ , let $u = g (x)$ (inner function). The derivative is $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ : differentiate outer function, multiply by derivative of inner function.

Example:

For $y = (x^{2} + 1)^{3}$ , $y^{'} = 3 (x^{2} + 1)^{2} \times 2 x = 6 x (x^{2} + 1)^{2}$

📐 Worked Example

Find the derivative of $y = \frac{x 2 x + 1}{( x ^{2} - 4 ) ^{2}}$ . 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
$\frac{dy}{dx} = \frac{-5x^3 - 3x^2 - 12x - 4}{(x^2 - 4)^3 \sqrt{2x+1}}$

5. Common Pitfalls

Wrong move:

Forgetting to multiply by the derivative of the inner function in chain rule

Why:

Only the outer function is differentiated, leading to an incorrect scaling factor for the derivative

Correct move:

Always identify the inner function first, differentiate it, and multiply its derivative to the derivative of the outer function

Wrong move:

Swapping the order of terms in the quotient rule numerator

Why:

Reversing the order gives all terms the wrong sign, leading to zero marks for the question

Correct move:

Use the mnemonic 'low d high minus high d low' to confirm the correct order before simplifying

Wrong move:

Differentiating a product by multiplying the derivatives of each factor

Why:

Confusing the derivative of a product with the derivative of a sum, leading to an incorrect result

Correct move:

Always use the full product rule: derivative of product is first d second plus second d first

Wrong move:

Incorrect negative exponent for reciprocal terms when applying power rule

Why:

Forgetting that $\frac{1}{x ^{n}} = x^{- n}$ leads to wrong exponent and wrong sign in the derivative

Correct move:

Rewrite all roots and reciprocals as explicit power terms before applying the power rule

Wrong move:

Forgetting to apply chain rule to inner composite functions when combining rules

Why:

Focusing on the outer product/quotient rule leads to missing the inner chain derivative

Correct move:

Work from the outside in: apply the outermost rule first, and apply chain rule to every composite term as you go

6. Quick Reference Cheatsheet

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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 1
Differentiate composite rational function
2022 · 1
Combine rules for complex derivative
2023 · 2
Differentiate for optimisation problem

Going deeper

practiceIB Past Bank Differentiation QuestionsFilter for differentiation questions by topic

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 implicit differentiation and related rates problems, 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.