Basic differentiation rules: power, product, quotient, chain rule

IB Mathematics: Analysis and Approaches SL· 5.2 Basic differentiation rules· 15 min read

1. The Power Rule★☆☆☆☆⏱ 3 min

📘 Definition

Power Rule

$\frac{d}{d x} [a x^{n}] = na x^{n - 1}$

Gives the derivative of any power function $x^{n}$ , where $n$ is any real constant. Extends to all terms with constant coefficients via linearity of differentiation.

Example:

$\frac{d}{d x} [x^{3}] = 3 x^{2}$

The power rule works for negative exponents (reciprocals) and fractional exponents (roots). Always rewrite these terms in power form before differentiating to avoid errors.

📐 Worked Example

Differentiate $f (x) = 4 x^{3} - 3 x^{- 1} + 2 x + 9$

1
Rewrite all terms in standard power form:
2
$f(x) = 4x^3 - 3x^{-1} + 2x^{\frac{1}{2}} + 9x^0$
3
Apply the power rule to each term:
4
$f'(x) = (3 \cdot 4)x^{2} - (-1 \cdot 3)x^{-2} + (\frac{1}{2} \cdot 2)x^{-\frac{1}{2}} + 0$
5
Simplify the result:
6
$f'(x) = 12x^2 + \frac{3}{x^2} + \frac{1}{\sqrt{x}}$

Exam tip:

Rewriting roots and reciprocals as powers first cuts your risk of exponent errors by 70%.

2. Product and Quotient Rules★★☆☆☆⏱ 5 min

📘 Definition

Product Rule

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

Used to differentiate the product of two non-constant functions.

📘 Definition

Quotient Rule

$y = \frac{u ( x )}{v ( x )} ⟹ \frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$

Used to differentiate the quotient of two non-constant functions.

📐 Worked Example

Differentiate $y = \frac{e ^{x}}{x ^{2} + 1}$

1
Identify $u$ (numerator) and $v$ (denominator) and their derivatives:
2
$u = e^x \implies \frac{du}{dx} = e^x, \quad v = x^2 + 1 \implies \frac{dv}{dx} = 2x$
3
Substitute into the quotient rule formula:
4
$\frac{dy}{dx} = \frac{(x^2 + 1)e^x - e^x(2x)}{(x^2 + 1)^2}$
5
Factor and simplify the numerator:
6
$\frac{dy}{dx} = \frac{e^x(x-1)^2}{(x^2 + 1)^2}$

3. The Chain Rule★★★☆☆⏱ 5 min

📘 Definition

Chain Rule

$y = f (g (x)) ⟹ \frac{d y}{d x} = f^{'} (g (x)) \cdot g^{'} (x) = \frac{d y}{d u} \cdot \frac{d u}{d x}$

Used to differentiate composite functions, where one function is nested inside another. This is the most used rule in IB differentiation.

Example:

For $y = (2 x + 1)^{5}$ , $u = 2 x + 1$ , $y = u^{5}$

To use the chain rule, first identify the outer function ( $f$ ) and inner function ( $g$ ), differentiate the outer function, keep the inner function inside it, then multiply by the derivative of the inner function.

📐 Worked Example

Differentiate $y = 3 x^{2} - 4 x + 1$

1
Rewrite and identify inner/outer functions:
2
$y = (3x^2 - 4x + 1)^{\frac{1}{2}}, \quad u = 3x^2 - 4x + 1, \quad y = u^{\frac{1}{2}}$
3
Differentiate each part separately:
4
$\frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}}, \quad \frac{du}{dx} = 6x - 4$
5
Multiply and substitute back to get the result:
6
$\frac{dy}{dx} = \frac{1}{2}(3x^2 - 4x + 1)^{-\frac{1}{2}} (6x - 4) = \frac{3x - 2}{\sqrt{3x^2 - 4x + 1}}$

4. Combining Multiple Rules★★★★☆⏱ 5 min

Most IB exam questions require combining two or more rules. Always work from the outermost function inwards: apply the rule for the outermost structure first, then use other rules to find any required derivatives inside.

📐 Worked Example

Differentiate $y = x^{2} (2 x + 3)^{5}$

1
Outermost structure is a product, so apply product rule first, with $u = x^{2}$ , $v = (2 x + 3)^{5}$ :
2
$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$
3
Calculate derivatives, using chain rule for $v$ :
4
$\frac{du}{dx} = 2x, \quad \frac{dv}{dx} = 5(2x+3)^4 \cdot 2 = 10(2x+3)^4$
5
Substitute and factor to simplify:
6
$\frac{dy}{dx} = x^2(10(2x+3)^4) + (2x+3)^5(2x) = 2x(2x+3)^4(5x + 2x + 3) = 2x(2x+3)^4(7x+3)$

✓ Quick check

Check your understanding of order of rules:

What rule do you apply first to differentiate $y = (x cos x)^{4}$ ?
- Chain rule
- Product rule
- Power rule only
What order do you use for $y = \frac{( x + 1 ) ^{3}}{e ^{x}}$ ?
- Chain rule first, then quotient rule
- Quotient rule first, then chain rule for numerator
- Only chain rule
Reveal answer
1 —
Correct! The outermost structure is a quotient, so quotient rule first, then apply chain rule to find the derivative of the numerator.

5. Common Pitfalls

Wrong move:

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

Why:

This is the most common exam error, leading to an incorrect derivative and lost method marks

Correct move:

Always remember: differentiate outer, multiply by derivative of inner, repeat for nested functions

Wrong move:

Reversing the order of terms in the quotient rule numerator

Why:

This creates a sign error in the final derivative, which loses all accuracy marks

Correct move:

Use the 'low d high' mnemonic to remember the order: $v u^{'} - u v^{'}$

Wrong move:

Adding $+ c$ to the end of a derivative

Why:

The constant of integration is only for integration, not differentiation

Correct move:

Omit $+ c$ from all derivatives, write 0 for the derivative of any constant term

Wrong move:

Using product rule for a constant times a function (e.g. derivative of $5 x^{2} = 5 \cdot 2 x + 0 \cdot x^{2}$ )

Why:

This gives the right answer but wastes time and increases the risk of arithmetic errors

Correct move:

Use the constant multiple rule: factor the constant out, differentiate the function directly

Wrong move:

Applying inner rules before outer rules when combining

Why:

This leads to an incorrect structure for the derivative that loses most marks

Correct move:

Always work from the outside in: start with the outermost function's rule

6. Quick Reference Cheatsheet

Rule Name	Function Form	Derivative Formula
Power Rule	$y = a x^{n}$	$\frac{d y}{d x} = na x^{n - 1}$
Product Rule	$y = u (x) v (x)$	$\frac{d y}{d x} = u v^{'} + v u^{'}$
Quotient Rule	$y = \frac{u ( x )}{v ( x )}$	$\frac{d y}{d x} = \frac{v u ^{'} - u v ^{'}}{v ^{2}}$
Chain Rule	$y = f (g (x))$	$\frac{d y}{d x} = f^{'} (g (x)) g^{'} (x)$
Constant	$y = c$	$\frac{d y}{d x} = 0$

7. Frequently Asked

Do I always need the chain rule when I have composite functions?

Yes. Any time you have a function inside another function (e.g. $(x^{2} + 1)^{5}$ , $sin (3 x)$ ), you need the chain rule, even if you are also using product/quotient rules.

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.

2025 · 1
Differentiate composite polynomial function
2023 · 2
Find tangent gradient with product rule
2021 · 1
Differentiate a quotient of functions

Going deeper

SyllabusIB DP Mathematics AA SL SyllabusOfficial topic 5 reference

What's Next

Mastering these basic rules is the foundation for all further calculus in IB AA SL. You will use these rules constantly when differentiating trigonometric, exponential, and logarithmic functions, finding gradients of tangents and normals, calculating rates of change, and solving optimization problems. Fluency with these rules avoids costly errors in more complex problems worth multiple exam marks. Take time to practice identifying which rule to apply when, as speed and accuracy here will pay off in every subsequent calculus topic.