Differentiation of products, quotients and composite functions

Q: Can I use the product rule instead of the quotient rule for fractions?

Yes, you can rewrite a quotient as and apply the product and chain rules. Both methods will give the same correct result, and both are accepted in mark schemes.

Edexcel International GCSE Further Pure Mathematics· S9B (2016 spec, first exam 2019)· 18 min read

1. 1. The Product Rule★★☆☆☆⏱ 4 min

✓ Calculator OK

📘 Definition

Product Rule

\frac{d}{dx}(uv) = u'v + uv'

Used to differentiate the product of two differentiable functions u(x) and v(x). You must recall this formula for your exam.

Example:

Differentiate $f (x) = x^{2} sin (3 x)$

To apply the product rule, first label the two functions being multiplied as u and v. Calculate their derivatives u' and v', then substitute into the product rule formula. Always simplify your final answer where possible.

📐 Worked Example

Differentiate $y = e^{2 x} cos (5 x)$ with respect to x.

1
Step 1: Label the two factors u and v: $u = e^{2 x}$ , $v = cos (5 x)$
2
Step 2: Calculate the derivatives of u and v:
$u' = 2e^{2x}, v' = -5\sin(5x)$
3
Step 3: Substitute into the product rule formula:
$\frac{dy}{dx} = (2e^{2x})(\cos 5x) + (e^{2x})(-5\sin 5x)$
4
Step 4: Simplify by factoring out common terms:
$\frac{dy}{dx} = e^{2x}(2\cos 5x - 5\sin 5x)$

Exam tip:

Always explicitly state u, v, u' and v' in your working to earn method marks even if you make an arithmetic error.

2. 2. The Quotient Rule★★★☆☆⏱ 5 min

✓ Calculator OK

📘 Definition

Quotient Rule

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

Used to differentiate the quotient of two differentiable functions u(x) (numerator) and v(x) (denominator). This formula is given on your exam formula sheet.

Example:

Differentiate $f (x) = \frac{3 x + 1}{x ^{2} - 2}$

When using the quotient rule, make sure you assign u to the numerator and v to the denominator. Pay close attention to the subtraction in the numerator of the formula, as sign errors are very common here.

📐 Worked Example

Differentiate $y = \frac{s i n ( 2 x )}{4 x + 1}$ with respect to x, leaving your answer as a single fraction.

1
Step 1: Assign u to the numerator and v to the denominator: $u = sin (2 x)$ , $v = 4 x + 1$
2
Step 2: Calculate derivatives: $u^{'} = 2 cos (2 x)$ , $v^{'} = 4$
3
Step 3: Substitute into the quotient rule formula:
$\frac{dy}{dx} = \frac{(2\cos 2x)(4x + 1) - (\sin 2x)(4)}{(4x + 1)^2}$
4
Step 4: Simplify the numerator by expanding the first term:
$\frac{dy}{dx} = \frac{2(4x + 1)\cos 2x - 4\sin 2x}{(4x + 1)^2}$

3. 3. The Chain Rule (Composite Functions)★★☆☆☆⏱ 4 min

✓ Calculator OK

📘 Definition

Chain Rule

If $y = f (g (x))$ , then $\frac{d y}{d x} = f^{'} (g (x)) \times g^{'} (x)$ (or $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ when substituting $u = g (x)$ )

Used to differentiate composite (function of a function) expressions, where one function is applied inside another. You must recall this formula for your exam.

Example:

Differentiate $f (x) = (x^{3} - 2 x)^{5}$

For simple composite functions at IGCSE FPM level, you can often apply the chain rule directly: multiply by the power, reduce the power by 1, then multiply by the derivative of the inner function. This works for powers, trigonometric and exponential composite functions.

📐 Worked Example

Differentiate $y = (cos (3 x))^{4}$ with respect to x.

1
Step 1: Identify the inner function $u = cos (3 x)$ , so $y = u^{4}$
2
Step 2: Calculate derivatives of y with respect to u, and u with respect to x:
$\frac{dy}{du} = 4u^3, \frac{du}{dx} = -3\sin(3x)$
3
Step 3: Apply the chain rule $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ :
$\frac{dy}{dx} = 4u^3 \times (-3\sin 3x)$
4
Step 4: Substitute back $u = cos (3 x)$ and simplify:
$\frac{dy}{dx} = -12\cos^3(3x) \sin(3x)$

Exam tip:

For simple powers of functions, you can use the shortcut: $\frac{d}{d x} [f (x)]^{n} = n [f (x)]^{n - 1} \times f^{'} (x)$ to save time in exams, as long as you show enough working.

4. 4. Combining Multiple Differentiation Rules★★★★☆⏱ 6 min

✓ Calculator OK

Most exam questions will require you to use more than one differentiation rule in the same problem. For example, a product where one term is a composite function, or a quotient where the numerator is a product. Always break the problem down step by step, identifying which rule to apply first.

📐 Worked Example

Differentiate $y = (x^{2} + 1)^{3} e^{4 x}$ with respect to x.

1
Step 1: Recognize this is a product of two functions: $u = (x^{2} + 1)^{3}$ , $v = e^{4 x}$
2
Step 2: Calculate v' directly: $v^{'} = 4 e^{4 x}$
3
Step 3: Calculate u' using the chain rule, since u is a composite function:
$u' = 3(x^2 + 1)^2 \times 2x = 6x(x^2 + 1)^2$
4
Step 4: Substitute u, v, u', v' into the product rule formula:
$\frac{dy}{dx} = [6x(x^2 + 1)^2]e^{4x} + [(x^2 + 1)^3]4e^{4x}$
5
Step 5: Factor out common terms to simplify:
$\frac{dy}{dx} = 2e^{4x}(x^2 + 1)^2(2x^2 + 3x + 2)$

✓ Quick check

When differentiating $y = \frac{s i n ( 2 x )}{x e ^{3 x}}$ , which rules do you need to use, in order?
- 1. Product rule, then quotient rule
- 1. Quotient rule, then product rule for the denominator derivative
- Only quotient rule
- Only chain rule
Reveal answer
1 —
The expression is a quotient, so you first apply the quotient rule. To differentiate the denominator $v = x e^{3 x}$ , you need to use the product rule.

5. Common Pitfalls

Wrong move:

Swapping the order of terms in the quotient rule numerator, writing $u v^{'} - u^{'} v$ instead of $u^{'} v - u v^{'}$

Why:

Sign error leads to the negative of the correct answer, losing accuracy marks

Correct move:

Always write u'v first, then subtract uv', and double-check the order before simplifying

Wrong move:

Forgetting to multiply by the derivative of the inner function when applying the chain rule, e.g. writing $\frac{d}{d x} [(2 x + 1)^{3}] = 3 (2 x + 1)^{2}$ instead of multiplying by 2

Why:

Misses a key step of the chain rule, leading to an incorrect derivative

Correct move:

After adjusting the power of the composite function, always multiply by the derivative of the inner function before simplifying

Wrong move:

Failing to label u, v, u', v' when applying product or quotient rules

Why:

Makes it harder to spot errors, and you may lose method marks if your final answer is wrong

Correct move:

Explicitly write down u, v, u', v' as your first step for every product or quotient rule question

Wrong move:

Overcomplicating simple problems by using the quotient rule when the product rule is easier, e.g. differentiating $\frac{x}{e ^{2 x}}$ as quotient instead of $x \times e^{- 2 x}$ with product rule

Why:

Increases the chance of sign or arithmetic errors

Correct move:

For simple denominators that are powers or exponentials, rewrite as a negative power and use the product + chain rules if you find it easier

6. Quick Reference Cheatsheet

Rule	Formula	Recall Required?	Use Case
Product Rule	$\frac{d}{d x} (uv) = u^{'} v + u v^{'}$	Yes	Product of two functions
Quotient Rule	$\frac{d}{d x} (\frac{u}{v}) = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$	No (given on formula sheet)	Fraction of two functions
Chain Rule	$\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) g^{'} (x)$	Yes	Composite (function of a function)

7. Frequently Asked

Do I need to memorize the quotient rule for Edexcel IGCSE FPM?

No, the quotient rule is provided on the official formula sheet given during your exam. You only need to memorize the product and chain rules.

Can I use the product rule instead of the quotient rule for fractions?

Yes, you can rewrite a quotient $\frac{u}{v}$ as $u \times v^{- 1}$ and apply the product and chain rules. Both methods will give the same correct result, and both are accepted in mark schemes.

Going deeper

official_resourceEdexcel 4PM1 Formula SheetContains the given quotient rule for exam reference

What's Next

Now that you have mastered the three core differentiation rules for Edexcel IGCSE FPM, you are ready to apply these to practical problems including finding gradients, tangents, normals, and stationary points. These applications form a large portion of the calculus section of your exam, and often carry 5-8 mark questions that require you to combine differentiation with algebraic manipulation. You should also practice past paper questions that combine these rules with trigonometric and exponential functions to build speed and accuracy before your exam.