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Additional Mathematics · Differentiation

Differentiating powers, products and quotients

You remember the rules, but products and fractions make you unsure which one applies.

The power rule differentiates single terms, the product rule handles two factors multiplied together, and the quotient rule handles one expression divided by another. Choosing the right rule is half the work.

This lesson opens SPM Additional Mathematics differentiation. Next comes applying the chain rule.

The power rule

If y = axⁿ, then dy/dx = anxⁿ⁻¹. Multiply by the power, then reduce the power by one.

Example: y = 4x³ − 5x² + 7x − 2 gives dy/dx = 12x² − 10x + 7. A constant term disappears, and x on its own becomes 1.

Roots and fractions must be rewritten first. For y = 3 ÷ x² + √x, write y = 3x⁻² + x^(1/2). Then dy/dx = −6x⁻³ + (1/2)x^(−1/2), which is −6 ÷ x³ + 1 ÷ (2√x).

The product rule

If y = uv, then dy/dx = u′v + uv′.

Example: y = x²(3x + 1). Let u = x² and v = 3x + 1, so u′ = 2x and v′ = 3. Then dy/dx = 2x(3x + 1) + x²(3) = 6x² + 2x + 3x² = 9x² + 2x.

Check by expanding: y = 3x³ + x², so dy/dx = 9x² + 2x. The two routes agree.

The quotient rule

If y = u ÷ v, then dy/dx = (u′v − uv′) ÷ v².

Example: y = (2x + 1) ÷ (x − 3). Let u = 2x + 1, v = x − 3, so u′ = 2 and v′ = 1.

dy/dx = (2(x − 3) − (2x + 1)(1)) ÷ (x − 3)² = (2x − 6 − 2x − 1) ÷ (x − 3)² = −7 ÷ (x − 3)².

Which rule should you choose?

The expression is Use
A sum of terms such as ax^n Power rule term by term
Two expressions multiplied, long to expand Product rule
One expression divided by another with x in both Quotient rule
A bracket raised to a power Chain rule

The mistake that costs marks

The common error is to differentiate each factor and multiply the answers.

Step Wrong Right
d/dx of x²(3x + 1) 2x × 3 = 6x 2x(3x + 1) + x² × 3 = 9x² + 2x

For quotients, the usual slip is reversing the numerator as uv′ − u′v, which changes the sign of the whole answer.

Check yourself

Differentiate y = (x² + 1) ÷ (x − 1), then find the gradient at x = 3.

Answer

u = x² + 1, v = x − 1, so u′ = 2x and v′ = 1.

dy/dx = (2x(x − 1) − (x² + 1)(1)) ÷ (x − 1)² = (2x² − 2x − x² − 1) ÷ (x − 1)² = (x² − 2x − 1) ÷ (x − 1)².

At x = 3: numerator = 9 − 6 − 1 = 2 and denominator = 2² = 4, so the gradient is 1/2.

What to study next

Continue with applying the chain rule. Use the polynomial differentiation tutor to practise the rules, then try the differentiation practice set.

If you would like a teacher to watch you choose between rules on new expressions, see online one-to-one Additional Mathematics tuition.

Common questions

How do I differentiate a fraction like 3 over x squared?

Rewrite it as a power: 3 over x² is 3x⁻². Then use the power rule: multiply by the power and subtract one from it. So the derivative is −6x⁻³, which can be written as −6 over x³.

When do I need the product rule?

Use it when two expressions containing x are multiplied and expanding first would be long. Write u and v, find u′ and v′, then use dy/dx = u′v + uv′. If expanding is short, expanding and using the power rule gives the same answer.

What is the quotient rule and how do I remember the order?

For y = u ÷ v, dy/dx = (u′v − uv′) ÷ v². The numerator starts with u′v. Reversing the subtraction gives the wrong sign, so write u′v first every time.

Why is the derivative of a product not u′ times v′?

Both factors change at once, and each change affects the product. The product rule adds the effect of changing u with v fixed and the effect of changing v with u fixed. Multiplying u′ and v′ ignores that.

If you mix up the product and quotient rules under time pressure, a one-to-one teacher can drill the choice on unseen expressions until it is automatic.

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