The product and quotient rules

Recognising which rule you need

• Product — two functions multiplied: $x^2\sin x$, $e^{3x}\ln x$, $(2x+1)\sqrt{x-3}$.

• Quotient — one function divided by another: $\dfrac{x^2}{x+1}$, $\dfrac{\sin x}{x}$.

• Neither — a sum or difference. $x^2 + \sin x$ differentiates term by term with no special rule.

• The 2025 report lists "did not recognise when it was necessary to apply the product and quotient rules" among the Not Achieved behaviours. Identifying the structure comes before applying anything.

The product rule

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

• Each part:

  • $u$, $v$ — the two factors
  • $u'$, $v'$ — their individual derivatives

• In words: derivative of the first times the second, plus the first times the derivative of the second.

• Note it is a sum, and the two terms are symmetric — if you swap $u$ and $v$ you get the same answer.

• The setting-out that avoids errors:

  • Write $u = \ldots$ and $v = \ldots$ on separate lines
  • Write $u' = \ldots$ and $v' = \ldots$ underneath
  • Substitute into the formula with every factor in brackets
  • Simplify only if asked

• Worked through for $y = x^3\sin x$:

  • $u = x^3$, $u' = 3x^2$
  • $v = \sin x$, $v' = \cos x$
  • $$\dfrac{dy}{dx} = 3x^2\sin x + x^3\cos x$$

The quotient rule

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

• Each part:

  • $u$ — the numerator (top)
  • $v$ — the denominator (bottom)
  • $v^2$ — the denominator squared

• The order matters and the sign matters. Unlike the product rule, this is a subtraction, so $u'v$ and $uv'$ cannot be swapped.

• A memory aid many students use, with $u$ = "hi" and $v$ = "lo":

  • "lo d-hi minus hi d-lo, all over lo squared"

• Worked through for $y = \dfrac{x^2}{x + 1}$:

  • $u = x^2$, $u' = 2x$
  • $v = x + 1$, $v' = 1$
  • $$\dfrac{dy}{dx} = \dfrac{2x(x+1) - x^2(1)}{(x+1)^2} = \dfrac{2x^2 + 2x - x^2}{(x+1)^2} = \dfrac{x^2 + 2x}{(x+1)^2}$$

Combining with the chain rule

• Most exam products and quotients need the chain rule inside as well.

• Worked through for $y = (2x+1)^3\cos 5x$:

  • $u = (2x+1)^3$, so $u' = 3(2x+1)^2 \times 2 = 6(2x+1)^2$ — chain rule
  • $v = \cos 5x$, so $v' = -5\sin 5x$ — chain rule
  • $$\dfrac{dy}{dx} = 6(2x+1)^2\cos 5x - 5(2x+1)^3\sin 5x$$

• Do the inner derivatives on scratch lines first, then assemble. Trying to do everything in one line is where factors vanish.

When you can avoid the rules

• Split a fraction with a single-term denominator rather than using the quotient rule:

  • $\dfrac{x^3 + 2x}{x} = x^2 + 2$, which differentiates to $2x$ in one step.

• Expand a short product rather than using the product rule:

  • $x^2(x+3) = x^3 + 3x^2$, giving $3x^2 + 6x$.

• Use a negative power when the numerator is a constant:

  • $\dfrac{5}{(x+2)^3} = 5(x+2)^{-3}$, a chain rule problem.

• Note that these shortcuts are not laziness — fewer steps means fewer opportunities for the algebra slips the assessors keep reporting.

Simplifying a quotient rule answer

• Quotient rule answers are messy by nature. If simplification is required:
  • Expand the numerator fully
  • Collect like terms
  • Factorise if a common factor appears
  • Leave the denominator as $v^2$ — do not expand it
• If the question says simplification is not required, stop after substituting.