Using the Product Rule | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Differentiate one factor at a time

Key idea

y=uv

, where both

u

and

v

depend on

x

, then

\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}

A prime is a shorter way to mean “differentiate with respect to $x$ ”. Thus $u'=du/dx$ and the same rule is $(uv)'=uv'+vu'$ .

Common mistake

(uv)'

is not

u'v'

. Each term keeps one original factor.

Apply the rule

Worked example

Differentiate two polynomial factors

y=(x^2+1)(x^3-2)

\begin{aligned}y'={}&2x(x^3-2)\\&+3x^2(x^2+1)\end{aligned}

y'=x(5x^3+3x-4)

Factor before solving

A stationary point has gradient 0. A factored derivative exposes every possible zero without unnecessary expansion.

Worked example

Product rule followed by a stationary equation

y=x^2(5-x)

y'=x^2(-1)+(5-x)(2x)

y'=x(10-3x)

x=0\quad\text{or}\quad x=\frac{10}{3}

Your turnIndependent check [4 marks]

Differentiate $y=(2x-3)(x^2+4)$ using the product rule and simplify the result.

Show worked answer

\begin{aligned}y'={}&2(x^2+4)\\&+(2x-3)(2x)\end{aligned}

Answer

y'=6x^2-6x+8

Differentiate one factor at a time

Key idea

y=uv

, where both

u

and

v

depend on

x

, then

\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}

A prime is a shorter way to mean “differentiate with respect to $x$ ”. Thus $u'=du/dx$ and the same rule is $(uv)'=uv'+vu'$ .

Common mistake

(uv)'

is not

u'v'

. Each term keeps one original factor.

Apply the rule

Worked example

Differentiate two polynomial factors

y=(x^2+1)(x^3-2)

\begin{aligned}y'={}&2x(x^3-2)\\&+3x^2(x^2+1)\end{aligned}

y'=x(5x^3+3x-4)

Factor before solving

A stationary point has gradient 0. A factored derivative exposes every possible zero without unnecessary expansion.

Worked example

Product rule followed by a stationary equation

y=x^2(5-x)

y'=x^2(-1)+(5-x)(2x)

y'=x(10-3x)

x=0\quad\text{or}\quad x=\frac{10}{3}

Your turnIndependent check [4 marks]

Differentiate $y=(2x-3)(x^2+4)$ using the product rule and simplify the result.

Show worked answer

\begin{aligned}y'={}&2(x^2+4)\\&+(2x-3)(2x)\end{aligned}

Answer

y'=6x^2-6x+8