Mathematics 9709 · Pure Mathematics 2

Product and Quotient Rule

The whole chapter is one idea. The derivative $\frac{dy}{dx}$ is the gradient of the curve (how steep it is at a point). A stationary point is where that gradient is zero.
P3 just adds more tools. Now things can be multiplied, divided, raised to $e$ , logged, put in trig, tangled together, or written through a parameter.
Product and quotient are the first two tools. Every later tool builds on them.

The chain rule runs through almost everything here. For exp, log, trig, implicit and parametric, you find the inside, then differentiate the outside, then multiply by the derivative of the inside. (Product and quotient are the only ones that stand on their own.) That last factor is called the “chain-rule tail”. It is the one people drop most. It is also the exact step examiners give the method mark for, so never leave it off.

The product rule

When $y=uv$ is two functions multiplied, call them $u$ and $v$ , differentiate each one, then combine.
You cannot just multiply the two derivatives together. That is the exact trap this rule stops.

$\dfrac{d}{dx}(uv)=u\,\dfrac{dv}{dx}+v\,\dfrac{du}{dx}$ . Say it in words: first × differentiate second, plus second × differentiate first.

Learn the words, not the symbols.
People who memorise just the symbols pair up the wrong terms. The sentence stops you doing that. (Key Point 4.1)

Worked example

Differentiate $y = x^2 e^{-x}$ , and read off its stationary points

Name the factors: $u=x^2$ , $v=e^{-x}$ .
Differentiate each (the $v$ needs the chain rule — derivative of the inside $-x$ is $-1$ ): $\dfrac{du}{dx}=2x$ , $\dfrac{dv}{dx}=-e^{-x}$ .
First × deriv of second, plus second × deriv of first: $\dfrac{dy}{dx}=x^2(-e^{-x})+e^{-x}(2x)$ .
Factorise $xe^{-x}$ out so the zeros can be read off: $\dfrac{dy}{dx}=xe^{-x}(2-x)$ .

Answer

\dfrac{dy}{dx}=xe^{-x}(2-x)

Set it to zero. Since $e^{-x}$ is never zero, the stationary points are $x=0$ and $x=2$ . Factorising first is what makes those zeros easy to see. (textbook Worked Example 4.1)

Worked example

Real question: $y = x e^{1-2x}$ has one stationary point, find it

9709/31 O/N 2021 Q3(a). A product ( $x$ times an exponential), then the usual follow-up: where is the gradient zero?

$u=x$ , $v=e^{1-2x}$ . The inside of $v$ is $1-2x$ , derivative $-2$ , so $\dfrac{dv}{dx}=-2e^{1-2x}$ .
Product rule: $\dfrac{dy}{dx}=x(-2e^{1-2x})+e^{1-2x}(1)=e^{1-2x}(1-2x)$ .
$e^{1-2x}$ is never zero, so the bracket gives it: $1-2x=0\Rightarrow x=\tfrac12$ .
Back-substitute for the $y$ -coordinate: $y=\tfrac12\,e^{1-1}=\tfrac12 e^{0}=\tfrac12$ .

Answer

\left(\tfrac12,\ \tfrac12\right)

The exponent landed on $0$ , so the messy $e$ -term became a clean answer. (9709/31 Oct/Nov 2021 Q3)

The quotient rule

When $y=\dfrac{u}{v}$ is one function divided by another, use:

$\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right)=\dfrac{v\,\frac{du}{dx}-u\,\frac{dv}{dx}}{v^{2}}$ . In words: (bottom × differentiate top) minus (top × differentiate bottom), all over the bottom squared.

The bottom is always squared ( $v^2$ , never $v$ ).
You cannot swap the order on top. Backwards gives you minus the right answer. (Key Point 4.2)

Worked example

Differentiate $y = \dfrac{\ln x}{x}$

Top $u=\ln x$ (deriv $\tfrac{1}{x}$ ), bottom $v=x$ (deriv $1$ ).
Bottom × deriv of top, minus top × deriv of bottom, over bottom squared: $\dfrac{dy}{dx}=\dfrac{x\cdot\frac{1}{x}-\ln x\cdot 1}{x^{2}}$ .
Tidy the top ( $x\cdot\tfrac1x=1$ ): $\dfrac{dy}{dx}=\dfrac{1-\ln x}{x^{2}}$ .

Answer

\dfrac{dy}{dx}=\dfrac{1-\ln x}{x^{2}}

It is zero when the top is zero: $1-\ln x=0\Rightarrow x=e$ . A clean exact stationary point. (9709-style)

A quotient is also a product

Any quotient $\dfrac{u}{v}$ can be written $u\,v^{-1}$ . Then you do it by the product rule and the chain rule. The textbook does $y=(x^2-5)(2x+1)^{-1}$ both ways to show they match. You can use this to check a messy answer. But pick one method per question and stick with it. Switching halfway is where slips happen. (textbook Explore 4.2)

What the derivative actually is: drag the tangent

The derivative at a point is the gradient of the tangentthere: the slope of the straight line that just touches the curve.
Slide the point along $y=xe^{-x}$ and watch the tangent tilt. Uphill means a positive gradient, downhill means negative, and the one flat spot in between is the stationary point.

move P

at x = 0.40·gradient dy/dx = +0.40·uphill

That flat tangent is why “set $\frac{dy}{dx}=0$ ” finds turning points. Every “find the stationary point” part just asks you to find where this line lies flat.

Where it shows up: stationary points

You rarely just get “differentiate”. The usual full question is a product or quotient, then find a stationary point.
Factorise the derivative fully so the zeros drop out.

Worked example

9709-style: $y = (2x-3)^2(x+5)^3$ , find where the gradient is $0$

Product rule with $u=(2x-3)^2$ , $v=(x+5)^3$ ; each derivative needs the chain rule: $\tfrac{du}{dx}=4(2x-3)$ , $\tfrac{dv}{dx}=3(x+5)^2$ .
Combine: $\tfrac{dy}{dx}=(2x-3)^2\cdot 3(x+5)^2+(x+5)^3\cdot 4(2x-3)$ .
Pull out the common factor $(2x-3)(x+5)^2$ : $\tfrac{dy}{dx}=(2x-3)(x+5)^2\big[3(2x-3)+4(x+5)\big]$ .
Tidy the bracket ( $6x-9+4x+20=10x+11$ ): $\tfrac{dy}{dx}=(2x-3)(x+5)^2(10x+11)$ .

Answer

x=\tfrac{3}{2},\ x=-5,\ x=-\tfrac{11}{10}

Three zeros, read straight off the brackets. You could not do that if you had left the derivative not factorised. (textbook Worked Example 4.2)

Worked example

Real question: minimum of $y = x e^{2x} - 5x$ sits at $a = \tfrac12 \ln\!\dfrac{5}{1+2a}$

9709/32 F/M 2024 Q7(a). It is a product plus a linear term, with a “show that” target, so show every step of the derivative.

Differentiate. The $xe^{2x}$ is a product (inside derivative $2$ ): $\dfrac{dy}{dx}=x(2e^{2x})+e^{2x}(1)-5=(2x+1)e^{2x}-5$ .
At the minimum the gradient is zero: $(2x+1)e^{2x}=5\Rightarrow e^{2x}=\dfrac{5}{2x+1}$ .
Take $\ln$ of both sides to free the exponent: $2x=\ln\!\dfrac{5}{2x+1}$ , so $x=\tfrac12\ln\!\dfrac{5}{2x+1}$ .

Answer

a=\tfrac12\ln\!\dfrac{5}{1+2a}

The $x$ is stuck on both sides because it cannot be untangled. That is exactly why the next part solves it by iteration, not algebra. (9709/32 Feb/March 2024 Q7)

Your turn— tap to reveal the worked answer (9709/32 Jun 2022 Q4)

$y=\cos^3 x\sin x$ has one stationary point in $0<x<\tfrac{\pi}{2}$ . Find its $x$ -coordinate (3 s.f.). Use the product rule, with $\tfrac{d}{dx}(\cos^3 x)=-3\cos^2 x\sin x$ (chain rule):

$\tfrac{dy}{dx}=\cos^3 x\cos x+\sin x(-3\cos^2 x\sin x)=\cos^2 x(\cos^2 x-3\sin^2 x)$ . Setting the bracket to zero gives $\tan^2 x=\tfrac13$ , so $x=\tfrac{\pi}{6}$ .

Answer

x=0.524\ \text{(3 s.f.)}

Common mistake

The three habits that lose marks here, in order:

Quotient subtraction backwards. It is bottom-times-deriv-of-top MINUS top-times-deriv-of-bottom. Swapping gives the negative of the answer.
Dropping the chain-rule tail on a bracket. $\tfrac{d}{dx}(3x-2)^3=9(3x-2)^2$ , not $3(3x-2)^2$ — the inside derivative is $3$ .
Not factorising fully on a stationary-point part, so you cannot read off the zeros.

PreviousExpressing a sin θ + b cos θ as R sin(θ ± α) or R cos(θ ∓ α)NextDifferentiating e^x, a^x and ln x

Product and Quotient Rule

The whole chapter is one idea. The derivative $\frac{dy}{dx}$ is the gradient of the curve (how steep it is at a point). A stationary point is where that gradient is zero.
P3 just adds more tools. Now things can be multiplied, divided, raised to $e$ , logged, put in trig, tangled together, or written through a parameter.
Product and quotient are the first two tools. Every later tool builds on them.

The product rule

When $y=uv$ is two functions multiplied, call them $u$ and $v$ , differentiate each one, then combine.
You cannot just multiply the two derivatives together. That is the exact trap this rule stops.

$\dfrac{d}{dx}(uv)=u\,\dfrac{dv}{dx}+v\,\dfrac{du}{dx}$ . Say it in words: first × differentiate second, plus second × differentiate first.

Learn the words, not the symbols.
People who memorise just the symbols pair up the wrong terms. The sentence stops you doing that. (Key Point 4.1)

Worked example

Differentiate $y = x^2 e^{-x}$ , and read off its stationary points

Name the factors: $u=x^2$ , $v=e^{-x}$ .
Differentiate each (the $v$ needs the chain rule — derivative of the inside $-x$ is $-1$ ): $\dfrac{du}{dx}=2x$ , $\dfrac{dv}{dx}=-e^{-x}$ .
First × deriv of second, plus second × deriv of first: $\dfrac{dy}{dx}=x^2(-e^{-x})+e^{-x}(2x)$ .
Factorise $xe^{-x}$ out so the zeros can be read off: $\dfrac{dy}{dx}=xe^{-x}(2-x)$ .

Answer

\dfrac{dy}{dx}=xe^{-x}(2-x)

Set it to zero. Since $e^{-x}$ is never zero, the stationary points are $x=0$ and $x=2$ . Factorising first is what makes those zeros easy to see. (textbook Worked Example 4.1)

Worked example

Real question: $y = x e^{1-2x}$ has one stationary point, find it

9709/31 O/N 2021 Q3(a). A product ( $x$ times an exponential), then the usual follow-up: where is the gradient zero?

$u=x$ , $v=e^{1-2x}$ . The inside of $v$ is $1-2x$ , derivative $-2$ , so $\dfrac{dv}{dx}=-2e^{1-2x}$ .
Product rule: $\dfrac{dy}{dx}=x(-2e^{1-2x})+e^{1-2x}(1)=e^{1-2x}(1-2x)$ .
$e^{1-2x}$ is never zero, so the bracket gives it: $1-2x=0\Rightarrow x=\tfrac12$ .
Back-substitute for the $y$ -coordinate: $y=\tfrac12\,e^{1-1}=\tfrac12 e^{0}=\tfrac12$ .

Answer

\left(\tfrac12,\ \tfrac12\right)

The exponent landed on $0$ , so the messy $e$ -term became a clean answer. (9709/31 Oct/Nov 2021 Q3)

The quotient rule

When $y=\dfrac{u}{v}$ is one function divided by another, use:

The bottom is always squared ( $v^2$ , never $v$ ).
You cannot swap the order on top. Backwards gives you minus the right answer. (Key Point 4.2)

Worked example

Differentiate $y = \dfrac{\ln x}{x}$

Top $u=\ln x$ (deriv $\tfrac{1}{x}$ ), bottom $v=x$ (deriv $1$ ).
Bottom × deriv of top, minus top × deriv of bottom, over bottom squared: $\dfrac{dy}{dx}=\dfrac{x\cdot\frac{1}{x}-\ln x\cdot 1}{x^{2}}$ .
Tidy the top ( $x\cdot\tfrac1x=1$ ): $\dfrac{dy}{dx}=\dfrac{1-\ln x}{x^{2}}$ .

Answer

\dfrac{dy}{dx}=\dfrac{1-\ln x}{x^{2}}

It is zero when the top is zero: $1-\ln x=0\Rightarrow x=e$ . A clean exact stationary point. (9709-style)

A quotient is also a product

What the derivative actually is: drag the tangent

The derivative at a point is the gradient of the tangentthere: the slope of the straight line that just touches the curve.
Slide the point along $y=xe^{-x}$ and watch the tangent tilt. Uphill means a positive gradient, downhill means negative, and the one flat spot in between is the stationary point.

move P

at x = 0.40·gradient dy/dx = +0.40·uphill

That flat tangent is why “set $\frac{dy}{dx}=0$ ” finds turning points. Every “find the stationary point” part just asks you to find where this line lies flat.

Where it shows up: stationary points

You rarely just get “differentiate”. The usual full question is a product or quotient, then find a stationary point.
Factorise the derivative fully so the zeros drop out.

Worked example

9709-style: $y = (2x-3)^2(x+5)^3$ , find where the gradient is $0$

Product rule with $u=(2x-3)^2$ , $v=(x+5)^3$ ; each derivative needs the chain rule: $\tfrac{du}{dx}=4(2x-3)$ , $\tfrac{dv}{dx}=3(x+5)^2$ .
Combine: $\tfrac{dy}{dx}=(2x-3)^2\cdot 3(x+5)^2+(x+5)^3\cdot 4(2x-3)$ .
Pull out the common factor $(2x-3)(x+5)^2$ : $\tfrac{dy}{dx}=(2x-3)(x+5)^2\big[3(2x-3)+4(x+5)\big]$ .
Tidy the bracket ( $6x-9+4x+20=10x+11$ ): $\tfrac{dy}{dx}=(2x-3)(x+5)^2(10x+11)$ .

Answer

x=\tfrac{3}{2},\ x=-5,\ x=-\tfrac{11}{10}

Three zeros, read straight off the brackets. You could not do that if you had left the derivative not factorised. (textbook Worked Example 4.2)

Worked example

Real question: minimum of $y = x e^{2x} - 5x$ sits at $a = \tfrac12 \ln\!\dfrac{5}{1+2a}$

9709/32 F/M 2024 Q7(a). It is a product plus a linear term, with a “show that” target, so show every step of the derivative.

Differentiate. The $xe^{2x}$ is a product (inside derivative $2$ ): $\dfrac{dy}{dx}=x(2e^{2x})+e^{2x}(1)-5=(2x+1)e^{2x}-5$ .
At the minimum the gradient is zero: $(2x+1)e^{2x}=5\Rightarrow e^{2x}=\dfrac{5}{2x+1}$ .
Take $\ln$ of both sides to free the exponent: $2x=\ln\!\dfrac{5}{2x+1}$ , so $x=\tfrac12\ln\!\dfrac{5}{2x+1}$ .

Answer

a=\tfrac12\ln\!\dfrac{5}{1+2a}

The $x$ is stuck on both sides because it cannot be untangled. That is exactly why the next part solves it by iteration, not algebra. (9709/32 Feb/March 2024 Q7)

Your turn— tap to reveal the worked answer (9709/32 Jun 2022 Q4)

$y=\cos^3 x\sin x$ has one stationary point in $0<x<\tfrac{\pi}{2}$ . Find its $x$ -coordinate (3 s.f.). Use the product rule, with $\tfrac{d}{dx}(\cos^3 x)=-3\cos^2 x\sin x$ (chain rule):

$\tfrac{dy}{dx}=\cos^3 x\cos x+\sin x(-3\cos^2 x\sin x)=\cos^2 x(\cos^2 x-3\sin^2 x)$ . Setting the bracket to zero gives $\tan^2 x=\tfrac13$ , so $x=\tfrac{\pi}{6}$ .

Answer

x=0.524\ \text{(3 s.f.)}

Common mistake

The three habits that lose marks here, in order:

Quotient subtraction backwards. It is bottom-times-deriv-of-top MINUS top-times-deriv-of-bottom. Swapping gives the negative of the answer.
Dropping the chain-rule tail on a bracket. $\tfrac{d}{dx}(3x-2)^3=9(3x-2)^2$ , not $3(3x-2)^2$ — the inside derivative is $3$ .
Not factorising fully on a stationary-point part, so you cannot read off the zeros.