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Product Rule

A LevelAQAEdexcelOCR

Product Rule

We use the product rule to find derivatives of functions which are (funnily enough), products of separate functions – we cannot simply differentiate our terms and multiply them together.

A LevelAQAEdexcelOCR
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Product Rule Formula

For a function y=f(x)=u(x)v(x)textcolor{limegreen}{y} = f(textcolor{blue}{x}) = u(textcolor{blue}{x})v(textcolor{blue}{x}), we have the derivative (with respect to xtextcolor{blue}{x}) given by

dydx=df(x)dx=u(x)dv(x)dx+v(x)du(x)dxdfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = dfrac{df(textcolor{blue}{x})}{dtextcolor{blue}{x}} = u(textcolor{blue}{x})dfrac{dv(textcolor{blue}{x})}{dtextcolor{blue}{x}} + v(textcolor{blue}{x})dfrac{du(textcolor{blue}{x})}{dtextcolor{blue}{x}}

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Extensions to the Formula

Let’s now say that we have y=u(x)v(x)w(x)textcolor{limegreen}{y} = u(textcolor{blue}{x})v(textcolor{blue}{x})w(textcolor{blue}{x}), and we want to find derivative with respect to xtextcolor{blue}{x}.

By setting u(x)v(x)=a(x)u(textcolor{blue}{x})v(textcolor{blue}{x}) = a(textcolor{blue}{x}), we have

dydx=d(a(x)w(x))dxdfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = dfrac{d(a(textcolor{blue}{x})w(textcolor{blue}{x}))}{dtextcolor{blue}{x}}

 

=a(x)dw(x)dx+da(x)dxw(x)= a(textcolor{blue}{x})dfrac{dw(textcolor{blue}{x})}{dtextcolor{blue}{x}} + dfrac{da(textcolor{blue}{x})}{dtextcolor{blue}{x}}w(textcolor{blue}{x})

 

=du(x)dxv(x)w(x)+u(x)dv(x)dxw(x)+u(x)v(x)dw(x)dx= dfrac{du(textcolor{blue}{x})}{dtextcolor{blue}{x}}v(textcolor{blue}{x})w(textcolor{blue}{x})+ u(textcolor{blue}{x})dfrac{dv(textcolor{blue}{x})}{dtextcolor{blue}{x}}w(textcolor{blue}{x}) + u(textcolor{blue}{x})v(textcolor{blue}{x})dfrac{dw(textcolor{blue}{x})}{dtextcolor{blue}{x}}

This technique can be repeated endlessly for nn functions, so we have a linear combination of nn terms, where each term is the product of one differentiated function and all other functions.

A LevelAQAEdexcelOCR
A LevelAQAEdexcelOCR

Example 1: Using the Product Rule

Say we have the function y=exsin⁡xtextcolor{limegreen}{y} = e^{textcolor{blue}{x}}sin textcolor{blue}{x}. Find dydxdfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}}.

[2 marks]

Let u(x)=exu(textcolor{blue}{x}) = e^textcolor{blue}{x} and v(x)=sin⁡xv(textcolor{blue}{x}) = sin textcolor{blue}{x}. Then

dydx=du(x)dxv(x)+u(x)dv(x)dxdfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = dfrac{du(textcolor{blue}{x})}{dtextcolor{blue}{x}}v(textcolor{blue}{x}) + u(textcolor{blue}{x})dfrac{dv(textcolor{blue}{x})}{dtextcolor{blue}{x}}

 

=exsin⁡x+excos⁡x=ex(sin⁡x+cos⁡x)= e^textcolor{blue}{x}sin textcolor{blue}{x} + e^textcolor{blue}{x}cos textcolor{blue}{x} = e^textcolor{blue}{x}(sin textcolor{blue}{x} + cos textcolor{blue}{x})

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Example 2: Using the Product Rule for Larger Functions

Let y=(x2−1)(ln⁡x)cos⁡xtextcolor{limegreen}{y} = (textcolor{blue}{x}^2 – 1)(ln textcolor{blue}{x})cos textcolor{blue}{x} where xtextcolor{blue}{x} is measured in radians. Find dydxdfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} and verify that there is a stationary point at the point (1,0)(1, 0).

[4 marks]

Let u(x)=x2−1u(textcolor{blue}{x}) = textcolor{blue}{x}^2 – 1, v(x)=ln⁡xv(textcolor{blue}{x}) = ln textcolor{blue}{x} and w(x)=cos⁡xw(textcolor{blue}{x}) = cos textcolor{blue}{x}. Then

dudx=2xdfrac{du}{dtextcolor{blue}{x}} = 2textcolor{blue}{x}, dvdx=1xdfrac{dv}{dtextcolor{blue}{x}} = dfrac{1}{textcolor{blue}{x}} and dwdx=−sin⁡xdfrac{dw}{dtextcolor{blue}{x}} = -sin textcolor{blue}{x}

This gives

dydx=(2xln⁡xcos⁡x)+((x2−1)cos⁡xx)−((x2−1)ln⁡xsin⁡x)dfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = (2textcolor{blue}{x}ln textcolor{blue}{x} cos textcolor{blue}{x}) + left( dfrac{(textcolor{blue}{x}^2 – 1)cos textcolor{blue}{x}}{textcolor{blue}{x}}right) – ((textcolor{blue}{x}^2 – 1)ln textcolor{blue}{x}sin textcolor{blue}{x})

When x=1textcolor{blue}{x} = 1,

dydx=(2(1)ln⁡1cos⁡1)+(0cos⁡11)−(0ln⁡1sin⁡1)dfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = (2(1)ln 1 cos 1) + left( dfrac{0cos 1}{1}right) – left( 0ln 1 sin 1right)

=0= 0

So, we can confirm that there is a stationary point at x=1textcolor{blue}{x} = 1.

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Product Rule Example Questions

Question 1: Using the product rule, show that the function y=x3y = x^3 has derivative dydx=3x2dfrac{dy}{dx} = 3x^2.

[2 marks]

A Level AQAEdexcelOCR

Let u(x)=xu(x) = x and v(x)=x2v(x) = x^2.

Then

dydx=1x2+x2xdfrac{dy}{dx} = 1x^2 + x2x

 

=x2+2x2= x^2 + 2x^2

 

=3x2= 3x^2

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Question 2: For f(x)=2sin⁡xcos⁡xf(x) = 2sin x cos x, use the product rule to find its derivative with respect to xx, and prove that 2sin⁡xcos⁡x=sin⁡2x2sin x cos x = sin 2x.

[4 marks]

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Let u(x)=2sin⁡xu(x) = 2sin x and v(x)=cos⁡xv(x) = cos x. Then

 

df(x)dx=2cos⁡xcos⁡x−2sin⁡xsin⁡xdfrac{df(x)}{dx} = 2cos x cos x – 2sin x sin x

 

=2cos⁡2x= 2cos 2x, by double angle formulae.

We also have

d(sin⁡2x)dx=2cos⁡2xdfrac{d(sin 2x)}{dx} = 2cos 2x

Therefore, d(sin⁡2x)dx=d(2sin⁡xcos⁡x)dxdfrac{d(sin 2x)}{dx} = dfrac{d(2sin x cos x)}{dx}

This is an example of the Uniqueness Theorem. You won’t necessarily need this, but it’s an interesting proof, all the same.

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Question 3: Find the derivative (w.r.t xx), of the function y=x33xtan⁡xy = x^3 3^xtan x where xx is measured in radians. Verify that there is a stationary point at the origin.

[4 marks]

 

A Level AQAEdexcelOCR

Set u=x3u = x^3, v=3xv = 3^x and w=tan⁡xw = tan x.

Then

u′=3x2u' = 3x^2, v′=3xln⁡3v' = 3^xln 3 and w′=sec⁡2xw' = sec ^2 x

 

Using the rule we learned for extended functions, we have

 

dydx=(3x23xtan⁡x)+(x33xln⁡3tan⁡x)+(x33xsec⁡2x)dfrac{dy}{dx} = (3x^2 3^x tan x) + (x^3 3^xln 3 tan x) + (x^3 3^x sec ^2 x)

 

Set x=0x = 0 to give

 

dydx=(0×1×tan⁡0)+(0×ln⁡3×tan⁡0)+(0×1×sec⁡20)dfrac{dy}{dx} = (0 times 1 times tan 0) + (0 times ln 3 times tan 0) + (0 times 1 times sec ^2 0)

 

=0= 0

So, we can confirm there is a stationary point at the origin.

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Additional Resources

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Exam Tips Cheat Sheet

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Formula Booklet

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Specification Points Covered

G4 – Differentiate using the product rule, the quotient rule and the chain rule, including problems involving connected rates of change and inverse functions

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