Product Rule

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Product Rule Revision

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.

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Product Rule Formula

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

\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \dfrac{df(\textcolor{blue}{x})}{d\textcolor{blue}{x}} = u(\textcolor{blue}{x})\dfrac{dv(\textcolor{blue}{x})}{d\textcolor{blue}{x}} + v(\textcolor{blue}{x})\dfrac{du(\textcolor{blue}{x})}{d\textcolor{blue}{x}}

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Extensions to the Formula

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

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

\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \dfrac{d(a(\textcolor{blue}{x})w(\textcolor{blue}{x}))}{d\textcolor{blue}{x}}


= a(\textcolor{blue}{x})\dfrac{dw(\textcolor{blue}{x})}{d\textcolor{blue}{x}} + \dfrac{da(\textcolor{blue}{x})}{d\textcolor{blue}{x}}w(\textcolor{blue}{x})


= \dfrac{du(\textcolor{blue}{x})}{d\textcolor{blue}{x}}v(\textcolor{blue}{x})w(\textcolor{blue}{x})+ u(\textcolor{blue}{x})\dfrac{dv(\textcolor{blue}{x})}{d\textcolor{blue}{x}}w(\textcolor{blue}{x}) + u(\textcolor{blue}{x})v(\textcolor{blue}{x})\dfrac{dw(\textcolor{blue}{x})}{d\textcolor{blue}{x}}

This technique can be repeated endlessly for n functions, so we have a linear combination of n 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 \textcolor{limegreen}{y} = e^{\textcolor{blue}{x}}\sin \textcolor{blue}{x}. Find \dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}}.

[2 marks]

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

\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \dfrac{du(\textcolor{blue}{x})}{d\textcolor{blue}{x}}v(\textcolor{blue}{x}) + u(\textcolor{blue}{x})\dfrac{dv(\textcolor{blue}{x})}{d\textcolor{blue}{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 \textcolor{limegreen}{y} = (\textcolor{blue}{x}^2 - 1)(\ln \textcolor{blue}{x})\cos \textcolor{blue}{x} where \textcolor{blue}{x} is measured in radians. Find \dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} and verify that there is a stationary point at the point (1, 0).

[4 marks]

Let u(\textcolor{blue}{x}) = \textcolor{blue}{x}^2 - 1, v(\textcolor{blue}{x}) = \ln \textcolor{blue}{x} and w(\textcolor{blue}{x}) = \cos \textcolor{blue}{x}. Then

\dfrac{du}{d\textcolor{blue}{x}} = 2\textcolor{blue}{x}, \dfrac{dv}{d\textcolor{blue}{x}} = \dfrac{1}{\textcolor{blue}{x}} and \dfrac{dw}{d\textcolor{blue}{x}} = -\sin \textcolor{blue}{x}

This gives

\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = (2\textcolor{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 \textcolor{blue}{x} = 1,

\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = (2(1)\ln 1 \cos 1) + \left( \dfrac{0\cos 1}{1}\right) - \left( 0\ln 1 \sin 1\right)

= 0

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

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Product Rule Example Questions

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


\dfrac{dy}{dx} = 1x^2 + x2x


= x^2 + 2x^2


= 3x^2

Let u(x) = 2\sin x and v(x) = \cos x. Then


\dfrac{df(x)}{dx} = 2\cos x \cos x - 2\sin x \sin x


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

We also have

\dfrac{d(\sin 2x)}{dx} = 2\cos 2x

Therefore, \dfrac{d(\sin 2x)}{dx} = \dfrac{d(2\sin 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.

Set u = x^3, v = 3^x and w = \tan x.


u' = 3x^2, v' = 3^x\ln 3 and w' = \sec ^2 x


Using the rule we learned for extended functions, we have


\dfrac{dy}{dx} = (3x^2 3^x \tan x) + (x^3 3^x\ln 3 \tan x) + (x^3 3^x \sec ^2 x)


Set x = 0 to give


\dfrac{dy}{dx} = (0 \times 1 \times \tan 0) + (0 \times \ln 3 \times \tan 0) + (0 \times 1 \times \sec ^2 0)


= 0

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

Additional Resources


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Formula Booklet

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