topic badge
New Zealand
Level 8 - NCEA Level 3

Product Rule


So far we have differentiated the sum or difference of functions involving terms which can be written in the form $y=ax^n$y=axn. The derivative of the sum of functions is the same as the sum of the derivatives of the parts. 

(Remember: we have to change reciprocals to negative indices and irrational terms to fractional powers).

Now for products: If we consider $y=x^5$y=x5 and its equivalent $y=x^2\times x^3$y=x2×x3

We know the derivative of $x^5$x5 is $5x^4$5x4, but can we differentiate the factors of the product and get this answer? Does $2x$2x times $3x^2$3x2 give us the same answer? No it gives us $6x^3$6x3, so it looks as though the derivative of a product is not the same as the product of the derivatives. Of course we will simplify the product if possible and then differentiate the terms, however, sometimes we can’t expand a product or it is too long and open to errors.

Let’s investigate this from first principles. Once we have derived the formula, or process, you will use the rule only.

Let us consider $y=u\times v$y=u×v where $u$u and $v$v are both functions of $x$x.

If we increase the $x$x values by a small amount, $\delta x$δx, then $u$u will change by a small amount $\delta u$δu and $v$v will change by a small amount $\delta v$δv.

Using our formula for differentiating from first principles:

$\frac{dy}{dx}$dydx $=$= $\lim_{\delta x\rightarrow0}\frac{\left(u+\delta u\right)\left(v+\delta v\right)-uv}{\delta x}$limδx0(u+δu)(v+δv)uvδx
  $=$= $\lim_{\delta x\rightarrow0}\frac{uv+u\delta v+v\delta u+\delta u\delta v-uv}{\delta x}$limδx0uv+uδv+vδu+δuδvuvδx
  $=$= $\lim_{\delta x\rightarrow0}\frac{u\delta v+v\delta u+\delta u\delta v}{\delta x}$limδx0uδv+vδu+δuδvδx

 Then dividing each term in the numerator by $\delta x$δx we get

$\frac{dy}{dx}$dydx $=$= $\lim_{\delta x\rightarrow0}\left(\frac{\delta u}{\delta x}v+u\frac{\delta v}{\delta x}\right)$limδx0(δuδxv+uδvδx)
  $=$= $\frac{du}{dx}v+u\frac{dv}{dx}$dudxv+udvdx

In abbreviated form this looks like:



Check with your teacher if you are expected to know and understand the proof. For some of you, knowing just the rule is sufficient.

Provided we know the derivative of both $u$u and $v$v, we can find the derivative of $y$y using the product rule. 

Product Rule

If $y=u\times v$y=u×v then $y'=u'v+uv'$y=uv+uv

If $y=u(x)v(x)$y=u(x)v(x) then $\frac{dy}{dx}=\frac{du}{dx}v+u\frac{dv}{dx}$dydx=dudxv+udvdx


In words you can remember this little rhythm.

The derivative of the first times the second plus the first times the derivative of the second


Let's look at a simple example $y=(x+2)(x-3)$y=(x+2)(x3)

Let $u=x+2$u=x+2 then $u'=1$u=1

Let $v=x-3$v=x3 then $v'=1$v=1

Using the product rule $y'=u'v+uv'$y=uv+uv


We can confirm this using our previous method of expansion and differentiating term by term.


So $y=x^2-x-6$y=x2x6 which means that $y'=2x-1$y=2x1

In the early days when you are learning the product rule, it often feels like we have made it harder by using a new rule. But when the functions get more difficult this rule really does help a lot!


Here is another example

Find the derivative of the function $y=x^4(2x^3-5)$y=x4(2x35)

Let $u=x^4$u=x4, then $u'=4x^3$u=4x3

Let $v=2x^3-5$v=2x35, then $v'=6x^2$v=6x2

Using the product rule







Note that this last step may not be necessary, sometimes fully factorised form is useful. 


When using the product rule, take it step by step. Jumping ahead is what leads most students to making errors. 

More Worked Examples


Differentiate the function $f\left(x\right)=\left(3x+2\right)\left(4x^2-5\right)$f(x)=(3x+2)(4x25).

You may use the substitution $u=3x+2$u=3x+2 and $v=4x^2-5$v=4x25 in your working.


Consider the function $y=\left(4x-3\right)\left(5x-2\right)$y=(4x3)(5x2).

  1. Differentiate $y$y.

  2. Now consider the function $f\left(x\right)=x^3\left(4x-3\right)\left(5x-2\right)$f(x)=x3(4x3)(5x2).

    Differentiate $f\left(x\right)$f(x).



Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods


Apply differentiation methods in solving problems

What is Mathspace

About Mathspace