11.09 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 such as $y=3x^4+5x^2$y=3x4+5x2 or $y=7x^6-x^4$y=7x6−x4. You may have noticed that the derivative of the sum of functions is the same as the sum of the derivatives of the parts. 

We have also learnt to differentiation composite functions like $y=(2x^4-4)^5$y=(2x4−4)5 using the chain rule.

Now, we are interested in the derivatives of products: two functions multiplied together. Of course we could simplify the product if possible and then differentiate the terms. However, sometimes we won't want to expand the product as it is too long and this method is prone to errors. For example, expanding $y=(3x-2)^2(7x^6-1)^7$y=(3x−2)2(7x6−1)7 would not be a simple process.

Let's 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. You might think that we could differentiate the two components $x^2$x2 and $x^3$x3 and multiply these answers together to get the derivative. But $2x$2x times $3x^2$3x2 gives us $6x^3$6x3. As is the case in many instances in mathematics, operations involving multiplication behave differently to those involving addition or subtraction. We saw earlier that the derivative of the sum of functions is the same as the sum of the derivatives of the parts but it is clear that the derivative of a product is not the product of the derivatives!

Proof of the product rule 

Let’s investigate this from first principles. Once we have derived the formula, or process, you will use the rule only. You will not need to know how to reproduce this proof for this course.

Let us consider two functions $u$u and $v$v both functions of $x$x. Hence, $u=u(x)$u=u(x) and $v=v(x)$v=v(x).

Now consider a function $y$y a product of the two functions $u$u and $v$v. Hence, $y=uv$y=uv.

Using our formula for differentiating from first principles:

$y'$y′	$=$=	$\lim_{h\rightarrow0}\frac{f(x+h))-f(x)}{h}$limh→0f(x+h))−f(x)h​
$y'$y′	$=$=	$\lim_{h\rightarrow0}\frac{u(x+h)v(x+h)-u(x)v(x)}{h}$limh→0u(x+h)v(x+h)−u(x)v(x)h​

We now subtract the term $u(x+h)v(x)$u(x+h)v(x)to the numerator to help us factorise later. We also need to add this term to the numerator so the fraction is not changed.

$y'$y′

$=$=

$\lim_{h\rightarrow0}\frac{u(x+h)v(x+h)-u(x+h)v(x)+u(x+h)v(x)-u(x)v(x)}{h}$limh→0u(x+h)v(x+h)−u(x+h)v(x)+u(x+h)v(x)−u(x)v(x)h​

Splitting the fraction and grouping terms together gives:

$y'$y′

$=$=

$\lim_{h\rightarrow0}\frac{u(x+h)[v(x+h)-v(x)]}{h}+\frac{v(x)[u(x+h)-u(x)]}{h}$limh→0u(x+h)[v(x+h)−v(x)]h​+v(x)[u(x+h)−u(x)]h​

Now we split the limit using the rule that the limit of an expression is equal to the product of the individual limits of its factors:

$y'$y′

$=$=

$\lim_{h\rightarrow0}\ u(x+h)\ \lim_{h\rightarrow0}\frac{v(x+h)-v(x)}{h}+\lim_{h\rightarrow0}\ v(x)\ \lim_{h\rightarrow0}\frac{u(x+h)-u(x)}{h}$limh→0 u(x+h) limh→0v(x+h)−v(x)h​+limh→0 v(x) limh→0u(x+h)−u(x)h​

 

If we look carefully we can now simplify this expression further. Let's look at the individual limits:

As $h\rightarrow0$h→0 you can just substitute into the function:

$\lim_{h\rightarrow0}\ u(x+h)=u(x)$limh→0 u(x+h)=u(x)

Recognising this limit is the formula for the derivative of $v(x)$v(x) using first principles:

$\lim_{h\rightarrow0}\frac{v(x+h)-v(x)}{h}=v'(x)$limh→0v(x+h)−v(x)h​=v′(x)

As $h$h isn't involved in this function:

$\lim_{h\rightarrow0}\ v(x)=v(x)$limh→0 v(x)=v(x)

Recognising this limit is the formula for the derivative of $u(x)$u(x) using first principles:

$\lim_{h\rightarrow0}\frac{u(x+h)-u(x)}{h}=u'(x)$limh→0u(x+h)−u(x)h​=u′(x)

Putting it all back together:

$y'=u(x)v'(x)+v(x)u'(x)$y′=u(x)v′(x)+v(x)u′(x)

Remembering that $u=u(x)$u=u(x) and $v=v(x)$v=v(x) we can rewrite as follows:

$y'=uv'+vu'$y′=uv′+vu′

 

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

 

Worked examples

Example 1

Find the derivative of the function $y=(x+2)(x-3)$y=(x+2)(x−3)

Think: This function can be thought of as the product of two functions.

Do:

Let $u=x+2$u=x+2 then $u'=1$u′=1

Let $v=x-3$v=x−3 then $v'=1$v′=1

Using the product rule $y'=uv'+vu'$y′=uv′+vu′

$y'=(x+2)\times1+(x-3)\times1=(x+2)+(x-3)=2x-1$y′=(x+2)×1+(x−3)×1=(x+2)+(x−3)=2x−1

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

$y=(x+2)(x-3)=x^2-x-6$y=(x+2)(x−3)=x2−x−6

And then differentiating each term, we get $y'=2x-1$y′=2x−1

 

Example 2

Find the equation of the tangent to $y=(x+2)^2(x-1)^2$y=(x+2)2(x−1)2 at the point $(3,100)$(3,100). 

Think: This function can be thought of as the product of two functions.

Do:

Let $u=(x+2)^2$u=(x+2)2, then $u'=2(x+2)$u′=2(x+2)

Let $v=(x-1)^2$v=(x−1)2, then $v'=2(x-1)$v′=2(x−1)

Using the product rule:

$y'$y′	$=$=	$uv'+vu'$uv′+vu′
$y'$y′	$=$=	$(x+2)^2\times2(x-1)+(x-1)^2\times2(x+2)$(x+2)2×2(x−1)+(x−1)2×2(x+2)
$y'$y′	$=$=	$2(x+2)^2(x-1)+2(x-1)^2(x+2)$2(x+2)2(x−1)+2(x−1)2(x+2)

When $x=3$x=3:

$y'$y′	$=$=	$2(3+2)^2(3-1)+2(3-1)^2(3+2)$2(3+2)2(3−1)+2(3−1)2(3+2)
$y'$y′	$=$=	$140$140

 

Now, to find the equation of the tangent:

$y-y_1$y−y1​	$=$=	$m(x-x_1)$m(x−x1​)
$y-100$y−100	$=$=	$140(x-3)$140(x−3)
$y$y	$=$=	$140x-420+100$140x−420+100
$y$y	$=$=	$140x-320$140x−320

 

Remember, the best way to give your final answer to a derivative is in fully factorised form. The two terms in your answer will generally have common factors because of the process of reducing powers when differentiating. 

 

Practice questions

Question 1

Question 2

Question 3