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. We have seen that the derivative of a sum of functions is the same as the sum of the derivatives of the parts.
We have also learnt to differentiation composite functions like $y=\left(2x^4-4\right)^5$y=(2x4−4)5 using the chain rule.
Now we are interested in the derivatives of products: two functions multiplied together. In some cases we could expand the product and then differentiate term by term. However, just as with cases using the chain rule this method can be inefficient and tedious. For example, expanding $y=\left(3x-2\right)^2\left(7x^6-1\right)^7$y=(3x−2)2(7x6−1)7 would not be a quick task and expanding $y=e^x\left(x^2+5x+1\right)$y=ex(x2+5x+1) will not make it simpler to differentiate the function.
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. It is tempting to 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. Here it is clear that the derivative of a product is not the product of the derivatives!
To deal with this situation the product rule was developed.
Proof of the product rule
Let’s investigate this from first principles. Once we have derived the formula we can apply it without the need to return to first principles.
Let us consider two functions $u$u and $v$v both functions of $x$x. Hence, $u=u\left(x\right)$u=u(x) and $v=v\left(x\right)$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.
Recall our definition for a derivative from first principles:
| $f'\left(x\right)$f′(x) | $=$= | $\lim_{h\rightarrow0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$limh→0f(x+h)−f(x)h |
If $y=f\left(x\right)$y=f(x), then $f\left(x\right)=u\left(x\right)v\left(x\right)$f(x)=u(x)v(x) and $f\left(x+h\right)=u\left(x+h\right)v\left(x+h\right)$f(x+h)=u(x+h)v(x+h), hence, we have:
| $y'$y′ | $=$= | $\lim_{h\rightarrow0}\frac{u\left(x+h\right)v\left(x+h\right)-u\left(x\right)v\left(x\right)}{h}$limh→0u(x+h)v(x+h)−u(x)v(x)h |
We now add and subtract the term $u\left(x+h\right)v\left(x\right)$u(x+h)v(x) to the numerator to help us factorise and simplify the formula later. As we have added and subtracted this term to the numerator the fraction is not changed.
If we look carefully we can now simplify this expression further. Let's look at the individual limits:
| $y'$y′ | $=$= | $\lim_{h\rightarrow0}\frac{u\left(x+h\right)v\left(x+h\right)-u\left(x+h\right)v\left(x\right)+u\left(x+h\right)v\left(x\right)-u\left(x\right)v\left(x\right)}{h}$limh→0u(x+h)v(x+h)−u(x+h)v(x)+u(x+h)v(x)−u(x)v(x)h |
Add and subtract $u\left(x+h\right)v\left(x\right)$u(x+h)v(x) to the numerator |
| $=$= | $\lim_{h\rightarrow0}\frac{u\left(x+h\right)[v\left(x+h\right)-v\left(x\right)]+v\left(x\right)\left[u\left(x+h\right)-u\left(x\right)\right]}{h}$limh→0u(x+h)[v(x+h)−v(x)]+v(x)[u(x+h)−u(x)]h |
Group and factorise the terms in two groups |
|
| $=$= | $\lim_{h\rightarrow0}\left(\frac{u\left(x+h\right)\left[v\left(x+h\right)-v\left(x\right)\right]}{h}+\frac{v\left(x\right)\left[u\left(x+h\right)-u\left(x\right)\right]}{h}\right)$limh→0(u(x+h)[v(x+h)−v(x)]h+v(x)[u(x+h)−u(x)]h) |
Split the fraction into the sum of two fractions |
|
| $=$= | $\lim_{h\rightarrow0}\frac{u\left(x+h\right)\left[v\left(x+h\right)-v\left(x\right)\right]}{h}+\lim_{h\rightarrow0}\frac{v\left(x\right)\left[u\left(x+h\right)-u\left(x\right)\right]}{h}$limh→0u(x+h)[v(x+h)−v(x)]h+limh→0v(x)[u(x+h)−u(x)]h |
Apply the limit to both fractions using limit properties |
|
| $=$= | $\lim_{h\rightarrow0}\ u\left(x+h\right)\ \lim_{h\rightarrow0}\frac{v\left(x+h\right)-v\left(x\right)}{h}+\lim_{h\rightarrow0}\ v\left(x\right)\ \lim_{h\rightarrow0}\frac{u\left(x+h\right)-u\left(x\right)}{h}$limh→0 u(x+h) limh→0v(x+h)−v(x)h+limh→0 v(x) limh→0u(x+h)−u(x)h |
Apply the limit to each term in the products using limit properties |
As $h\rightarrow0$h→0 we can substitute into the function for the limit:
$\lim_{h\rightarrow0}\ u\left(x+h\right)=u\left(x\right)$limh→0 u(x+h)=u(x)
Recognising this limit as the definition for the derivative of $v\left(x\right)$v(x) using first principles:
$\lim_{h\rightarrow0}\frac{v\left(x+h\right)-v\left(x\right)}{h}=v'\left(x\right)$limh→0v(x+h)−v(x)h=v′(x)
Since $v\left(x\right)$v(x) is not dependent on $h$h:
$\lim_{h\rightarrow0}\ v\left(x\right)=v\left(x\right)$limh→0 v(x)=v(x)
Recognising this limit as the definition for the derivative of $u\left(x\right)$u(x) using first principles:
$\lim_{h\rightarrow0}\frac{u\left(x+h\right)-u\left(x\right)}{h}=u'\left(x\right)$limh→0u(x+h)−u(x)h=u′(x)
Hence, putting it all back together:
$y'=u\left(x\right)v'\left(x\right)+v\left(x\right)u'\left(x\right)$y′=u(x)v′(x)+v(x)u′(x)
Remembering that $u=u\left(x\right)$u=u(x) and $v=v\left(x\right)$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.
If $y=uv$y=uv , where $u$u and $v$v are functions of $x$x, then:
$y'=uv'+vu'$y′=uv′+vu′
Alternative notation, if $y=uv$y=uv then:
$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$dydx=udvdx+vdudx
Think: First function $\times$× derivative of second $+$+ second function $\times$× derivative of first
Worked examples
Example 1
Find the derivative of the function $y=\left(5x+2\right)\left(x-3\right)$y=(5x+2)(x−3)
Think: This function can be thought of as the product of two functions.
Do:
Let $u=5x+2$u=5x+2 then $u'=5$u′=5
Let $v=x-3$v=x−3 then $v'=1$v′=1
Using the product rule:
| $y'$y′ | $=$= | $uv'+vu'$uv′+vu′ |
| $=$= | $\left(5x+2\right)\times1+\left(x-3\right)\times5$(5x+2)×1+(x−3)×5 | |
| $=$= | $10x-13$10x−13 |
We can confirm this using our previous method of expansion and differentiating term by term.
| $y$y | $=$= | $\left(5x+2\right)\left(x-3\right)$(5x+2)(x−3) |
| $=$= | $5x^2-13x-6$5x2−13x−6 |
And then differentiating each term, we get $y'=10x-13$y′=10x−13 as before.
Example 2
Find the equation of the tangent to $y=\left(x+2\right)^2\left(x-1\right)^2$y=(x+2)2(x−1)2 at the point $\left(3,100\right)$(3,100).
Think: This function can be thought of as the product of two functions. Use the product rule to differentiate, then evaluate the derivative at $x=3$x=3 and use this together with the given point of contact to find the tangent equation.
Do:
Let $u=\left(x+2\right)^2$u=(x+2)2, then $u'=2\left(x+2\right)$u′=2(x+2)using the chain rule
Let $v=\left(x-1\right)^2$v=(x−1)2, then $v'=2\left(x-1\right)$v′=2(x−1)using the chain rule
Using the product rule:
| $y'$y′ | $=$= | $uv'+vu'$uv′+vu′ |
| $y'$y′ | $=$= | $\left(x+2\right)^2\times2\left(x-1\right)+\left(x-1\right)^2\times2\left(x+2\right)$(x+2)2×2(x−1)+(x−1)2×2(x+2) |
| $y'$y′ | $=$= | $2\left(x+2\right)^2\left(x-1\right)+2\left(x-1\right)^2\left(x+2\right)$2(x+2)2(x−1)+2(x−1)2(x+2) |
Gradient: When $x=3$x=3:
| $y'$y′ | $=$= | $2\left(3+2\right)^2\left(3-1\right)+2\left(3-1\right)^2\left(3+2\right)$2(3+2)2(3−1)+2(3−1)2(3+2) |
| $y'$y′ | $=$= | $140$140 |
Now, to find the equation of the tangent using the given point of contact and the gradient found:
| $y-y_1$y−y1 | $=$= | $m\left(x-x_1\right)$m(x−x1) |
| $y-100$y−100 | $=$= | $140\left(x-3\right)$140(x−3) |
| $y$y | $=$= | $140x-420+100$140x−420+100 |
| $y$y | $=$= | $140x-320$140x−320 |
Note: While not required for this calculation the derivative $y'=2\left(x+2\right)^2\left(x-1\right)+2\left(x-1\right)^2\left(x+2\right)$y′=2(x+2)2(x−1)+2(x−1)2(x+2) can be written in a fully factorised form by taking out the common factors of $2$2, $\left(x+2\right)$(x+2) and $\left(x-1\right)$(x−1), as follows:
| $y'$y′ | $=$= | $2\left(x+2\right)^2\left(x-1\right)+2\left(x-1\right)^2\left(x+2\right)$2(x+2)2(x−1)+2(x−1)2(x+2) |
| $=$= | $2\left(x+2\right)\left(x-1\right)\left[x+2+x-1\right]$2(x+2)(x−1)[x+2+x−1] | |
| $=$= | $2\left(x+2\right)\left(x-1\right)\left(2x+1\right)$2(x+2)(x−1)(2x+1) |
Factorised form will be very useful when finding stationary points of the function.
Practice questions
Question 1
Differentiate the function $f\left(x\right)=\left(3x+2\right)\left(4x^2-5\right)$f(x)=(3x+2)(4x2−5).
You may use the substitution $u=3x+2$u=3x+2 and $v=4x^2-5$v=4x2−5 in your working.
Question 2
Differentiate $y=x^3\left(5x+3\right)^7$y=x3(5x+3)7 using the product rule. Express your answer in factorised form.
You may let $u=x^3$u=x3 and $v=\left(5x+3\right)^7$v=(5x+3)7.
Question 3
Differentiate $y=8x^5\sqrt{8x+3}$y=8x5√8x+3 using the product rule. Give your final answer in surd form.
You may use the substitutions $u=8x^5$u=8x5 and $v=\sqrt{8x+3}$v=√8x+3.
Question 4
Differentiate $y=8x\left(5+8x\right)^{\frac{7}{4}}-3$y=8x(5+8x)74−3. Express the derivative in factorised form.
You may let $u=8x$u=8x and $v=\left(5+8x\right)^{\frac{7}{4}}$v=(5+8x)74 in your working if necessary.