iGCSE (2021 Edition)

# 11.06 Product rule

Lesson

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=7x6x4. 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=(2x44)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=(3x2)2(7x61)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!

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 $y'=uv'+vu'$y=uv+vu

Or,

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

#### Worked examples

##### Example 1

Find the derivative of the function $y=(x+2)(x-3)$y=(x+2)(x3)

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=x3 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+(x3)×1=(x+2)+(x3)=2x1

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)(x3)=x2x6

And then differentiating each term, we get $y'=2x-1$y=2x1

##### Example 2

Find the equation of the tangent to $y=(x+2)^2(x-1)^2$y=(x+2)2(x1)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=(x1)2, then $v'=2(x-1)$v=2(x1)

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

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 2

Differentiate $y=8x^5\sqrt{8x+3}$y=8x58x+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 3

Differentiate $y=8x\left(5+8x\right)^{\frac{7}{4}}-3$y=8x(5+8x)743. 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.

### Outcomes

#### 0606C14.4

Differentiate products and quotients of functions.