NZ Level 8 (NZC) Level 3 (NCEA) [In development] Chain Rule
Lesson

Let's consider a function of the form $y=\left(3x^2-2\right)^5$y=(3x22)5, where we have an expression raised to a power.

How can we find $\frac{dy}{dx}$dydx?

To use what we know, we'd need to expand the function and differentiate term by term, but that would be really inefficient.

Let's use substitution to break the function up into parts that we can differentiate.

Example
$y=\left(3x^2-2\right)^5$y=(3x22)5
Let $u=3x^2-2$u=3x22 We get $y=u^5$y=u5
$\frac{du}{dx}=6x$dudx=6x $\frac{dy}{du}=5u^4$dydu=5u4

Generalising
$y=f(x)^n$y=f(x)n
Let $u=f(x)$u=f(x) We get $y=u^n$y=un
$\frac{du}{dx}=f'(x)$dudx=f(x) $\frac{dy}{du}=nu^{n-1}$dydu=nun1

Remember that we want to find $\frac{dy}{dx}$dydx.

How can we use $\frac{dy}{du}$dydu and $\frac{du}{dx}$dudx to get the derivative we want?

We can create a 'chain'.

Chain Rule

If $y=f(u)$y=f(u) and $u=g(x)$u=g(x)

then

$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$dydx=dydu×dudx

Algebraically this looks OK, but it needs ideas from university level maths to prove, so we won't go into that here. At the moment we can just accept the statement.

So moving forward with our example:

Example
$y=\left(3x^2-2\right)^5$y=(3x22)5
$\frac{dy}{dx}$dydx $=$= $\frac{dy}{du}\times\frac{du}{dx}$dydu×dudx   Set up the chain rule.
$=$= $5u^4\times6x$5u4×6x   Substitute the derivatives.
$=$= $30x\left(3x^2-2\right)^4$30x(3x22)4   Rewrite $\frac{dy}{dx}$dydx in terms of $x$x.

Special case of the Chain Rule

If $y=f(x)^n$y=f(x)n, its derivative is

$\frac{dy}{dx}=nf\left(x\right)^{n-1}$dydx=nf(x)n1$\times$×$f'(x)$f(x)

We can remember this as "take the derivative of the outside, times the derivative of the inside”.

#### Example

##### Example 1

Find the derivative of $y=(x^5+3x+1)^3$y=(x5+3x+1)3

Let $u=x^5+3x+1$u=x5+3x+1, then$\frac{du}{dx}=5x^4+3$dudx=5x4+3 (this is the derivative of the inside piece)

So $y=u^3$y=u3, then $\frac{dy}{du}=3u^2$dydu=3u2 (this is the derivative of the outside piece)

$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$dydx=dydu·dudx

$\frac{dy}{dx}=(5x^4+3)\cdot3u^2$dydx=(5x4+3)·3u2 (now we substitute in the value of $u$u, and simplify algebraically)

$\frac{dy}{dx}=(5x^4+3)\cdot3(x^5+3x+1)^2$dydx=(5x4+3)·3(x5+3x+1)2

$\frac{dy}{dx}=3(5x^4+3)(x^5+3x+1)^2$dydx=3(5x4+3)(x5+3x+1)2

#### More Worked Examples

##### QUESTION 1

We want to differentiate the function $y=\left(2x^4+6\right)^5$y=(2x4+6)5 using the substitution $u=2x^4+6$u=2x4+6.

1. Determine $\frac{du}{dx}$dudx.

2. Express $y$y as a function of $u$u

3. Determine $\frac{dy}{du}$dydu.

4. Hence determine $\frac{dy}{dx}$dydx.

##### QUESTION 2

Find the derivative of $y=\sqrt{8x+5}$y=8x+5 using the chain rule. Give your answer in surd form.

##### QUESTION 3

Consider the function $f\left(x\right)=\left(5x^3-4x^2+3x-5\right)^7$f(x)=(5x34x2+3x5)7.

Redefine the function as composite functions $f\left(u\right)$f(u) and $u\left(x\right)$u(x), where $u\left(x\right)$u(x) is a polynomial.

1.  $u\left(x\right)=\editable{}$u(x)= $f\left(u\right)=\left(\editable{}\right)^{\editable{}}$f(u)=()

### Outcomes

#### M8-11

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

#### 91578

Apply differentiation methods in solving problems