11.05 Chain rule

So far in this chapter you have learned how to differentiate functions involving multiple terms raised to powers. 

Let's consider a function of the form $y=\left(3x^2-2\right)^5$y=(3x2−2)5, where we have an expression raised to a power. How can we find $\frac{dy}{dx}$dydx​?

So far with what we know about finding derivatives, we'd need to expand the function and differentiate term by term. This would be very inefficient.

There are two methods we can use. The first method uses substitution to break the function up into parts that we can differentiate. Here, we will let the function found inside the brackets be $u$u.

Exploration

Our original function is:

$y$y

$=$=

$(3x^2-2)^5$(3x2−2)5

Let $u=3x^2-2$u=3x2−2, therefore:

$y$y

$=$=

$u^5$u5

And:

$\frac{du}{dx}$dudx​

$=$=

$6x$6x

Differentiating $y$y with respect to $u$u gives:

$\frac{dy}{du}$dydu​

$=$=

$5u^4$5u4

Substituting $u=3x^2-2$u=3x2−2 into $\frac{dy}{du}$dydu​ gives:

$\frac{dy}{du}$dydu​

$=$=

$5(3x^2-2)^4$5(3x2−2)4

 

Remember that we want to find $\frac{dy}{dx}$dydx​. To find this we can use $\frac{dy}{du}$dydu​ and $\frac{du}{dx}$dudx​ to create a chain as follows:

$\frac{dy}{dx}$dydx​

$=$=

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

Combining out two results above gives:

$\frac{dy}{dx}$dydx​

$=$=

$5(3x^2-2)^4\times6x$5(3x2−2)4×6x

Which gives:

$\frac{dy}{dx}$dydx​

$=$=

$(30x)(3x^2-2)^4$(30x)(3x2−2)4

 

 

Practice questions 

Question 1

Question 2

 

Simplifying the chain rule 

Substituting is clearly faster than expanding out these questions with brackets. However, there is an even quicker way to solve these problems. You might have noticed this pattern when working through the examples above. 

 

Worked examples

Example 1

Differentiate $y=(7x^3-1)^4$y=(7x3−1)4

Think: This function is in the form that allows us to use the chain rule for differentiation.

 

$y$y	$=$=	$f(x)^n$f(x)n
$y$y	$=$=	$(7x^3-1)^4$(7x3−1)4

 

Do: Differentiating the outside function:

$n\ (f(x))^{n-1}$n (f(x))n−1

$=$=

$4(7x^3-1)^3$4(7x3−1)3

 

Differentiating the inside function:

$f'(x)$f′(x)

$=$=

$21x^2$21x2

Putting it all together:

$\frac{dy}{dx}$dydx​	$=$=	$21x^2\times4(7x^3-1)^3$21x2×4(7x3−1)3
$\frac{dy}{dx}$dydx​	$=$=	$84x^2(7x^3-1)^3$84x2(7x3−1)3

Example 2

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

Think: We first need to find the derivative of the function.

Do:

$\frac{dy}{dx}$dydx​	$=$=	$2(3)(3x+2)$2(3)(3x+2)
$\frac{dy}{dx}$dydx​	$=$=	$6(3x+2)$6(3x+2)
$\frac{dy}{dx}$dydx​	$=$=	$18x+12$18x+12

When $x=-2$x=−2

$\frac{dy}{dx}$dydx​	$=$=	$18(-2)+12$18(−2)+12
$\frac{dy}{dx}$dydx​	$=$=	$-24$−24

 

The equation of the tangent is found using:

$y-y_1=m(x-x_1)$y−y1​=m(x−x1​)

when $x_1=-2,y_1=16$x1​=−2,y1​=16

$y-16$y−16	$=$=	$-24(x+2)$−24(x+2)
$y$y	$=$=	$-24x-32$−24x−32

 

 

Practice questions

Question 3

Question 4