11.08 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 |
If $y=g(u)$y=g(u) and $u=f(x)$u=f(x)
then
$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$dydx=dydu×dudx
Practice questions
Question 1
Find the derivative of $y=\left(4x+3\right)^9$y=(4x+3)9 using the chain rule. Give your answer fully factorised.
Question 2
Let $y=\sqrt{5+x^2}$y=√5+x2 be defined as a composition of the functions $y=\sqrt{u}$y=√u and $u=5+x^2$u=5+x2.
Determine $\frac{dy}{du}$dydu, expressing your answer in surd form.
Determine $\frac{du}{dx}$dudx.
Hence determine $\frac{dy}{dx}$dydx, expressing your answer in surd form.
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.
If $y=f(x)^n$y=f(x)n then:
$\frac{dy}{dx}\ =\ n\ f'(x)\ [f(x)]^{n-1}$dydx = n f′(x) [f(x)]n−1
Or sometimes written:
$\frac{d(f(x)^n)}{dx}\ =\ n\ f'(x)\ [f(x)]^{n-1}$d(f(x)n)dx = n f′(x) [f(x)]n−1
We can remember this as take the derivative of the outside, times the derivative of the inside.
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 |
Watch out for negatives inside the brackets! Some functions inside the brackets might look like their derivative is $1$1 and will not affect the chain, but in fact is equal to $-1$−1.
For example, if $y=(2-x)^4$y=(2−x)4, then $\frac{dy}{dx}=-1\times4\times(2-x)^3$dydx=−1×4×(2−x)3.
Practice questions
Question 3
Find the $x$x-coordinate(s) of the point(s) at which $f\left(x\right)=\left(x+2\right)^3$f(x)=(x+2)3 has a gradient of $48$48.
If there is more than one solution, write all the solutions on the same line, separated by commas.
Question 4
Find the gradient of $f\left(x\right)=\left(x-5\right)^3$f(x)=(x−5)3 at the point $\left(8,27\right)$(8,27).
Denote this gradient by $f'\left(8\right)$f′(8).