C.1 Chain Rule - Developing a Conceptual Understanding

Calculus students often hear that "the Chain Rule" is scary or difficult. Let's break it down, and learn how to find the most complicated derivatives. On this screen we'll develop a conceptual understanding of the Chain Rule, and then on the next screen we'll present the rule and work some initial practice problems.

You might have noticed that, so far in this "Calculating Derivatives" chapter, we have only found the derivative of simple functions like 𝑥2, sin⁡𝑥, 𝑒𝑥, and the product and quotient of such functions. We have not yet found the derivative of a composite (or compound) function like sin⁡(2𝑥), or 𝑒𝑥2. The reason is that we need a new tool to be able to do so! That tool is called "the Chain Rule," which we will spend this entire Section exploring because it's that important. We'll of course also provide lots of practice so you'll confidently have its use comfortably in your toolkit.

Why we need a new rule

To begin, let's quickly consider two examples to illustrate why we need a new rule at all.

First, consider the function 𝑓(𝑥) =(2𝑥)3. Viewing this function as 𝑓(𝑥) =8𝑥3, we know from the Power Rule that its derivative is 𝑓′(𝑥) =24𝑥2.

Naively, looking at 𝑓(𝑥) =(2𝑥)3 to find the derivative you might simply bring the 3 down in front of the parentheses and change the power to 2: 𝑓′(𝑥)=[(2𝑥)3]′?⏞=3(2𝑥)2[Does this naive approach work??]?⏞=3(4𝑥2)?⏞=12𝑥2≠24𝑥2✗[No![(2𝑥)3]′≠3(2𝑥)2] Ack: our naive approach gives a result that is off by a factor of 2 from the correct answer.

As a second example, consider the function 𝑝(𝑥) =(3𝑥 −1)2. If we view this function as 𝑝(𝑥) =9𝑥2 −6𝑥 +1, then we know immediately from the Power Rule that its derivative is 𝑝′(𝑥) =18𝑥 −6.

Again thinking naively, you might simply bring the power of 2 down in front of the parentheses: 𝑝′(𝑥)=[(3𝑥−1)2]′?⏞=2(3𝑥−1)[Does this naive approach work??]?⏞=6𝑥−2≠18𝑥−6✗[No!6𝑥−2≠18𝑥−6] Again the naive approach doesn't work, and this time we're off by a factor of 3. (Hmmm.)

As you'll see below, the Chain Rule resolves this discrepancy, and will let us — easily, with practice! — find the derivatives of functions that are quite complicated.

In order to understand the Chain Rule, we first need to make sure we're clear about compound functions.

Compound (Composite) Functions Review

Recall that a compound function, also known as a composite function, is a function comprised of one or more functions inside it.

For instance, (𝑥2+1)7 is comprised of the inner function 𝑥2 +1 inside the outer function (⋯)7.

As another example, 𝑒sin⁡𝑥 is comprised of the inner function sin⁡𝑥 inside the outer function 𝑒⋯.

As yet another example, ln⁡(𝑡3−2𝑡2+5) is comprised of the inner function 𝑡3 −2𝑡2 +5 inside the outer function ln⁡(⋯).

"How can I tell what the inner and outer functions are?"

Here's a foolproof method to determine the inner and outer functions: Imagine calculating the numerical output of the function for a particular input value of x and identify the steps you would take, because you'll always automatically start with the inner function and work your way out to the outer function.

For example, imagine computing (𝑥2+1)7 for 𝑥 =3. Without thinking about it, you would first calculate 𝑥2 +1 (which equals 32 +1 =10), so that's the inner function, guaranteed. Then you would next calculate 107, and so (⋯)7 is the outer function.

This imaginary computational process works every time to identify correctly what the inner and outer functions are.

(𝑓 ∘𝑔)(𝑥) also means
"f of g of x, 𝑓(𝑔(𝑥))"

Alternate notation: You may see the compound (or composite) function 𝑓(𝑔(𝑥)) written instead as (𝑓 ∘𝑔)(𝑥). These mean exactly the same thing, and both are said as "f of g of x." Most students prefer the former notation, as do we, and so we'll almost always use it as we did above. However, if you encounter (𝑓 ∘𝑔)(𝑥), please know it's simply a different way of writing 𝑓(𝑔(𝑥)).

Now please set aside this quick review of compound functions. It'll be imporant again in a bit, but first we're going to develop an intuitive understanding of the Chain Rule before we present it formally.

Developing a Conceptual Understanding of the Chain Rule: A balloon ascends and cools

Before we present the official Chain Rule, let's consider an example situation to illustrate the basic idea.

Imagine a balloon that travels straight upward at a rate given in m/s.

As you may know, as you move upward away from the Earth's surface, the temperature of the air decreases. Specifically, the air around it gets cooler and cooler at a rate given in ∘F/m. For simplicity, let's imagine the balloon's temperature always matches that of the surrounding air.

To keep our focus on the key point here, we're going to pretend that the two rates are constant. Specifically, let's imagine that the balloon ascends such that its elevation, E, changes at the constant rate with respect to time: 𝑑( elevation )𝑑( time )=𝑑𝐸𝑑𝑡=0.004ms Hence, for instance in 1 second the balloon ascends 0.004 meters.

Let's also imagine that as the balloon travels upward from the Earth's surface, its temperature, T, changes at the constant rate with respect to elevation: 𝑑( temperature )𝑑( elevation )=𝑑𝑇𝑑𝐸=−0.01∘Fm For instance, when the balloon gains 1 meter in elevation, its temperature changes by −0.01∘F.

Here's the question that gets to the core of the Chain Rule:

What is 𝑑( temperature )𝑑( time )?
For instance, in 1 second, how much does the balloon's temperature change?

[Do you have an answer in mind? If not, please stop and develop one for yourself. In particular, imagine what happens over 1 second: the balloon travels upward ___ m, which means its temperature changes by . . . . ]

If your instinct was simply to multiply the two rates, then great! Hold onto that intuition, because it is perfectly correct and is at the core of the Chain Rule.

If not, then think about what happens over the course of a 1-second time change. Given 𝑑𝐸/𝑑𝑡 =0.004 ms, over 1 second the balloon's elevation increases by 0.004 meters.

Now shift focus. When the balloon's elevation increases by 0.004 meters, what temperature change does it undergo? Since the temperature rate of change is 𝑑𝑇/𝑑𝐸 = −0.01∘F/m, then over the 1-second elevation increase of 0.004 meters the balloon's temperature changes by Δ𝑇=(0.004 m )(−0.01∘Fm)=−0.00004∘F So for a time change of 1 second, our temperature changes by −.00004∘F.

Having focused on what happens in 1 second, let's return to the time-rate at which the balloon's temperature T changes, 𝑑𝑇𝑑𝑡. To find the small change in temperature, dT, relative to a small change in time, dt, we simply multiply the rate at which elevation changes with time (dE/dt) by the rate at which the temperature changes with elevation (dT/dE): 𝑑𝑇𝑑𝑡=𝑑𝑇𝑑𝐸⋅𝑑𝐸𝑑𝑡=(0.004ms)(−0.01∘Fm)=−0.00004∘Fs Notice that the units cancel in this calculation, as we would expect.

If that all makes sense, you have the fundamental idea behind the Chain Rule.

Recasting the balloon scenario in terms of functions

To see how this scenario relates to compound functions, let's recast the balloon's temperature change in function notation.

We know that variations in time result in corresponding variations in elevation, so elevation is a function of time. We denote this functional dependence by writing 𝐸 =𝐸(𝑡). (That equation is read as "elevation E is a function of time t," and that's all it means.)

Similarly, variations in elevation result in corresponding variations in temperature, so temperature is a function of elevation. We denote this by writing 𝑇 =𝑇(𝐸).

Putting the pieces together, we can write temperature T as a function of time t as the composition of the functions T and 𝐸, written as 𝑇(𝑡)=𝑇(𝐸(𝑡)) So really, from the beginning we could have said this scenario considers a composition of the functions T and 𝐸. 𝐸(𝑡) is just a pesky intermediate function that we had to go through to see what the relationship between the balloon's temperature and time.

Returning to the question of 𝑓(𝑔(𝑥)), which for the moment we'll write as 𝑦(𝑢(𝑥)). Your quick calculation above shows that if you are interested in the derivative of y with respect to x, but there's a pesky intermediate function u between y and x, you can still find the derivative — easily! — by taking the derivative of the outside function with respect to the inside function, 𝑑𝑦𝑑𝑢, and then multiplying by the derivative of the inside function with respect to the input variable x, 𝑑𝑢𝑑𝑥. That is, 𝑑𝑦𝑑𝑥=𝑑𝑦𝑑𝑢⋅𝑑𝑢𝑑𝑥 This is exactly the process you probably landed on intuitively when you multiplied the two rates and computed 𝑑𝑇𝑑𝑡 =𝑑𝑇𝑑𝐸 ⋅𝑑𝐸𝑑𝑡 above.

The key thing to notice: to find the rate-of-change of the overall function 𝑇(𝑡) =𝑇(𝐸(𝑡)) with respect to the inner variable t, you automatically multiplied the rate-of-change of the outer function by the rate-of-change of the inner function. That is the Chain Rule, which we will formalize on the next screen. We'll also work some initial practice problems, and easily resolve the discrepancy we saw at the beginning of this screen when finding the derivative of 𝑓(𝑥) =(2𝑥)3 and 𝑝(𝑥) =(3𝑥 −1)2.

The Upshot

1. A compound (or composite) function is comprised of an outer function and an inner function.
2. Typically, without much thought, to find the rate-of-change of a compound function in an everyday scenario you would automatically multiply the rate-of-change of the outer function with the rate-of-change of the inner function. That is the Chain Rule.