“Mathematical intuition is the ability to see the truth without first having gone through a formal process of reasoning.”
— Henri Poincaré

The goal of this post is to force you to think about the chain rule a bit deeply.

The only prerequisite for reading this is understanding the power rule(i.e $\dfrac{d}{dx}[x^n] = nx^{n-1}$). Understanding differentiation intuitively is recommended.

The chain rule is a differentiation method for composite functions. It is defined as follows for a function $f(u)$ where $u = g(x)$ $$\dfrac{d}{dx}[f(u)] = \dfrac{d}{du}f(u)*\dfrac{d}{dx}(u)$$ which implies differentiate $f(u)$ with respect to $u$ treating $u$ as an independent variable, and differentiate $g(x)$ with respect to $x$ and then multiply both.

For example, given the function $f(x) = (2x^2 + 10)^2$, we can differentiate it as follows: we would set $g(x) = 2x^2 + 10$ and $u = g(x)$ so we can express $f(g(x))$ as $f(u) = u^2$.

Now, let’s use the chain rule: $\dfrac{d}{du}f(u) = 2u$, $\dfrac{d}{dx}(u) = 4x$ and $\dfrac{d}{dx}[f(u)] = 2u*4x$. Recall that $u = g(x) = 2x^2 + 10$, so we have $\dfrac{d}{dx}[f(u)] = 2(2x^2 + 10)*4x = 16x^3 + 80x$.

To understand the chain rule, you must first understand how function compositions work.

If we have a functions $f(x)$ and $g(x)$, what does $f(g(x))$ mean?

It means that for the function $f(x)$, evaluate it at $g(x)$ instead of $x$. We want to study the effect of this change.

How does $f(g(x))$ change with respect to a change $x$? Another way to put it is this: how does $f(g(x))$ change with respect to $g(x)$ and $g(x)$ with respect to $x$?

Let’s explore the behaviour of change in a function composition.

Example 1: For $g(x) = 2x$ and $f(x) = x^2$ for an interval $x \in [-4, 4]$.

Let’s see the graph:

Let’s build the intuition for this by looking at the table of values of $g(x)$, $f(x)$ $f(g(x))$ and $\dfrac{d}{dx}[f(g(x))]$ on the interval.

The graph shows that the maximum value for $f(x)$ is $16$ while that of $f(g(x))$ is $64$. That’s a scale-up.

Can you see a pattern? Every value of $f(g(x))$ is $4$ times the value of $f(x)$.

We can show this formally by picking any two unique pairs of $f(x)$ and $f(g(x))$ (i.e at the same $x$) and finding the slope(i.e $\dfrac{y_2 - y_1}{x_2 - x_1}$). For example, using the pairs ($16$, $64$) and ($4$, $16$), we have $\dfrac{64 - 16}{16 - 4} = \dfrac{48}{12} = 4$. Let’s use another two pairs ($9$, $36$) and ($1$, $4$), we have $\dfrac{4 - 36}{1 - 9} = \dfrac{-32}{-8} = 4$

This turns out to be easy to see because $f(g(x))$ is simply $4$ times $f(x)$ so it makes sense that this is the case. Other places you can notice is as follows:

• The values of $\dfrac{d}{dx}[f(g(x))]$ is 4 times the values of $\dfrac{d}{dx}[f(x)]$
• Every value of $f(g(x))$ and $\dfrac{d}{dx}[f(g(x))]$ is divisible by $4$
• The difference between any two values of $f(g(x))$ and $\dfrac{d}{dx}[f(g(x))]$ is divisible by $4$ The means that for every change in the value of $f(x)$ there’s a 4x change in the value of $f(g(x))$. This sounds weird, right? we just represented the rate of change between two somewhat independent functions as if they were dependent on each other. If this relationship was a function it would be $h(x) = 4x$ so that $\dfrac{d}{dx}[h(x))] = 4$. This works because $f(g(x))$ is a composition of $f(x)$ and $g(x)$ and $f(x) = x^2$.

Let’s try this for $x$ and $g(x)$. But then, you might ask: Is $g(x)$ a composition of $x$? Let’s see.

If we created a function $v(x) = x$, then $g(v(x))$ equals $g(x)$. This is true for every function of $x$. Having set this foundation, we can see that for every change in $x$, $g(x)$ by a factor of $2$.

Note: You won’t always have a constant factor between $f(x)$ and $f(g(x))$ or between their derivatives. But most times $g(x)$ would affect the behaviour of change in the $f(g(x))$ compared to $f(x)$. It either makes it faster, slower, scale-up, scale-down and even combination of these along certain intervals.

The examples below show some of these other behaviours of change.

Example 2: For $g(x) = -x$ and $f(x) = x^3$ for an interval $x \in [-3, 3]$.

Example 3: For $g(x) = x + 1$ and $f(x) = x^2+x$ for an interval $x \in [-2, 2]$.

Chain rule answers the question: how do you express a function dependent on a variable as the rate of change in respect to that variable when the variable is a function?

Why is $g(x)$ treated like a variable?

In the chain rule why are we differentiating in respect to $g(x)$ as if it was a variable? This is simply because it is.

The intuition is the definition of the behaviour of change in a composite function: as $g(x)$ changes $f(g(x))$ and as $x$ changes $g(x)$.

$g(x)$ is a variable in $f(g(x))$ so it’s treated as such when represent the change in $f(g(x))$

Why is the change in $g(x)$ multiplied by the change in $f(g(x))$?

This is simply because as $x$ changes $g(x)$ changes and that’s a factor affecting the change in $f(g(x))$.

Why Multiplication(and Not Addition)?

Why is the chain rule $\dfrac{d}{dx}[f(u)] = \dfrac{d}{du}f(u)*\dfrac{d}{dx}(u)$ and not $\dfrac{d}{dx}[f(u)] = \dfrac{d}{du}f(u)+\dfrac{d}{dx}(u)$?

We would look two ways to look at this: the first is a bit hand-wavy and the other is more intuitive.

For the first one, the intuition lies is in the definition: as $g(x)$ changes $f(g(x))$ and as $x$ changes $g(x)$. For clarity: Let’s break this statement into three part:

• as $g(x)$ changes $f(x)$ - $g(x)$ is treated as a variable
• $x$ changes $g(x)$
• and - from binary operations, we know that and implies multiplication. just like or implies addition.

Let’s look the more intuitive way version: currency conversions!

I have Nigerian Naira (NGN) and I want to convert it to Pounds sterling(GBP). But, there’s a little challenge: there are only two exchanges available; NGN/USD and USD/GBP. We would have to convert from NGN to USD and then, from USD to GBP. The rates are 1 USD = NGN1,500 and 1 GBP = 1.25 USD respectively. How do we convert NGN20,000 to GBP?

First, we would convert to USD. NGN20,000 to USD is 20,000/1500 = 13.33 USD. Then, USD to GBP is 13.33 / 1.25 = 10.67 GBP. That is, NGN20,000 equals 10.67 GBP. To get the rate of NGN to GBP, we multiply the both rates(.i.e 1/1500 x 1/1.25 = 1/1875).

The interesting thing is we can convert these individual conversions to functions. The function for converting NGN to USD would be $g(x) = \dfrac{x}{1500}$ and the one for converting from USD to GBP would be $f(x) = \dfrac{x}{1.25}$. The most interesting part of this is that the function for converting from NGN to GBP is function composition of $f(x)$ and $g(x)$! That is, $f(g(x)) = \dfrac{\dfrac{x}{1500}}{1.25} = \dfrac{x}{1875}$.

If we apply the chain rule to the composite function $f(u)$ where $u = g(x)$, we have $\dfrac{d}{dx}[f(u)] = \dfrac{d}{du}f(u)*\dfrac{d}{dx}(u) = \dfrac{1}{1.25} * \dfrac{1}{1500} = \dfrac{1}{1875}$. That is our expected rate!

But, if we change multiplication to addition to we have $\dfrac{d}{dx}[f(u)] = \dfrac{d}{du}f(u)+\dfrac{d}{dx}(u) = \dfrac{1}{1.25}+\dfrac{1}{1500}=\dfrac{1201}{1500}$. This is very wrong!

I employ you to think of others ways asides currency conversions whereby this can be shown!

I hope you were able to think wide and far about function compositions and the chain rule!