Maths — Pure Mathematics 1

Chain mapping

Building blocksIntroducing1.14.2

Chain mapping

Excerpt: Students are led to an understanding of what the chain rule means and why it works, using mapping diagrams. It builds on two earlier mapping diagrams resources.

Warm-up

In Mapping a function, we explored mapping diagrams for linear functions. What happens if we now compose two functions?

How could we draw a mapping diagram to show the composition of the function $f(x)=\frac{3}{2}x$ with the function $g(x)=2x$ to get the function $h(x)=g(f(x))$ ?
We described linear functions as scalings. How are the scalings represented by the three functions $f$ , $g$ and $h$ related to each other?

Warm-up ideas

In Mapping a function, we explored mapping diagrams for linear functions. What happens if we now compose two functions?

How could we draw a mapping diagram to show the composition of the function $f(x)=\frac{3}{2}x$ with the function $g(x)=2x$ to get the function $h(x)=g(f(x))$ ?

We can represent this composite by using three numberlines, with arrows from the left to middle one representing $f(x)$ , and then arrows from the middle to the right representing $g(x)$ , like this:

Here, we have centred the left-hand numberline on $2$ ; the numberlines are centred in such a way that the arrow starting from $2$ is horizontal, as before.

We can then show the composite function $h(x)=g(f(x))$ by joining the starts to the ends of the arrow chains, like this:

This shows how the function $h$ takes values from the left-hand numberline to values on the right-hand numberline.

As in Mapping a function, we can also extend the arrows beyond the numberlines. For clarity, we only do this for the composite function $h$ :

The lines all meet in a point, showing that $h(x)$ is a scaling with a scale factor of $3$ . (This is the scaling factor from the left-hand to right-hand numberline.) As $h(x)=g(f(x))=g\bigl(\frac{3}{2}x\bigr)=3x$ , this is to be expected.

We can also check that the formula we worked out in Mapping a function for the location of the focal point gives us the expected location: that formula gives the location as $\frac{1}{1-3}=-\frac{1}{2}$ times the distance between the numberlines. This matches what we see in the diagram, because the numberlines of interest here are the left-hand and right-hand ones.

We described linear functions as scalings. How are the scalings represented by the three functions $f$ , $g$ and $h$ related to each other?

$f(x)=\frac{3}{2}x$ is a scaling with scale factor $\frac{3}{2}$ and $g(x)=2x$ is a scaling with scale factor $2$ . The composite function $h(x)=3x$ is a scaling with scale factor $3$ .

In general, if we compose a scaling with scale factor $s$ together with a scaling with scale factor $t$ , we will get a scaling with scale factor $st$ . In our particular case, we have $\frac{3}{2}\times 2=3$ .

Here, we used linear functions with no constant term and centred our mapping diagram on $0$ . Would we have obtained the same results if we had started with the functions $f(x)=\frac{3}{2}x+4$ and $g(x)=2x-3$ , still with $h(x)=g(f(x))$ , and centred our mapping diagram on $7$ (as input value)?

Problem

In Mapping a derivative, we saw that we can think of a derivative as a local scaling. What happens if we now want to find the derivative of a composite of two functions?

What does the mapping diagram of the composite $h(x)=g(f(x))$ look like if $f(x)=x^2+3$ and $g(x)=\sqrt{x}$ (so that $h(x)=\sqrt{x^2+3}$ ), and we centre our mapping diagram at $1$ on the input numberline?
How does the diagram change as we zoom in?
What does this tell us about the derivative $h'(1)$ ?
Generalising this, given any composite function $h(x)=g(f(x))$ , how can we find $h'(a)$ for a given $a$ ?

You can download blank composite mapping diagrams with one on a page or two on a page, or you can use the interactivity to explore this.

Things you might have noticed

If we draw the mapping diagram at a scale of $1$ unit per tick mark, we obtain the following, where the non-linear behaviour of $f$ is clear, though that of $g$ is less so:

We can now draw the composite arrows and extend them into lines:

We see that the green lines fail to meet at a point. If we zoom in, things now look very different:

What is going on here?

The composite function $h(x)$ appears to be a local scaling, as all of the lines meet at a point. From the diagram, the scale factor appears to be $\frac{1}{2}$ .
The function $f(x)=x^2+3$ is locally a scaling with scale factor $2$ . We can see this either by looking at the mapping diagram carefully, or by noting that the derivative of $f(x)$ is $f'(x)=2x$ , so that the scale factor at $x=1$ is $f'(1)=2$ .
The function $g(x)=\sqrt{x}$ is also locally a scaling when we zoom in. The scale factor is a little harder to make out reliably from the diagram, but we can use what we know about differentiation to calculate it. We have $g'(x)=\dfrac{1}{2\sqrt{x}}$ . The function $g$ is centred on $x=f(1)=4$ in this diagram, so the local scale factor is $g'(f(1))=g'(4)=\dfrac{1}{2\sqrt{4}}=\dfrac{1}{4}$ .
Composing these, we obtain a local scaling with scale factor $2\times\frac{1}{4}=\frac{1}{2}$ , multiplying the scale factors as we saw in the Warm-up section. This calculated composite scale factor agrees with what we observed about the function $h(x)$ .

So we conclude that $h'(1)=\frac{1}{2}$ .

Here is a somewhat generic sketch of a composite mapping diagram for $h(x)=g(f(x))$ , where we have shown the arrows and lines for $h(x)$ but not for the individual functions $f(x)$ or $g(x)$ :

The function $f(x)$ gives a local scaling with scale factor $f'(a)$ , the derivative of $f$ at $a$ .

The function $g(x)$ is centred on $f(a)$ , so it gives rise to a local scaling with scale factor $g'(f(a))$ , the derivative of $g$ at $f(a)$ .

Since $h(x)$ is locally the composition of these two scalings, we see that $h'(a)$ is also a local scaling. The scale factor of the composition is the product of the scale factors of $f$ and $g$ , so $h'(a)=f'(a)\times g'(f(a))$ .

This is the chain rule, which can therefore be written: $\text{if }h(x)=g(f(x))\text{, then }h'(x)=g'(f(x)).f'(x)\text{.}$

There is another common way of writing the chain rule, which we can get by setting $u=f(x)$ and $y=h(x)$ , so that $y=g(f(x))=g(u)$ . We then have $f'(x)=\dfrac{du}{dx}$ and $g'(f(x))=g'(u)$ , which we can write as $\dfrac{dy}{du}$ . Substituting this into the previous version then gives us: $\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}.$

Interactivity

To enter the function $x^2$ into this applet, type x^2. To enter $\sqrt{x}$ , type sqrt(x).

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/R8RFvC2q/width/600/height/493/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

Chain mapping

Excerpt: Students are led to an understanding of what the chain rule means and why it works, using mapping diagrams. It builds on two earlier mapping diagrams resources.

Warm-up

In Mapping a function, we explored mapping diagrams for linear functions. What happens if we now compose two functions?

How could we draw a mapping diagram to show the composition of the function $f(x)=\frac{3}{2}x$ with the function $g(x)=2x$ to get the function $h(x)=g(f(x))$ ?
We described linear functions as scalings. How are the scalings represented by the three functions $f$ , $g$ and $h$ related to each other?

Warm-up ideas

In Mapping a function, we explored mapping diagrams for linear functions. What happens if we now compose two functions?

How could we draw a mapping diagram to show the composition of the function $f(x)=\frac{3}{2}x$ with the function $g(x)=2x$ to get the function $h(x)=g(f(x))$ ?

We can represent this composite by using three numberlines, with arrows from the left to middle one representing $f(x)$ , and then arrows from the middle to the right representing $g(x)$ , like this:

Here, we have centred the left-hand numberline on $2$ ; the numberlines are centred in such a way that the arrow starting from $2$ is horizontal, as before.

We can then show the composite function $h(x)=g(f(x))$ by joining the starts to the ends of the arrow chains, like this:

This shows how the function $h$ takes values from the left-hand numberline to values on the right-hand numberline.

As in Mapping a function, we can also extend the arrows beyond the numberlines. For clarity, we only do this for the composite function $h$ :

We described linear functions as scalings. How are the scalings represented by the three functions $f$ , $g$ and $h$ related to each other?

$f(x)=\frac{3}{2}x$ is a scaling with scale factor $\frac{3}{2}$ and $g(x)=2x$ is a scaling with scale factor $2$ . The composite function $h(x)=3x$ is a scaling with scale factor $3$ .

Problem

In Mapping a derivative, we saw that we can think of a derivative as a local scaling. What happens if we now want to find the derivative of a composite of two functions?

What does the mapping diagram of the composite $h(x)=g(f(x))$ look like if $f(x)=x^2+3$ and $g(x)=\sqrt{x}$ (so that $h(x)=\sqrt{x^2+3}$ ), and we centre our mapping diagram at $1$ on the input numberline?
How does the diagram change as we zoom in?
What does this tell us about the derivative $h'(1)$ ?
Generalising this, given any composite function $h(x)=g(f(x))$ , how can we find $h'(a)$ for a given $a$ ?

You can download blank composite mapping diagrams with one on a page or two on a page, or you can use the interactivity to explore this.

Things you might have noticed

If we draw the mapping diagram at a scale of $1$ unit per tick mark, we obtain the following, where the non-linear behaviour of $f$ is clear, though that of $g$ is less so:

We can now draw the composite arrows and extend them into lines:

We see that the green lines fail to meet at a point. If we zoom in, things now look very different:

What is going on here?

The composite function $h(x)$ appears to be a local scaling, as all of the lines meet at a point. From the diagram, the scale factor appears to be $\frac{1}{2}$ .
The function $f(x)=x^2+3$ is locally a scaling with scale factor $2$ . We can see this either by looking at the mapping diagram carefully, or by noting that the derivative of $f(x)$ is $f'(x)=2x$ , so that the scale factor at $x=1$ is $f'(1)=2$ .
The function $g(x)=\sqrt{x}$ is also locally a scaling when we zoom in. The scale factor is a little harder to make out reliably from the diagram, but we can use what we know about differentiation to calculate it. We have $g'(x)=\dfrac{1}{2\sqrt{x}}$ . The function $g$ is centred on $x=f(1)=4$ in this diagram, so the local scale factor is $g'(f(1))=g'(4)=\dfrac{1}{2\sqrt{4}}=\dfrac{1}{4}$ .
Composing these, we obtain a local scaling with scale factor $2\times\frac{1}{4}=\frac{1}{2}$ , multiplying the scale factors as we saw in the Warm-up section. This calculated composite scale factor agrees with what we observed about the function $h(x)$ .

So we conclude that $h'(1)=\frac{1}{2}$ .

Here is a somewhat generic sketch of a composite mapping diagram for $h(x)=g(f(x))$ , where we have shown the arrows and lines for $h(x)$ but not for the individual functions $f(x)$ or $g(x)$ :

The function $f(x)$ gives a local scaling with scale factor $f'(a)$ , the derivative of $f$ at $a$ .

The function $g(x)$ is centred on $f(a)$ , so it gives rise to a local scaling with scale factor $g'(f(a))$ , the derivative of $g$ at $f(a)$ .

This is the chain rule, which can therefore be written: $\text{if }h(x)=g(f(x))\text{, then }h'(x)=g'(f(x)).f'(x)\text{.}$

Interactivity

To enter the function $x^2$ into this applet, type x^2. To enter $\sqrt{x}$ , type sqrt(x).

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/R8RFvC2q/width/600/height/493/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file