The Chain Rule (VCE SSCE Mathematical Methods): Revision Notes

The Chain Rule

Introduction to composite functions

Sometimes we need to differentiate functions that are built from simpler functions combined together. For example, consider $q(x) = (x^3 + 1)^2$. We could expand this and differentiate term by term, but this becomes very tedious for something like $q(x) = (x^3 + 1)^{30}$.

Instead, we can recognise that $q(x) = (x^3 + 1)^2$ is made up of two simpler functions "chained" together:

Let $u = g(x) = x^3 + 1$ (the inside function)

Let $y = f(u) = u^2$ (the outside function)

This creates a chain: $x \xrightarrow{g} u \xrightarrow{f} y$

We write this as $q(x) = f(g(x))$ or $q = f \circ g$, which is called a composite function. The chain rule gives us an efficient method to differentiate such functions.

The chain rule formula

If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then the composite function $q(x) = f(g(x))$ is differentiable at $x$ and:

$$q'(x) = f'(g(x)) \cdot g'(x)$$

Using Leibniz notation, where $u = g(x)$ and $y = f(u)$:

$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$

This form makes it easy to remember - the $du$ terms appear to "cancel".

Understanding the chain rule

The chain rule works because when we have a composite function, the rate of change of $y$ with respect to $x$ depends on:

• How fast $y$ changes with respect to $u$ (which is $\frac{dy}{du}$)
• How fast $u$ changes with respect to $x$ (which is $\frac{du}{dx}$)

Worked examples

Using technology to verify your work

You can use a calculator to check your chain rule calculations:

TI-Nspire calculator:

• Define your functions $g(x)$ and $h(x)$ separately
• Define the composite function $f(x) = g(h(x))$
• Use menu > Calculus > Derivative to find $\frac{d}{dx}(f(x))$

Casio ClassPad calculator:

• Define $g(x)$ and $h(x)$ separately
• Define $f(x) = g(h(x))$
• Use the differentiation function to find $\frac{d}{dx}(f(x))$

The calculator will apply the chain rule automatically and give you the derivative in simplified form.