The chain rule differentiates composite functions by multiplying the derivative of the outer function by the derivative of the inner function.

What this dot point is asking

A composite function is a "function of a function" - one expression nested inside another, like $(3x+1)^5$, $e^{x^2}$ or $\ln(\cos x)$. The chain rule tells you how to differentiate these, and it is the single most-used differentiation rule in SACE Stage 2 Mathematical Methods because it sits inside almost every calculus question in Topics 1, 3 and 4.

The rule

If $y=f(u)$ and $u=g(x)$, then

Equivalently, for $y=f(g(x))$,

The intuition: differentiate the outside function treating the inside as a single block, then multiply by the derivative of the inside. The Leibniz form makes this look like cancelling fractions, which is a useful memory aid even though $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$ are not literally fractions.

Standard chain-rule results

These come up constantly and are worth knowing by sight:

• $\dfrac{d}{dx}\big[g(x)\big]^n=n\big[g(x)\big]^{n-1}\,g'(x)$
• $\dfrac{d}{dx}e^{g(x)}=g'(x)\,e^{g(x)}$
• $\dfrac{d}{dx}\ln\big(g(x)\big)=\dfrac{g'(x)}{g(x)}$
• $\dfrac{d}{dx}\sin\big(g(x)\big)=g'(x)\cos\big(g(x)\big)$
• $\dfrac{d}{dx}\cos\big(g(x)\big)=-g'(x)\sin\big(g(x)\big)$

Each of these is just the chain rule applied to a known derivative. If you can spot the inner function $g(x)$ and write down $g'(x)$, every one of these follows mechanically.

Combining with other rules

In practice the chain rule rarely appears alone. Watch for it inside products and quotients.

For $y=x\,e^{3x}$, use the product rule with $u=x$ and $v=e^{3x}$, where $v'=3e^{3x}$ comes from the chain rule:

For a quotient such as $y=\dfrac{\ln(2x)}{x}$, the numerator $\ln(2x)$ needs the chain rule to differentiate to $\dfrac{2}{2x}=\dfrac{1}{x}$, after which the quotient rule assembles the result. The chain rule is the inner gear that the product and quotient rules turn.

A nested chain

Sometimes there are three layers, for example $y=\ln\!\big(\cos(2x)\big)$. Work from the outside in: the derivative of $\ln(\square)$ is $\dfrac{1}{\square}$ times the derivative of $\square$; the derivative of $\cos(2x)$ is $-2\sin(2x)$. Multiplying gives

Each layer contributes one factor; you simply keep multiplying by the derivative of the next inner function until you reach $x$.

Common errors

Why it matters

The chain rule unlocks differentiation of exponential and logarithmic models, which dominate Topic 4, and it is essential for the optimisation problems where a quantity is expressed through an intermediate variable. Combined with the product and quotient rules, it lets you differentiate essentially any function you will meet in Stage 2, and it reappears in reverse as integration by substitution in the calculus you meet later. Examiners reward clear identification of the inner function and the explicit $g'(x)$ factor, so always show that step.