Apply the product, quotient and chain rules to differentiate composite functions involving exponential, logarithmic, polynomial and trigonometric pieces, including logarithmic differentiation and the differentiation of inverse functions

What this dot point is asking

QCAA wants you to extend the Unit 3 differentiation toolkit to handle composite functions involving exponentials, logarithms, polynomials, trigonometric functions, and their inverses. Logarithmic differentiation and the inverse-function rule are the new techniques. The dot point appears in PSMT modelling contexts and in the External Assessment short-response.

The Unit 3 toolkit (review)

From Unit 3 you have:

• Power rule. $\frac{d}{dx}(x^n) = n x^{n-1}$.
• Exponential. $\frac{d}{dx}(e^{kx}) = k e^{kx}$.
• Logarithmic. $\frac{d}{dx}(\ln x) = \frac{1}{x}$ for $x > 0$.
• Sum rule, product rule, quotient rule, chain rule.

Unit 4 builds on these for more complex composites.

Logarithmic differentiation

When the function has the form $y = f(x)^{g(x)}$ (variable base, variable exponent), or is a product / quotient of powers, the standard rules cannot be applied directly. Logarithmic differentiation is the technique.

Procedure

1. Take the natural logarithm of both sides. $\ln y = \ln(f(x))$.
2. Simplify using log rules. $\ln(a b) = \ln a + \ln b$; $\ln(a/b) = \ln a - \ln b$; $\ln(a^k) = k \ln a$.
3. Differentiate both sides with respect to $x$. Use the chain rule on the left ($\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}$) and the product / quotient rules on the right.
4. **Multiply both sides by $y$** and substitute $y$ back.

Worked example

$y = x^x$ for $x > 0$.

Take ln. $\ln y = x \ln x$.

Differentiate. $\frac{1}{y} \frac{dy}{dx} = \ln x + 1$ (product rule on $x \ln x$).

Multiply. $\frac{dy}{dx} = y (\ln x + 1) = x^x (\ln x + 1)$.

When to use it

Logarithmic differentiation is the right tool when:

• The function has variable in both base and exponent.
• The function is a product / quotient of multiple factors raised to powers (it splits into a sum / difference of logs).
• The function is a nasty quotient with multiple roots in the numerator and denominator.

For ordinary polynomial or single-factor functions, the regular rules are faster.

Derivatives of inverse functions

If $y = f^{-1}(x)$, then $f(y) = x$. Differentiating both sides with respect to $x$:

In words: the derivative of the inverse function at $x$ is the reciprocal of the derivative of the original function evaluated at $y = f^{-1}(x)$.

Worked example. Inverse of IMATH_26

$f(x) = x^3 + x$. Find the derivative of $f^{-1}$ at $x = 2$.

$f(1) = 1 + 1 = 2$, so $f^{-1}(2) = 1$.

$f'(x) = 3 x^2 + 1$. $f'(1) = 4$.

$(f^{-1})'(2) = \frac{1}{f'(1)} = \frac{1}{4}$.

This technique is useful when you can compute $f^{-1}(x)$ at a specific point but the inverse function does not have a closed form.

Derivatives of inverse trig functions (boundary topic)

QCAA Methods Unit 4 may or may not include derivatives of $\arcsin$, $\arccos$, $\arctan$ depending on syllabus revision. The current QCAA Mathematical Methods General Senior Syllabus excludes these (they belong to Specialist Mathematics). Check your school's interpretation.

If included, the standard derivatives are:

Higher derivatives and applications

The second derivative $f''(x) = \frac{d^2 y}{dx^2}$ is the derivative of the derivative. Used to:

• Classify stationary points (concavity): if $f'(a) = 0$ and $f''(a) > 0$, $x = a$ is a local minimum; if $f''(a) < 0$, a local maximum.
• Identify points of inflection. If $f''(c) = 0$ and the sign of $f''$ changes across $x = c$, then $x = c$ is a point of inflection.

The third and higher derivatives are rarely required in QCE Methods.

Worked applications

Optimisation with logarithmic differentiation

The revenue from selling $x$ items is $R(x) = x \cdot p(x)$ where $p(x) = 100 - x^{0.5}$. Find $x$ that maximises $R$.

$R(x) = x (100 - x^{0.5}) = 100 x - x^{1.5}$.

$R'(x) = 100 - 1.5 x^{0.5}$.

Set $R'(x) = 0$: $x^{0.5} = 100 / 1.5 = 66.67$. $x = 4444.4$.

$R''(x) = -0.75 x^{-0.5} < 0$ at this $x$, confirming a maximum.

The student does not need logarithmic differentiation here; standard chain rule suffices. Logarithmic differentiation is needed when the variable appears as an exponent or in nested products and powers.

Implicit derivative for shape constraints

If a curve is defined implicitly by $x^2 + y^2 + xy = 10$, find $dy/dx$ at the point $(1, 2)$ (which satisfies the equation: $1 + 4 + 2 = 7$ - let me re-check: $(1)^2 + (2)^2 + (1)(2) = 1 + 4 + 2 = 7 \neq 10$. Use $(2, 1)$: $4 + 1 + 2 = 7$, still not. Skip the worked specific point; use general approach).

Differentiate both sides with respect to $x$, treating $y = y(x)$:

$2x + 2y \frac{dy}{dx} + \left(y + x \frac{dy}{dx}\right) = 0$.

Collect $\frac{dy}{dx}$ terms: $(2y + x) \frac{dy}{dx} = -(2x + y)$.

$\frac{dy}{dx} = -\frac{2x + y}{2y + x}$.

Implicit differentiation is treated in the related-rates dot point separately.

Common errors

Forgetting the chain rule inside. Differentiating $\ln(x^2 + 1)$ requires the chain rule: $\frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}$.

Logarithmic differentiation without final $y$ substitution. After computing $\frac{1}{y} \frac{dy}{dx}$, you must multiply by $y$ and replace $y$ with the original expression.

Confusing $\frac{d}{dx}(\ln x)$ and $\frac{d}{dx}(\log_{10} x)$. The natural log has derivative $1/x$. The base-10 log has derivative $1 / (x \ln 10)$. QCAA Methods uses natural log throughout.

Power rule applied to variable exponent. $\frac{d}{dx}(x^x)$ is NOT $x \cdot x^{x-1} = x^x$. The power rule applies only to constant exponents. Use logarithmic differentiation.

Sign error in inverse function rule. The derivative of $f^{-1}$ is $1 / f'(y)$, not $1 / f'(x)$. The $y$ is the value of the inverse function at $x$.

In one sentence

Further differentiation in QCE Methods Unit 4 extends the Unit 3 toolkit with logarithmic differentiation (for variable exponents and complex products / quotients via $\ln y$ first), the inverse-function rule ($\frac{d}{dx} f^{-1}(x) = 1 / f'(y)$), and applications including optimisation and higher-derivative classification of stationary points.