Further differentiation and applications: QCE Maths Methods Unit 4

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 $f(x) = x^3 + x$

$f(x) = x^3 + x$. 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 the power rule

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.

You do not need logarithmic differentiation here; the power rule is enough. 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 $x^2 + y^2 + xy = 7$, find $dy/dx$ at the point $(1, 2)$. The point is on the curve because $1^2 + 2^2 + (1)(2) = 1 + 4 + 2 = 7$.

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

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

At $(1, 2)$: $\frac{dy}{dx} = -\frac{2(1) + 2}{2(2) + 1} = -\frac{4}{5}$.

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