What are other bases?

For a > 0 and a ≠ 1, use the change of base a^x = e^x ln a, which gives

QLDMath MethodsSyllabus dot point

Topic 1: Further differentiation and applications

Differentiate exponential and logarithmic functions, including compositions of the form $e^{f(x)}$ and $\ln(f(x))$ , and apply the derivatives to model and analyse rates of change

A focused answer to the QCE Mathematical Methods Unit 3 dot point on differentiating exponential and logarithmic functions. Covers the derivatives of $e^x$ , $a^x$ , $\ln x$ and $\log_a x$ , the chain rule generalisations $e^{f(x)}$ and $\ln(f(x))$ , and the application to rates of change, with worked Paper 1 and Paper 2 examples.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

QCAA wants you to differentiate exponential and logarithmic functions, including compositions like $e^{f(x)}$ and $\ln(f(x))$ , and apply those derivatives to model rates of change in context. Exponential and logarithmic derivatives appear in Paper 1 (technology-free), in Paper 2 modelling questions, and in PSMTs that involve growth, decay, or any process where the rate is proportional to the current quantity.

The answer

The base case

The exponential function with base $e$ is its own derivative.

\frac{d}{dx}(e^x) = e^x

This is the defining property of $e$ and is the reason $e^x$ appears in every growth and decay model in Unit 3.

The natural logarithm is the inverse of $e^x$ and differentiates to the reciprocal.

\frac{d}{dx}(\ln x) = \frac{1}{x}, \quad x > 0

Chain-rule generalisations

For any differentiable inner function $f(x)$ :

\frac{d}{dx}\bigl(e^{f(x)}\bigr) = f'(x) e^{f(x)}

\frac{d}{dx}\bigl(\ln(f(x))\bigr) = \frac{f'(x)}{f(x)}, \quad f(x) > 0

These two formulas cover almost every Methods question. Memorise them in this form.

Other bases

For $a > 0$ and $a \neq 1$ , use the change of base $a^x = e^{x \ln a}$ , which gives

\frac{d}{dx}(a^x) = (\ln a) \, a^x.

Similarly, $\log_a x = \dfrac{\ln x}{\ln a}$ , so

\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}.

QCAA frequently rewards students who recognise that $a^x$ is not differentiated by the power rule. The power rule applies to $x^n$ (variable base, constant exponent), not $a^x$ (constant base, variable exponent).

Logarithm laws first

Before differentiating a complicated logarithm, simplify using logarithm laws. For example,

\ln\bigl(x^2 \sqrt{x + 1}\bigr) = 2 \ln x + \frac{1}{2} \ln(x + 1),

which differentiates to $\dfrac{2}{x} + \dfrac{1}{2(x + 1)}$ in one step, with no quotient rule needed.

Proportional rate of change

The reason $e^{kt}$ models growth and decay is that its derivative is proportional to itself: if $Q(t) = Q_0 e^{kt}$ then $Q'(t) = kQ_0 e^{kt} = kQ$ . So the rate of change at any instant is exactly $k$ times the current amount, which is the defining feature of exponential change. A positive $k$ gives growth (the rate rises as the quantity rises) and a negative $k$ gives decay (the rate of loss shrinks as the quantity shrinks). Recognising $Q' = kQ$ as the signature of an exponential model is a recurring QCAA theme and connects this dot point to differential equations.

Combining with the product and quotient rules

Exponential and logarithmic derivatives rarely appear alone; they are usually one factor in a product or quotient. For $y = x^2 e^{x}$ the product rule gives $y' = 2x e^x + x^2 e^x = xe^x(2 + x)$ , and for $y = \dfrac{\ln x}{x}$ the quotient rule gives $y' = \dfrac{(1/x)x - \ln x}{x^2} = \dfrac{1 - \ln x}{x^2}$ . The exponential or logarithm is differentiated with its own rule, and the surrounding structure dictates whether the product or quotient rule wraps around it. Factorising the result is often worth a final mark.

Worked example

Five differentiation tasks on exponential and logarithmic functions, each showing how to identify the inner function and apply the appropriate rule.

Step 1: Apply the chain rule to $y = e^{-2x}$

The function has the form $e^{f(x)}$ , so we use $\frac{d}{dx} e^{f(x)} = f'(x) e^{f(x)}$ . The inner function is $f(x) = -2x$ , which has derivative $f'(x) = -2$ .

\frac{dy}{dx} = -2 e^{-2x}

Step 2: Differentiate a logarithm of a polynomial, $y = \ln(x^2 + 4)$

The function has the form $\ln(f(x))$ , so we use $\frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}$ . The inner function is $f(x) = x^2 + 4$ , giving $f'(x) = 2x$ .

\frac{dy}{dx} = \frac{2x}{x^2 + 4}

Step 3: Handle a non- $e$ exponential, $y = 5^x$

The power rule does not apply here because the base is constant and the exponent is variable. We recognise $5^x = e^{x \ln 5}$ and apply the chain rule, bringing $\ln 5$ out as the constant factor.

\frac{dy}{dx} = (\ln 5) \cdot 5^x \approx 1.6094 \cdot 5^x

Step 4: Use logarithm laws before differentiating $y = \ln\!\left( \dfrac{x^3}{x^2 + 1} \right)$

Differentiating this quotient inside a logarithm directly would require the quotient rule inside the chain rule. It is much cleaner to split using $\ln(a/b) = \ln a - \ln b$ first.

y = 3 \ln x - \ln(x^2 + 1)

Now each term differentiates immediately.

\frac{dy}{dx} = \frac{3}{x} - \frac{2x}{x^2 + 1}

Step 5: Apply the derivative in a modelling context

A quantity decays according to $Q(t) = 200 e^{-0.05 t}$ milligrams, with $t$ in days. The chain rule gives the rate of change, where the inner function $-0.05t$ contributes a factor of $-0.05$ .

Q'(t) = 200 \times (-0.05) \, e^{-0.05 t} = -10 e^{-0.05 t}

Substituting $t = 10$ gives the instantaneous rate.

Q'(10) = -10 e^{-0.5} \approx -6.07 \text{ mg/day}

Final answer: the quantity is decreasing at approximately $6.07$ mg per day when $t = 10$ days. The negative sign confirms decay and its magnitude gives the current rate of loss.

Common traps

Treating $a^x$ like $x^n$: $\frac{d}{dx}(3^x) \neq x \cdot 3^{x-1}$ . The correct answer is $(\ln 3) \cdot 3^x$ .
Forgetting the chain rule factor: $\frac{d}{dx}(e^{2x}) = 2 e^{2x}$ , not $e^{2x}$ . The factor of $2$ comes from $f'(x) = 2$ .
Missing the domain on $\ln x$: The natural log is only defined for $x > 0$ . If a domain crosses zero, you may need $\ln|x|$ instead. In Paper 1, state the domain when QCAA asks for it.
Skipping logarithm laws: Differentiating $\ln(uv)$ directly with the chain rule still works but is slower and more error-prone than splitting first to $\ln u + \ln v$ .
Forgetting the sign in decay: A decay model $Q(t) = A e^{-kt}$ with $k > 0$ has derivative $-k A e^{-kt}$ , which is negative. The negative sign carries through and is part of the answer.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

QCAA 20233 marksPaper 1 (technique). Differentiate

y = e^{3x^2 + 2}

with respect to

x

Show worked answer →

This is a composition, so use the chain rule generalisation $\frac{d}{dx} e^{f(x)} = f'(x) e^{f(x)}$ .

Let $f(x) = 3x^2 + 2$ , so $f'(x) = 6x$ .

$\frac{dy}{dx} = 6x \cdot e^{3x^2 + 2}.$

Markers reward explicit identification of $f(x)$ and $f'(x)$ , application of the rule, and a tidy final answer with the exponential factor on the right.

QCAA 20224 marksPaper 2 (complex familiar). A population of bacteria is modelled by

N(t) = 500 e^{0.15 t}

where

t

is hours after measurement. (a) Determine the rate of change of the population at

t = 6

hours. (b) Interpret this rate in context.

Show worked answer →

A 4-mark answer needs the derivative, the substituted value, and a contextual interpretation.

(a) $N'(t) = 500 \cdot 0.15 \cdot e^{0.15 t} = 75 e^{0.15 t}$ .

At $t = 6$ : $N'(6) = 75 e^{0.9} \approx 75 \times 2.4596 \approx 184.5$ bacteria per hour.

(b) Six hours after measurement, the population is growing at about 184 bacteria per hour. Because the model is exponential, the rate of growth itself increases with time; this rate would be roughly 220 per hour an hour later.

Markers reward the chain-rule derivative, an accurate numerical value (3 significant figures is fine on Paper 2), and a sentence interpreting the rate in context with correct units.