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

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.

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.

Chain-rule generalisations

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

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

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

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,

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

Worked examples

Direct chain rule

Differentiate $y = e^{-2x}$.

Let $f(x) = -2x$, so $f'(x) = -2$. Then $\dfrac{dy}{dx} = -2 e^{-2x}$.

Logarithm of a polynomial

Differentiate $y = \ln(x^2 + 4)$.

$\dfrac{dy}{dx} = \dfrac{2x}{x^2 + 4}$.

Non-$e$ exponential

Differentiate $y = 5^x$.

$\dfrac{dy}{dx} = (\ln 5) \cdot 5^x \approx 1.6094 \cdot 5^x$.

Logarithm laws save work

Differentiate $y = \ln\!\left( \dfrac{x^3}{x^2 + 1} \right)$.

Rewrite first: $y = 3 \ln x - \ln(x^2 + 1)$.

$\dfrac{dy}{dx} = \dfrac{3}{x} - \dfrac{2x}{x^2 + 1}$.

Modelling context

A quantity decays according to $Q(t) = 200 e^{-0.05 t}$ milligrams, with $t$ in days. The rate of decay at $t = 10$ is

$Q'(t) = -10 e^{-0.05 t}, \quad Q'(10) = -10 e^{-0.5} \approx -6.07 \text{ mg/day}.$

The negative sign confirms decay; the magnitude is 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.

In one sentence

The exponential function $e^x$ is its own derivative and the natural log differentiates to $\frac{1}{x}$, and applying the chain rule turns these into $\frac{d}{dx} e^{f(x)} = f'(x) e^{f(x)}$ and $\frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}$, with non-$e$ bases handled by writing $a^x = e^{x \ln a}$.