WAMath MethodsSyllabus dot point

How do we differentiate, integrate and apply exponential and logarithmic functions?

Differentiate and integrate exponential and natural logarithm functions and apply them to growth, decay and other modelling contexts

WACE Year 12 Mathematics Methods Unit 3 exponential and logarithmic functions: derivatives and integrals of e^x and ln x, the chain rule with exponentials, and growth and decay modelling with worked SCSA-style examples.

Generated by Claude Opus 4.79 min answerUpdated 2026-05-29

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

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

Jump to a section

What this dot point is asking
Derivatives
Integrals
Growth and decay modelling
Logarithm laws for solving

What this dot point is asking

SCSA Unit 3 develops calculus of the natural exponential function $e^{x}$ and the natural logarithm $\ln x$ . The defining feature of $e^{x}$ is that it is its own derivative, which makes it the natural model for any quantity whose rate of change is proportional to its current size. This dot point appears in both examination sections.

Derivatives

Worked derivative

For $y = e^{2x}\sin x$ apply the product rule with the chain rule on $e^{2x}$ :

\frac{dy}{dx} = 2e^{2x}\sin x + e^{2x}\cos x = e^{2x}(2\sin x + \cos x).

Integrals

Antidifferentiation reverses the derivatives above.

The absolute value in $\ln|x|$ matters because $1/x$ is defined for negative $x$ as well, but $\ln x$ is not.

Growth and decay modelling

A quantity whose rate of change is proportional to its size satisfies $\dfrac{dN}{dt} = kN$ , and the solution is the exponential model

N(t) = N_0 e^{kt},

where $N_0$ is the initial amount, $k>0$ gives growth and $k<0$ gives decay. Logarithms are used to solve for the time $t$ at which a target value is reached.

Radioactive decay

A sample decays with $N(t) = 80 e^{-0.1t}$ grams. When does it reach $20$ g?

Set $N(t) = 20$ .

80 e^{-0.1t} = 20 \;\Rightarrow\; e^{-0.1t} = \frac{20}{80} = 0.25.

Take the natural logarithm of both sides.

-0.1t = \ln(0.25) \;\Rightarrow\; t = \frac{\ln(0.25)}{-0.1} = \frac{-1.3863}{-0.1} \approx 13.9.

The sample reaches $20$ g after about $13.9$ time units. The decay rate at any time is $N'(t) = -0.1 \times 80 e^{-0.1t} = -8e^{-0.1t}$ , which is itself proportional to $N(t)$ .

Logarithm laws for solving

When solving exponential equations you will use:

\ln(ab) = \ln a + \ln b, \quad \ln\!\left(\frac{a}{b}\right) = \ln a - \ln b, \quad \ln(a^{k}) = k\ln a.

These let you bring an unknown out of an exponent before solving.