Year 12 / Exponential and logarithmic calculus

A rate proportional to itself.

The natural exponential $e^{x}$ has the distinctive property that its derivative is itself. For a composite exponent $g(x)$ , the chain rule adds a multiplier $g'(x)$ . Thus $\frac{\mathrm d}{\mathrm dx}e^{3x}$ = $3e^{3x}$ .

In $N(t)$ = $N_0e^{kt}$ , the derivative is kN(t). Positive k models growth and negative k models decay. The continuous rate k differs from a percentage multiplier over a whole time period: the one-period multiplier is $e^{k}$ .

INTERACTIVE MODEL

Try it. Watch it change.

f(x)

e^{x}

tangent

f'(x)

FunctionTangent point x1

f′(1) = 2.7183 · f″(1) = 2.7183

The graph is calculated from the displayed rule. Drag the highlighted point or use the sliders.

Explore: Select $e^{x}$ and move the tangent point. Compare the function value with its slope.

Chain rule for exponentials

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

WORKED EXAMPLE

Find the instantaneous growth rate of $N(t)$ = $200e^{0.04t}$ at t = 0.

$N'(t)$ = 200(0.04)e^(0.04t) = $8e^{0.04t}$ .
$N'(0)$ = 8 units per unit time.

Assumed knowledge

Chain rule, exponential models and units of rates.

Learning checkpoints & source

YOUR LEARNING CHECKPOINT

Differentiate natural exponential functions.
Interpret exponential growth and decay rates.

QCAA Mathematical Methods 2025 v1.3 · p. 24 ↗

READY TO TRY IT?

A rate proportional to itself.

Find the instantaneous growth rate of N(t)N(t)N(t) = 200e0.04t200e^{0.04t}200e0.04t at t = 0.

Put the idea to work.

Find the instantaneous growth rate of $N(t)$ = $200e^{0.04t}$ at t = 0.