QCE Mathematical Methods - Unit 3 - Differentiation of exponential and logarithmic functions

Calculus of Exponential Functions | QCE Mathematical Methods

Learn QCE Mathematical Methods exponential calculus, including why $e$ matters, how to differentiate $e^x$ and how to read exponential growth rates.

Updated 2026-05-18 - 4 min read

QCAA official coverage - Mathematical Methods 2025 v1.3

Exact syllabus points covered

  1. Estimate the limit $\frac{a^h - 1}{h}$ as $h \to 0$, using technology, for various values of $a > 0$.
  2. Recognise that $e$ is the unique number $a$ for which the above limit is $1$.
  3. Recognise and determine the qualitative features of the graph of $y=e^x$, including asymptote and intercept.
  4. Use the rules $\frac{d}{dx}e^x=e^x$ and $\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$.

Exponential functions are used when a quantity changes by a constant percentage or factor rather than by a constant amount. In Methods, the special exponential function $e^x$ matters because its rate of change is itself. That is the cleanest growth model you will meet: the height of the graph and the gradient of the graph are the same number at each $x$ value.

Exponential and logarithmic inverse relationship

Original Sylligence diagram for exponential log inverse.

Exponential and logarithmic inverse relationship

Why e is special

For a general exponential function $a^x$, the first-principles derivative leads to the limit:

$ \frac{a^h-1}{h}\quad\text{as }h\to 0 $

Different bases give different limiting gradients. The number $e$ is the base where that limit is exactly $1$, so:

$ \frac{d}{dx}e^x=e^x $

That is not just a formula to memorise. It means the instantaneous growth rate of $e^x$ is equal to the current value of $e^x$.

The first-principles calculation is worth understanding even if you rarely do it by hand. For:

$ f(x)=a^x, $

the derivative definition gives:

$ f'(x)=\lim_{h\to0}\frac{a^{x+h}-a^x}{h}. $

Factor out $a^x$:

$ f'(x)=a^x\lim_{h\to0}\frac{a^h-1}{h}. $

That limit is a constant depending only on the base $a$. Technology can estimate it for different bases. The special base $e$ is the one that makes the limit equal $1$, so the derivative has no extra multiplier.

For a base other than $e$, the derivative is:

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

This comes from rewriting:

$ a^x=e^{x\ln a}. $

Then the chain rule gives:

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

So all exponential functions have exponential derivatives; $e^x$ is just the one where the coefficient is exactly $1$.

The graph of $y=e^x$

The graph has these features:

  • domain: all real numbers
  • range: $y>0$
  • horizontal asymptote: $y=0$
  • $y$-intercept: $(0,1)$
  • always increasing
  • gradient is always positive

For sketching, remember that $e^0=1$, $e^1=e$, and $e^{-1}=\frac{1}{e}$. The curve gets close to the $x$-axis on the left but never crosses it.

Transformations behave like other functions. In:

$ y=Ae^{k(x-h)}+c, $

$c$ moves the horizontal asymptote to $y=c$. The sign of $A$ decides whether the curve sits above or below that asymptote, and the sign of $k$ decides whether the curve grows or decays as $x$ increases. This is useful in modelling because the asymptote often represents a limiting value rather than just a graph feature.

Differentiating e to a function

Most exam questions do not stop at $e^x$. They use a function in the exponent:

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

This is the chain rule. Differentiate the exponent, then multiply by the original exponential expression.

Worked example

Interpreting exponential derivatives

If $P(t)=1200e^{0.08t}$ models a population after $t$ years, then:

$ P'(t)=1200(0.08)e^{0.08t}=96e^{0.08t} $

This derivative is the instantaneous rate of growth, in people per year. Notice that $P'(t)=0.08P(t)$. The rate is $8\%$ of the current population, so the growth rate increases as the population increases.

For decay, the exponent coefficient is negative. If:

$ M(t)=500e^{-0.12t}, $

then:

$ M'(t)=-60e^{-0.12t}=-0.12M(t). $

The negative sign says the amount is decreasing. The magnitude of the rate is still proportional to the current amount, so the loss slows down as the quantity becomes smaller.

Common traps

Other traps:

  • treating $e$ as a variable instead of a constant
  • forgetting the chain rule multiplier
  • using degrees/radians settings for an exponential-only question, which is irrelevant
  • writing a final derivative without simplifying constants

Quick check

Sources