Year 11 SAC HSC: Differentiate (a) $f(x) = e^{3x^2 - 2}$ and (b) $g(x) = \ln(5x + 1)$.

Q: Year 11 SAC HSC: Differentiate (a) $f(x) = e^{3x^2 - 2}$ and (b) $g(x) = \ln(5x + 1)$.

**(a) Chain rule.** Let $u = 3x^2 - 2$. $du/dx = 6x$. $\frac{d}{dx} e^u = e^u \cdot du/dx$. $f'(x) = 6x \cdot e^{3x^2 - 2}$. **(b) Chain rule on $\ln$.** $\frac{d}{dx} \ln u = (1/u) \cdot du/dx$. $g'(x) = \dfrac{5}{5x + 1}$. Markers reward chain-rule structure, $e^u$ unchanged on differentiation, and the $1/u$ form for $\ln u$.

← Unit 2: Functions, calculus and probability

VICMath MethodsSyllabus dot point

How are exponential and logarithmic functions differentiated?

Differentiate exponential ($e^x$, $a^x$) and logarithmic ($\ln x$, $\log_b x$) functions, including composite functions via the chain rule

A focused answer to the VCE Maths Methods Unit 2 dot point on derivatives of exponential and logarithmic functions. States $\frac{d}{dx} e^x = e^x$, $\frac{d}{dx} \ln x = 1/x$, the chain-rule extensions, and works the VCAA SAC-style continuous-decay derivative problem.

Generated by Claude OpusReviewed by Better Tuition Academy4 min answerUpdated 2026-05-19

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

What this dot point is asking

VCAA wants you to differentiate exponential and logarithmic functions, including composite functions via the chain rule.

Standard derivatives

DMATH_0
DMATH_1
DMATH_2

\frac{d}{dx} \log_b x = \frac{1}{x \ln b}

$e^x$ is the unique function (up to scaling) whose derivative equals itself.

Chain rule extensions

DMATH_4

\frac{d}{dx} \ln |u(x)| = \frac{u'(x)}{u(x)}

The absolute value extends the natural log to negative arguments (the derivative is the same form on both sides of zero).

Applications

For continuous decay $N = N_0 e^{-\lambda t}$ :

\frac{dN}{dt} = -\lambda N_0 e^{-\lambda t} = -\lambda N

The decay rate is proportional to the current amount; this is the differential-equation definition of exponential decay.

For population growth $P = P_0 e^{kt}$ :

\frac{dP}{dt} = k P

The growth rate is proportional to current population.

Worked example

Differentiate $h(x) = x^2 e^{2x}$ .

Product rule: $h'(x) = 2x \cdot e^{2x} + x^2 \cdot 2 e^{2x} = 2x e^{2x}(1 + x)$ .

Common traps

Treating $e^{u(x)}$ as $e^x \cdot u(x)$ . It is a composite function, not a product.

Forgetting the chain factor on $\ln u$ . $\frac{d}{dx} \ln(3x) = \dfrac{3}{3x} = \dfrac{1}{x}$ , not $\dfrac{1}{3x}$ .

Confusing $a^x$ and $x^a$ . $a^x$ is an exponential; its derivative is $a^x \ln a$ . $x^a$ is a power; its derivative is $a x^{a-1}$ .

In one sentence

$\frac{d}{dx} e^x = e^x$ , $\frac{d}{dx} \ln x = 1/x$ , $\frac{d}{dx} a^x = a^x \ln a$ , $\frac{d}{dx} \log_b x = 1/(x \ln b)$ , with chain-rule extensions $\frac{d}{dx} e^u = u' e^u$ and $\frac{d}{dx} \ln u = u'/u$ .

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

Year 11 SAC4 marksDifferentiate (a) $f(x) = e^{3x^2 - 2}$ and (b) $g(x) = \ln(5x + 1)$.

Show worked answer →

(a) Chain rule. Let $u = 3x^2 - 2$ . $du/dx = 6x$ . $\frac{d}{dx} e^u = e^u \cdot du/dx$ .

$f'(x) = 6x \cdot e^{3x^2 - 2}$ .

(b) Chain rule on $\ln$ . $\frac{d}{dx} \ln u = (1/u) \cdot du/dx$ .

$g'(x) = \dfrac{5}{5x + 1}$ .

Markers reward chain-rule structure, $e^u$ unchanged on differentiation, and the $1/u$ form for $\ln u$ .