Derivatives of exponential and logarithmic functions (VCE Maths Methods Unit 2)

Syllabus dot point

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

7 min readUpdated 2 Jun 2026

Quick answer

Key results: $\frac{d}{dx} e^x = e^x$ , $\frac{d}{dx} a^x = a^x \ln a$ , $\frac{d}{dx} \ln x = 1/x$ , $\frac{d}{dx} \log_b x = 1/(x\ln b)$ ; for composites the chain rule gives $\frac{d}{dx} e^u = u'e^u$ and $\frac{d}{dx} \ln u = u'/u$ .

Jump to a section

What this dot point is asking
Standard derivatives
Chain rule extensions
Applications
Combining rules: product, quotient and chain
Logarithmic functions and their domain
In one sentence
Examples in context
Try this
Exam-style questions

What this dot point is asking

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

Standard derivatives

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

\frac{d}{dx} a^x = a^x \ln a

\frac{d}{dx} \ln x = \frac{1}{x} \quad (x > 0)

\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

\frac{d}{dx} e^{u(x)} = u'(x) \cdot e^{u(x)}

\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.

Combining rules: product, quotient and chain

Most exam derivatives mix an exponential or logarithm with another function, so you must choose the right combination of rules. A product such as $x^2 e^{2x}$ needs the product rule; a quotient such as $\dfrac{e^x}{x}$ needs the quotient rule; a composite such as $e^{3x^2 - 2}$ needs the chain rule. When two rules are needed (for example $x e^{\sin x}$ ), apply the product rule first, then the chain rule on the inner composite. A useful habit is to factor the common exponential out of the final answer, since $e^{u}$ is never zero and factoring makes solving $f'(x) = 0$ for stationary points immediate.

Logarithmic functions and their domain

The derivative $\dfrac{d}{dx}\ln x = \dfrac{1}{x}$ holds for $x > 0$ , matching the domain of $\ln x$ . Using $\ln|x|$ extends the rule to $x \ne 0$ with the same derivative $\dfrac{1}{x}$ . For a composite $\ln(u(x))$ , the chain rule gives $\dfrac{u'(x)}{u(x)}$ , and the result is only valid where $u(x) > 0$ . VCAA expects you to state the implied domain when differentiating logarithmic functions, because the gradient function inherits the restriction.

Differentiating a product with an exponential

Step 1: Identify the rule needed

For $h(x) = x^2 e^{2x}$ , the function is the product of two separate parts: $u = x^2$ (a polynomial) and $v = e^{2x}$ (an exponential composite). Because neither factor can be absorbed into the other, we must use the product rule, which states that the derivative of a product is the first factor times the derivative of the second, plus the second factor times the derivative of the first:

h'(x) = u'v + uv'

Step 2: Differentiate each factor separately

We differentiate $u$ and $v$ in turn before combining them. For $u = x^2$ , the power rule gives $u' = 2x$ . For $v = e^{2x}$ , the argument $2x$ is itself a function of $x$ , so the chain rule applies: the derivative of $e^{f(x)}$ is $f'(x) \cdot e^{f(x)}$ , and the inner derivative of $2x$ is $2$ :

u' = 2x, \qquad v' = 2e^{2x}

Step 3: Combine using the product rule

Substituting both derivatives and both original factors into $h' = u'v + uv'$ :

h'(x) = 2x \cdot e^{2x} + x^2 \cdot 2e^{2x} = 2xe^{2x} + 2x^2 e^{2x}

Step 4: Factor the result

Both terms share the factor $2xe^{2x}$ . Factoring this out simplifies the expression and immediately reveals where the derivative equals zero. Since $e^{2x}$ is never zero, the stationary points come only from the polynomial factor:

h'(x) = 2xe^{2x}(1 + x)

The stationary points are at $x = 0$ and $x = -1$ .

Final answer: $h'(x) = 2xe^{2x}(1 + x)$ .

Common traps

Treating $e^{u(x)}$ as $e^x \cdot u(x)$: It is a composite function, so use the chain rule, not the product rule.
Forgetting the chain factor on $\ln u$: $\dfrac{d}{dx}\ln(3x) = \dfrac{3}{3x} = \dfrac{1}{x}$ ; the inner derivative must appear in the numerator first.
Confusing $a^x$ and $x^a$: $a^x$ is exponential with derivative $a^x \ln a$ ; $x^a$ is a power with derivative $a x^{a-1}$ .
Dropping the domain on logarithmic derivatives: State $x > 0$ (or $u(x) > 0$ ) where it applies.

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 $\frac{d}{dx} \ln u = u'/u$ .

Examples in context

Example 1. Rate of cooling. A cup of coffee cools according to $T(t) = 20 + 60e^{-0.05t}$ degrees Celsius, $t$ minutes after pouring. The rate of cooling is $T'(t) = 60 \times (-0.05) e^{-0.05t} = -3e^{-0.05t}$ degrees per minute. At $t = 10$ : $T'(10) = -3e^{-0.5} = -3 \times 0.6065 = -1.82$ degrees per minute (still cooling, but slowing).

Example 2. Marginal log-utility. An economic model uses $U(x) = 40\ln(x + 1)$ for the satisfaction from $x$ units of a good. The marginal utility is $U'(x) = \frac{40}{x + 1}$ . At $x = 3$ , $U'(3) = \frac{40}{4} = 10$ units of satisfaction per extra good; at $x = 9$ , $U'(9) = \frac{40}{10} = 4$ , illustrating diminishing returns as the chain factor shrinks the rate.

Try this

Q1. Differentiate $f(x) = e^{4x}$ and $g(x) = \ln(2x + 5)$ . [2+2 marks]

Cue. $f'(x) = 4e^{4x}$ ; $g'(x) = \frac{2}{2x + 5}$ .

Q2. Find the gradient of $y = xe^{x}$ at $x = 1$ . [3 marks]

Cue. Product rule: $y' = e^x + xe^x = e^x(1 + x)$ ; at $x = 1$ , $y' = 2e$ .

Q3. A radioactive sample has mass $M = 50e^{-0.1t}$ grams. (a) Find $\frac{dM}{dt}$ . (b) Find the rate of decay at $t = 5$ . [2+2 marks]

Cue. (a) $\frac{dM}{dt} = -5e^{-0.1t}$ . (b) At $t = 5$ , $-5e^{-0.5} = -3.03$ g per unit time.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

Differentiate (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 $\ln u$ .

Practice5 marks

Let

f(x) = x^2 e^{-x}

for

x \ge 0

. (a) Find

f'(x)

. (b) Determine the coordinates of any stationary points and classify them. (c) State the maximum value of

f

[0, \infty)

Show worked answer →

(a) Product rule with $u = x^2$ , $v = e^{-x}$ : $f'(x) = 2x e^{-x} + x^2(-e^{-x}) = e^{-x}(2x - x^2) = x(2 - x)e^{-x}$ .

(b) Stationary points where $f'(x) = 0$ . Since $e^{-x} > 0$ , set $x(2 - x) = 0$ , giving $x = 0$ and $x = 2$ . At $x = 0$ , $f(0) = 0$ ; at $x = 2$ , $f(2) = 4e^{-2} \approx 0.541$ .

Sign of $f'$ : for $0 < x < 2$ , $x(2 - x) > 0$ so $f$ increases; for $x > 2$ , $x(2 - x) < 0$ so $f$ decreases. Hence $(2, 4e^{-2})$ is a local maximum and $(0, 0)$ is a minimum at the endpoint.

(c) The maximum value on $[0, \infty)$ is $f(2) = 4e^{-2} \approx 0.541$ .

Markers reward the product rule, factoring out $e^{-x}$ , locating both stationary points, and classifying via the sign of $f'$ .

Derivatives of exponential and logarithmic functions (VCE Maths Methods Unit 2)

What this dot point is asking

Standard derivatives

Chain rule extensions

Applications

Combining rules: product, quotient and chain

Logarithmic functions and their domain

Step 1: Identify the rule needed

Step 2: Differentiate each factor separately

Step 3: Combine using the product rule

Step 4: Factor the result

In one sentence

Examples in context

Try this

Exam-style questions

Practise this

Sources & how we know this