Derivatives of exponential and log functions (SACE Stage 2 Mathematical Methods)

Syllabus dot point

“The derivative of e^x is itself, the derivative of ln x is 1/x, and the chain rule extends both to composite functions.”

8 min readUpdated 2 Jun 2026

Quick answer

$\dfrac{d}{dx}e^x=e^x$ and $\dfrac{d}{dx}\ln x=\dfrac{1}{x}$ . With the chain rule these extend to $\dfrac{d}{dx}e^{f(x)}=f'(x)\,e^{f(x)}$ and $\dfrac{d}{dx}\ln\big(f(x)\big)=\dfrac{f'(x)}{f(x)}$ .

Jump to a section

What this dot point is asking
The base results
Chain-rule extensions
Combining with the product and quotient rules
Using the log laws to simplify first
Common errors
Tangents to exponential and log curves
Rates of change in models
Why it matters
Exam-style questions

What this dot point is asking

The exponential function $e^x$ is special: it is its own derivative. The natural logarithm $\ln x$ differentiates to $\tfrac1x$ . These two results, combined with the chain rule, let you differentiate every exponential and logarithmic model in Stage 2. They are also the reason the constant $e$ matters at all: $e$ is precisely the base for which the exponential curve has gradient exactly equal to its height at every point, which is why calculus is built around base $e$ rather than base $10$ or base $2$ .

The base results

Derivatives of e^x and ln x

\frac{d}{dx}e^x=e^x, \qquad \frac{d}{dx}\ln x=\frac{1}{x}\ (x>0).

Chain-rule extensions

When the exponent or the argument is itself a function of $x$ , multiply by its derivative:

Composite exponential and log derivatives

\frac{d}{dx}e^{f(x)}=f'(x)\,e^{f(x)}, \qquad \frac{d}{dx}\ln\big(f(x)\big)=\frac{f'(x)}{f(x)}.

Differentiating an exponential model

A population is modelled by $P(t)=500\,e^{0.08t}$ . Find the rate of growth at $t=10$ .

Step 1: Identify the composite structure

The function is of the form $A\,e^{f(t)}$ where the inner function is $f(t)=0.08t$ . The chain rule for exponentials says the derivative is $f'(t)$ times the original exponential, so we need $f'(t)$ first.

Differentiating the inner function gives $f'(t)=0.08$ .

Step 2: Apply the composite exponential rule

The rule $\dfrac{d}{dt}\bigl(A\,e^{f(t)}\bigr)=A\cdot f'(t)\,e^{f(t)}$ gives:

P'(t)=500\times 0.08\,e^{0.08t}=40\,e^{0.08t}.

The factor $40$ comes from multiplying the constant $500$ by the inner derivative $0.08$ . The exponential part $e^{0.08t}$ stays unchanged.

Step 3: Substitute $t=10$

Now substitute the given time to find the instantaneous rate:

P'(10)=40\,e^{0.8}\approx 40\times 2.2255\approx 89.0.

Final answer: The population is growing at approximately $89$ per unit time at $t=10$ .

Differentiating a logarithm

Differentiate $y=\ln(3x^2+1)$ .

Step 1: Identify the inner function

The function is a composite: the outer function is $\ln(\cdot)$ and the inner function is $f(x)=3x^2+1$ . The chain rule for logarithms states $\dfrac{d}{dx}\ln\bigl(f(x)\bigr)=\dfrac{f'(x)}{f(x)}$ , so the key task is finding $f'(x)$ .

Differentiating $f(x)=3x^2+1$ gives $f'(x)=6x$ .

Step 2: Apply the logarithm chain rule

Place the derivative of the inner function in the numerator and the inner function itself in the denominator:

\frac{dy}{dx}=\frac{f'(x)}{f(x)}=\frac{6x}{3x^2+1}.

The denominator $3x^2+1$ is the original argument of the $\ln$ ; it does not simplify further.

Final answer: $\dfrac{dy}{dx}=\dfrac{6x}{3x^2+1}$ .

Combining with the product and quotient rules

These derivatives frequently sit inside products and quotients.

A product needing the chain rule inside

Differentiate $y=x^2 e^{x}$ and find its stationary points.

Step 1: Recognise the product structure

The function is a product of two functions of $x$ : set $u=x^2$ and $v=e^x$ . Because neither factor is a constant, the product rule $(uv)'=u'v+uv'$ is required. Compute the individual derivatives: $u'=2x$ and $v'=e^x$ (since $e^x$ differentiates to itself).

Step 2: Apply the product rule

Substitute into $(uv)'=u'v+uv'$ :

\frac{dy}{dx}=2x\,e^x+x^2 e^x.

Step 3: Factorise the derivative

Both terms share the factors $e^x$ and $x$ , so factor these out to make setting the derivative to zero straightforward:

\frac{dy}{dx}=e^x\bigl(2x+x^2\bigr)=x\,e^x(x+2).

Step 4: Find the stationary points

Stationary points occur where $\dfrac{dy}{dx}=0$ . The factor $e^x$ is always positive (it is never zero for any real $x$ ), so the equation reduces to $x(x+2)=0$ . This gives $x=0$ and $x=-2$ .

Final answer: The derivative is $\dfrac{dy}{dx}=x\,e^x(x+2)$ and the stationary points are at $x=0$ and $x=-2$ .

Using the log laws to simplify first

Before differentiating a complicated log, expand it with the log laws - this is far easier than the quotient or product rule.

For $y=\ln\!\left(\dfrac{x^3}{2x+1}\right)$ , rewrite as $y=3\ln x-\ln(2x+1)$ , then

\frac{dy}{dx}=\frac{3}{x}-\frac{2}{2x+1}.

Common errors

Treating ln(f(x)) like a power or forgetting the chain rule

A common slip is differentiating $\ln(3x^2+1)$ as $\dfrac{1}{3x^2+1}$ without the $f'(x)=6x$ on top - the chain rule requires the numerator to be the derivative of the argument. Likewise $\dfrac{d}{dx}e^{2x}=2e^{2x}$ , not $e^{2x}$ ; the exponent's derivative multiplies the result.

Expand logs before you differentiate

If a log contains a product, quotient or power, use the log laws to split it into a sum of simple logs first. Differentiating $3\ln x-\ln(2x+1)$ term by term is far cleaner and less error-prone than applying the quotient rule inside a single logarithm.

Tangents to exponential and log curves

A common application is finding the equation of a tangent. The gradient at a point comes from the derivative, and the tangent line is then $y-y_1=m(x-x_1)$ . For $y=e^x$ at $x=0$ , the gradient is $e^0=1$ and the point is $(0,1)$ , so the tangent is $y=x+1$ . For $y=\ln x$ at $x=1$ , the gradient is $\dfrac{1}{1}=1$ and the point is $(1,0)$ , so the tangent is $y=x-1$ . These two tangents are reflections of each other in $y=x$ , which is the graphical echo of the exponential and logarithm being inverse functions.

Rates of change in models

Because $\dfrac{d}{dx}e^{kx}=k e^{kx}$ , the derivative of an exponential model is proportional to the model itself. This is the defining feature of exponential growth and decay: the rate of change at any instant is a fixed multiple $k$ of the current amount. A positive $k$ gives growth and a negative $k$ gives decay. When SACE asks for the rate of growth of a population or the rate of cooling of an object, you differentiate the model and substitute the given time, exactly as in the worked population example above.

Why it matters

These derivatives are the basis of every exponential growth-and-decay and logarithmic application in Stage 2, and they connect directly to the antidifferentiation results in Topic 3 (since $\int e^x\,dx=e^x+C$ and $\int \tfrac1x\,dx=\ln|x|+C$ are simply these rules reversed).

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2023-style2 marks

Calculator-free. Find

\dfrac{dy}{dx}

for

y = e^{x^2}

. There is no need to simplify your answer.

Show worked answer →

This is the exponential chain rule $\dfrac{d}{dx}e^{g(x)} = g'(x)e^{g(x)}$ . The inner function is $g(x) = x^2$ with $g'(x) = 2x$ .

So $\dfrac{dy}{dx} = 2x\,e^{x^2}$ .

Marks: one for keeping the exponential factor $e^{x^2}$ , one for multiplying by $g'(x) = 2x$ . Writing $e^{x^2}$ alone and omitting the factor $2x$ is the frequent error.

SACE 2022-style2 marks

Calculator-free. Find

\dfrac{dy}{dx}

y = x e^x

Show worked answer →

This is a product of $u = x$ and $v = e^x$ , so use the product rule $(uv)' = u'v + uv'$ .

Here $u' = 1$ and $v' = e^x$ .

\frac{dy}{dx} = (1)(e^x) + (x)(e^x) = e^x + x e^x = e^x(1 + x).

Marks: one for the product-rule set-up, one for the simplified derivative $e^x(1 + x)$ . Factoring $e^x$ is the form needed if the next part asks for stationary points.

SACE 2021-style3 marks

Calculator-assumed. For

y = \ln(x^2 + 4)

, find

\dfrac{dy}{dx}

and hence the

x

-coordinate of the stationary point.

Show worked answer →

Use the log chain rule $\dfrac{d}{dx}\ln f(x) = \dfrac{f'(x)}{f(x)}$ with $f(x) = x^2 + 4$ , $f'(x) = 2x$ :

\frac{dy}{dx} = \frac{2x}{x^2 + 4}.

Stationary point where $\dfrac{dy}{dx} = 0$ : the denominator is never zero, so $2x = 0$ , giving $x = 0$ .

Marks: one for the chain-rule structure, one for $f'(x) = 2x$ on top, and one for solving to $x = 0$ .

Practise this

Sources & how we know this

Stage 2 Mathematical Methods subject outline (2025 onwards) — SACE Board
Stage 2 Mathematical Methods: external assessment — SACE Board

What this dot point is asking

The base results

Chain-rule extensions

Step 1: Identify the composite structure

Step 2: Apply the composite exponential rule

Step 3: Substitute t=10t=10t=10

Step 1: Identify the inner function

Step 2: Apply the logarithm chain rule

Combining with the product and quotient rules

Step 1: Recognise the product structure

Step 2: Apply the product rule

Step 3: Factorise the derivative

Step 4: Find the stationary points

Using the log laws to simplify first

Common errors

Tangents to exponential and log curves

Rates of change in models

Why it matters

Exam-style questions

Practise this

Sources & how we know this

Step 3: Substitute $t=10$