Mathematical Methods · Units 3 & 4

Derivatives of Exponential and Logarithmic Functions

Learn to differentiate exponential and logarithmic functions the easy way, with plain English intuition, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

Money is the friendliest place to meet these functions. Leave cash in an account that grows continuously and the balance follows an exponential curve, while the time it takes to double follows a logarithm. The remarkable thing about the exponential is that its rate of growth at any instant is exactly equal to its current size, which is why $e^x$ is the one function in all of mathematics that is its own derivative. Once you believe that, every exponential and log derivative in the course becomes a short recipe.

The two base facts

Everything in this lesson grows from two results you should know cold. They are the starting points, and the chain, product and quotient rules do the rest.

\frac{d}{dx}\left(e^{x}\right) = e^{x}, \qquad \frac{d}{dx}\left(\log_e x\right) = \frac{1}{x}

The first says the exponential is unchanged by differentiation. The second says the natural log turns into a simple reciprocal. Notice the log result only makes sense for $x > 0$ , because you can only take the log of a positive number in the first place.

For y = e^x, the tangent at x = 1 (dashed) has slope e, exactly equal to the height of the curve there, since the exponential is its own derivative.

Exponentials with a multiplier inside

The plain idea first. When the power is not just $x$ but something like $3x$ or $-2x$ , the inside is changing faster or slower than $x$ on its own, so that extra rate has to be carried through. This is the chain rule doing its job.

For $y = e^{kx}$ , let the inside be $u = kx$ . The outside differentiates to itself and the inside contributes a factor of $k$ , so

\frac{d}{dx}\left(e^{kx}\right) = k\,e^{kx}

The same reasoning handles a base other than $e$ . Writing $a^x = e^{x \log_e a}$ turns any exponential into a natural one, which gives

\frac{d}{dx}\left(a^{x}\right) = a^{x} \log_e a

The single most common mistake is dropping the inside factor and writing $e^{3x}$ for the derivative of $e^{3x}$ . The factor of $k$ is the whole point of the chain rule, and examiner reports list its omission as a recurring error.

Logs with a multiplier inside

Here is a result that surprises almost everyone. The derivative of $\log_e(kx)$ does not keep the $k$ at all. The cleanest way to see why is to split the log before you differentiate.

\log_e(kx) = \log_e k + \log_e x

The first term is a constant, so it differentiates to zero, and the second term gives $\frac{1}{x}$ . Therefore

\frac{d}{dx}\left(\log_e (kx)\right) = \frac{1}{x}

Split the log first, then differentiate. Students who skip that step often write the wrong answer of one over $kx$ , which is the most documented log error in the examiner reports.

Combining with the other rules

Real exam questions rarely hand you a bare $e^x$ or $\log_e x$ . They wrap it inside a product or a quotient, so you choose the right rule and use the two base facts as building blocks.

For a product such as $y = x^2 e^{2x}$ , set $u = x^2$ and $v = e^{2x}$ , then apply $\frac{dy}{dx} = u'v + uv'$ , remembering the chain rule gives $v' = 2e^{2x}$ . For a quotient such as $y = \frac{e^x}{x}$ , set $u = e^x$ and $v = x$ , then apply

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

Two warnings the examiner reports raise every year. First, keep brackets around each product so the result reads as a genuine product and not a stray difference of two terms. Second, name your answer correctly as $f'(x)$ , never as $f(x)$ , because the marker is reading your notation as carefully as your algebra.

Learn it, then fade the help

1 · I do — fully worked

Differentiate $y = x^2 e^{2x}$ with the product rule. Set $u = x^2$ and $v = e^{2x}$ , so $u' = 2x$ and (chain rule) $v' = 2e^{2x}$ . Then $\dfrac{dy}{dx} = u'v + uv' = 2x\,e^{2x} + x^2 \cdot 2e^{2x}$ . Take out the common factor $2xe^{2x}$ to get $\dfrac{dy}{dx} = 2xe^{2x}(1 + x)$ .

2 · We do — your turn, guided

Same steps, your turn. Differentiate $y = x^3 e^{2x}$ . Set $u = x^3$ , $v = e^{2x}$ , so $u' = 3x^2$ and $v' = 2e^{2x}$ . Fill the gaps: $\dfrac{dy}{dx} = 3x^2 \cdot e^{2x} + x^3 \cdot \,?\, = x^2 e^{2x}(\,?\,)$ .

Check your step

$\dfrac{dy}{dx} = 3x^2 e^{2x} + 2x^3 e^{2x} = x^2 e^{2x}(3 + 2x)$ , taking out the common factor $x^2 e^{2x}$ .

3 · You do — solo

No scaffolding now: differentiate $y = x e^{3x}$ and simplify by taking out a common factor.

Reveal solution

With $u = x$ , $v = e^{3x}$ : $u' = 1$ and $v' = 3e^{3x}$ . So $\dfrac{dy}{dx} = 1 \cdot e^{3x} + x \cdot 3e^{3x} = e^{3x}(1 + 3x)$ .

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

What is

\dfrac{d}{dx}(e^{x})

What is

\dfrac{d}{dx}(\log_e x)

, and for what values of

x

Differentiate

e^{kx}

. Where does the

k

go?

Why does the multiplier in

\log_e(kx)

vanish when you differentiate?

Differentiate

a^{x}

for a base

a \ne e

What is the single most common mistake when differentiating

e^{3x}

Recall · Chain Rule

State the chain rule using

\dfrac{dy}{du}

and

\dfrac{du}{dx}

Recall · Antidifferentiation

What is the antiderivative of

\dfrac{1}{x}

See these recipes in action in the Worked Examples tab, then test yourself in Try It.

Worked Example 1Cooling model with an exponential, from a real exam

A model for the temperature in a room is given by $g(t) = 22 - 10e^{-6t}$ , $t \ge 0$ . (i) Find the derivative $g'(t)$ . (ii) Find the value of $t$ for which $g'(t) = 10$ , correct to three decimal places.

1
Differentiate using the chain rule. The constant $22$ vanishes, and $e^{-6t}$ brings down a factor of $-6$ .
$g'(t) = -10 \cdot (-6)e^{-6t} = 60e^{-6t}$
2
For part (ii), set the derivative equal to $10$ and isolate the exponential.
$60e^{-6t} = 10 \implies e^{-6t} = \tfrac{1}{6}$
3
Take the natural log of both sides and solve for $t$ , then give a decimal as required.
$t = \frac{\ln 6}{6} \approx 0.299$

Answer

g'(t) = 60e^{-6t}; \quad t = \dfrac{\ln 6}{6} \approx 0.299

VCAA 2024 Mathematical Methods Exam 2, Section B Q2c

Worked Example 2Differentiating a base-two exponential, from a real exam

For $g(x) = 2^x + 5$ , the derivative $g'(x)$ can be expressed in the form $g'(x) = k \times 2^x$ . Find the real number $k$ .

1
Rewrite the base-two exponential as a natural one using $2^x = e^{x\log_e 2}$ .
$g(x) = e^{x\log_e 2} + 5$
2
Differentiate. The constant $5$ disappears, and the chain rule brings down the factor $\log_e 2$ .
$g'(x) = \log_e 2 \cdot e^{x\log_e 2} = \log_e 2 \cdot 2^x$
3
Compare with $k \times 2^x$ . The examiner report notes some students omitted the base of the logarithm.
$k = \log_e 2$

Answer

k = \log_e 2

VCAA 2023 Mathematical Methods Exam 2, Section B Q3b

Worked Example 3An exponential with a multiplier inside

Differentiate $y = e^{3x}$ .

1
This is a chain rule job. The outside is $e^{u}$ and the inside is $u = 3x$ .
$y = e^{u}, \quad u = 3x$
2
The exponential differentiates to itself, and the inside has rate of change $3$ .
$\frac{dy}{du} = e^{u}, \quad \frac{du}{dx} = 3$
3
Multiply the two pieces, then put the inside back.
$\frac{dy}{dx} = e^{u} \times 3 = 3e^{3x}$

Answer

\frac{dy}{dx} = 3e^{3x}

Worked Example 4A natural log of a linear inside

Differentiate $y = \log_e(5x)$ .

1
Split the log first using a log law, since $\log_e(5x) = \log_e 5 + \log_e x$ .
$y = \log_e 5 + \log_e x$
2
The term $\log_e 5$ is just a number, so its derivative is $0$ .
$\frac{d}{dx}\left(\log_e 5\right) = 0$
3
Differentiate $\log_e x$ to get $\frac{1}{x}$ . The multiplier inside has vanished.
$\frac{dy}{dx} = 0 + \frac{1}{x} = \frac{1}{x}$

Answer

\frac{dy}{dx} = \dfrac{1}{x}

Worked Example 5Product rule with an exponential

Differentiate $y = x^2 e^{2x}$ .

1
Two functions multiplied, so use the product rule with $u = x^2$ and $v = e^{2x}$ .
$u = x^2, \quad v = e^{2x}$
2
Differentiate each. The exponential needs the chain rule, giving a factor of $2$ .
$u' = 2x, \quad v' = 2e^{2x}$
3
Apply $u'v + uv'$ and keep brackets around each product.
$\frac{dy}{dx} = 2x \cdot e^{2x} + x^2 \cdot 2e^{2x}$
4
Take out the common factor $2xe^{2x}$ to tidy up.
$\frac{dy}{dx} = 2xe^{2x}(1 + x)$

Answer

\frac{dy}{dx} = 2xe^{2x}(1 + x)

Practice test

Try it yourself

9 questions, 12 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.Consider the function $h(x) = a\log_e(bx)$ , where $a, b \in R \setminus \{0\}$ . Given that its derivative $h'(x)$ has range $(0, \infty)$ , which of the following must be true?

1mark

Need a hint?

Differentiate and watch what happens to

b

. Then think about the sign

x

must take for the function to be defined.

VCAA 2025 Mathematical Methods Exam 2, Section A Q16

Q2.Let $f(x) = \log_e\left(x^3 - 3x + 2\right)$ . Find $f'(3)$ .

2marks

Work this on paper. The worked solution appears once you submit.

VCAA 2024 Mathematical Methods Exam 1, Q1b

Q3.Let $h(x) = f(x) - g(x)$ , where $f(x) = \tfrac{x}{2} + 7$ and $g(x) = Ae^{kx}$ . Write down an expression for the derivative of $h(x)$ .

1mark

Work this on paper. The worked solution appears once you submit.

VCAA 2025 Mathematical Methods Exam 2, Section B Q2di

Q4.The derivative of $y = e^{4x}$ is:

1mark

Need a hint?

Use

\frac{d}{dx}(e^{kx}) = ke^{kx}

. What is

k

here, and does the exponential ever change form?

Q5.The derivative of $y = \log_e(7x)$ is:

1mark

Need a hint?

Split the log first using

\log_e(7x) = \log_e 7 + \log_e x

, then ask what happens to the constant term.

Q6.If $f(x) = e^{-2x}$ , then $f'(x)$ equals:

1mark

Need a hint?

Apply

\frac{d}{dx}(e^{kx}) = ke^{kx}

with a negative

k

. Watch the sign carefully.

Q7.The derivative of $y = x\,\log_e x$ is:

1mark

Need a hint?

Two functions are multiplied, so reach for the product rule with

u = x

and

v = \log_e x

Q8.The derivative of $y = \dfrac{e^{x}}{x}$ is:

1mark

Need a hint?

A function over a function calls for the quotient rule. Mind the order of subtraction in the numerator.

Q9.Let $f(x) = x^2 \log_e(3x)$ . Find $f'(x)$ , expressing your answer in a fully simplified form.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

Why is e to the x its own derivative?

The exponential function is defined so that its rate of growth at any point is exactly equal to its current value. That special property is what makes the derivative of e to the x equal to e to the x, with no change at all.

Why does the number in log of kx disappear when you differentiate?

Because of the log laws, the log of kx splits into the log of k plus the log of x. The log of k is just a constant, so it differentiates to zero, leaving only one over x. The multiplier never survives.

Do I need the chain rule to differentiate e to the kx?

Yes. The inside function is kx, so its derivative k must be carried out the front, giving k times e to the kx. Forgetting that factor of k is the single most common mistake in this topic.

Derivatives of Exponential and Logarithmic Functions

Learn

The two base facts

Exponentials with a multiplier inside

Logs with a multiplier inside

Combining with the other rules

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions