Your weak spots

Insights load after your first practice round.

Module 4 · L9 of 15 ~35 min ⚡ +95 XP available

Differentiating $e^x$

The function $e^x$ is unique — it is the only function equal to its own derivative. Like a feedback loop where output feeds directly back as input, $e^x$ regenerates itself under differentiation. Once you see why, calculus will never look the same again.

Today's hook — Radioactive decay, bacterial growth, compound interest, population dynamics — they all share one equation: the rate of change equals the current value. Is there a function where the output and the rate of change are permanently, identically equal? And if so, what is it?

0/5QUESTS

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall — your gut answer first

+5 XP warm-up

Is there a non-zero function whose derivative is itself? Without looking it up — take a guess at what it might look like and why such a function would even exist.

auto-saved

The two moves

+5 XP to read

Differentiating any exponential involving $e$ comes down to one rule applied in two ways. Lock the core identity into muscle memory — the chain rule does the rest.

For $e^x$ the derivative is itself. For any composite like $e^{u(x)}$, use the chain rule: multiply by $u'(x)$. That's every case covered.

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

$e^{kx}$ shortcut

$\dfrac{d}{dx}(e^{kx}) = ke^{kx}$ — the constant $k$ comes down in front. Worth memorising.

Product rule reminder

When $e^x$ is in a product: $(e^x \cdot f)' = e^x f + e^x f' = e^x(f + f')$. Factorise out $e^x$.

Quotient rule reminder

$\left(\dfrac{e^x}{v}\right)' = \dfrac{e^x v - e^x v'}{v^2} = \dfrac{e^x(v - v')}{v^2}$. Always simplify by factorising $e^x$.

What you'll master

Know

Key facts

$\dfrac{d}{dx}(e^x) = e^x$
$\dfrac{d}{dx}(e^{kx}) = ke^{kx}$
Chain rule formula $\dfrac{d}{dx}(e^{u}) = e^{u} \cdot u'$

Understand

Concepts

Why $e$ is the unique base for this property
How the chain rule extends to composite exponentials
Why $e^x$ is always positive and never zero

Can do

Skills

Differentiate products and quotients involving $e^x$
Find gradients and stationary points of exponential functions
Factorise derivatives to simplify answers

Key terms

Derivative of $e^x$$\dfrac{d}{dx}(e^x) = e^x$. The natural exponential is the unique function that is its own derivative.

Chain rule for exponentials$\dfrac{d}{dx}(e^{u}) = e^{u} \cdot \dfrac{du}{dx}$ where $u$ is a function of $x$.

Natural number $e$Euler's number $e \approx 2.718$, the unique base for which $\dfrac{d}{dx}(a^x) = a^x$.

Stationary pointA point where $\dfrac{dy}{dx} = 0$. Since $e^x > 0$ always, stationary points occur only when the other factor is zero.

The self-replicating function

core concept

The derivative of $e^x$ is itself: $\dfrac{d}{dx}(e^x) = e^x$. This is the defining property of $e$. No other base has this exact property (though $\dfrac{d}{dx}(a^x) = a^x \ln a$, and $\ln e = 1$, which explains everything).

$$\frac{d}{dx}(e^x) = e^x \qquad \frac{d}{dx}(e^{kx}) = ke^{kx} \qquad \frac{d}{dx}(e^{u}) = e^{u} \cdot \frac{du}{dx}$$

For composite exponentials like $e^{x^2 + 3}$, the chain rule gives: $\dfrac{d}{dx}(e^{x^2+3}) = e^{x^2+3} \cdot 2x$. The exponential is never changed — only multiplied by the derivative of the exponent.

Why this matters for differential equations. The equation $\dfrac{dy}{dx} = ky$ (rate proportional to current value) has solution $y = Ce^{kx}$. This underpins radioactive decay ($k < 0$), compound interest, bacterial growth, and Newton's law of cooling — all solved by differentiating $e^{kx}$.

Core rule: $\dfrac{d}{dx}(e^x) = e^x$ — the only function equal to its own derivative; Chain rule extension: $\dfrac{d}{dx}(e^{kx}) = ke^{kx}$; bring down derivative of exponent

Pause — copy the rule $\dfrac{d}{dx}(e^x) = e^x$ and its chain rule extension $\dfrac{d}{dx}(e^{kx}) = ke^{kx}$ into your book.

Did you get this? True or false: $\dfrac{d}{dx}(e^{3x}) = e^{3x}$ (without any extra factor).

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · CHAIN RULE

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

Band 3 Apply

$u = 3x \Rightarrow u' = 3$

Identify the exponent and differentiate it.

PROBLEM 2 · PRODUCT RULE

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

Band 4 Apply

$u = x^2,\; v = e^x \Rightarrow u' = 2x,\; v' = e^x$

Two functions multiplied — product rule required. Differentiate each factor.

PROBLEM 3 · QUOTIENT RULE + GRADIENT

Differentiate $y = \dfrac{e^x}{x + 1}$ and find the gradient at $x = 0$.

Band 4 Apply

$u = e^x,\; v = x + 1 \Rightarrow u' = e^x,\; v' = 1$

Function divided by function — quotient rule. Identify numerator and denominator.

Quick check: What is $\dfrac{d}{dx}(e^{x^2})$?

Common errors · the 3 traps that cost marks

Trap 01

Forgetting the chain rule factor for $e^{kx}$

$\dfrac{d}{dx}(e^{2x}) = 2e^{2x}$, not $e^{2x}$. The derivative of the exponent ($2x$ gives $2$) must be multiplied. Missing this factor is the most common error when differentiating exponentials.

Trap 02

Writing $\dfrac{d}{dx}(e^{x^2}) = e^{x^2}$ without the chain rule

$\dfrac{d}{dx}(e^{x^2}) = 2x \cdot e^{x^2}$. The exponent is $x^2$, not $x$, so the chain rule adds a factor of $2x$. Always differentiate the exponent and multiply — no exceptions.

Trap 03

Not factorising $e^x$ from the derivative

Leaving $2xe^x + x^2 e^x$ unfactorised costs a mark. HSC markers expect $xe^x(x + 2)$. Factorise $e^x$ first, then look for further common factors.

Fill in the blank: To find the stationary point of $y = xe^{-x}$, we set $\dfrac{dy}{dx} = e^{-x}(1 - x) = 0$. Since $e^{-x} \neq 0$, we solve [?] to get $x = 1$.

Work mode · how are you completing this lesson?

Quick-fire practice · 5 problems

Differentiate $y = e^{5x}$.

Differentiate $y = 3e^{-2x}$.

Differentiate $y = xe^x$.

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

Find the gradient of $y = e^{x^2}$ at $x = 1$.

Think-then-look: What is the derivative of $y = e^{5x}$? Write your answer before revealing.

Reveal answer

$\dfrac{dy}{dx} = 5e^{5x}$. Chain rule: derivative of exponent $5x$ is $5$, so multiply $e^{5x}$ by $5$.

Revisit your thinking

Earlier you were asked: is there a function equal to its own derivative? The answer is $y = e^x$ — and the constant $e$ is defined precisely so this works. For $e^{kx}$, the chain rule pulls down a factor of $k$, which makes exponentials the natural solutions to differential equations like $\dfrac{dy}{dx} = ky$. Every radioactive decay curve and population model is built on this foundation.

auto-saved

Odd one out: Three of these derivatives are correct. Which one is wrong?

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 42 marks

Q1. Differentiate natural exponentials. Find $\dfrac{dy}{dx}$ for $y = e^{4x} + e^{-x}$. Show working. (2 marks)

auto-saved

ApplyBand 43 marks

Q2. Product with exponential. Differentiate $y = x^3 e^{2x}$. Fully factorise your answer. (3 marks)

auto-saved

AnalyseBand 54 marks

Q3. Stationary point with exponential. Find the stationary point of $y = xe^{-x}$ and determine its nature. (4 marks)

auto-saved

Comprehensive answers (click to reveal)

Drill 1: $5e^{5x}$ · 2: $-6e^{-2x}$ · 3: $e^x(1+x)$ · 4: $\dfrac{e^{2x}(2x-1)}{x^2}$ · 5: $2e$ (at $x=1$: $2 \cdot 1 \cdot e^{1} = 2e$)

Q1 (2 marks): Chain rule on each term. $\dfrac{dy}{dx} = 4e^{4x} - e^{-x}$ [2].

Q2 (3 marks): Product rule: $u = x^3$, $v = e^{2x}$ [0.5]. $u' = 3x^2$, $v' = 2e^{2x}$ [0.5]. $\dfrac{dy}{dx} = 3x^2 e^{2x} + 2x^3 e^{2x} = x^2 e^{2x}(3 + 2x)$ [2].

Q3 (4 marks): Product rule: $\dfrac{dy}{dx} = e^{-x} + x(-e^{-x}) = e^{-x}(1-x)$ [1]. Set $= 0$: since $e^{-x} \neq 0$, $x = 1$ [1]. $y(1) = e^{-1} = \frac{1}{e}$, so stationary point $\left(1, \frac{1}{e}\right)$ [0.5]. $\dfrac{d^2y}{dx^2} = e^{-x}(x-2)$; at $x = 1$: $e^{-1}(-1) < 0$ so local maximum [1.5].

Boss battle · Module 4

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

Enter the arena

Science Jump · platform challenge

Climb platforms by answering Module 4 questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 8 Lesson 10 · Differentiating $a^x$ →

Module overview · Maths Advanced