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hscscience Maths Adv · Y11
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Module 4 · L9 of 15 ~35 min ⚡ +95 XP available

Differentiating $e^x$

The function $e^x$ is unique in a precise way: it is the only function equal to its own derivative that also passes through $(0, 1)$. Every $Ce^x$ satisfies $f' = f$, and $e^x$ is the one normalised by $f(0) = 1$. Like a feedback loop where output feeds directly back as input, it 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
1

Get oriented

Set up your goals and key terms for differentiating e to the x.

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
01
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.

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02
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 $e^x$.
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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
04
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.
2

See where the rule comes from

Learn why the exponential function is its own derivative.

05
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: $\dfrac{d}{dx}(e^x) = e^x$, and more generally $\dfrac{d}{dx}(Ce^x) = Ce^x$, so $e^x$ is the solution of $f' = f$ with $f(0) = 1$; 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).

Interactive · Tangents on $y = e^{kx}$

Try this: drag the point along $y = e^x$ and press Mark (x, m) a few times, and watch every mark land back on the curve. Then set $k = 2$, mark again, and switch on both candidate curves to see which one the marks sit on. Finish with Find gradient 3.

Use the explorer. Press Find gradient 3 and move the point until the tangent gradient reads 3.0. Read the panel and give each value to the nearest whole number: the height $y$ there is and $m \div y$ is . The $x$-value you found is $\ln 3$.

Use the explorer. Set $k = 2$, place four marks, then switch on both candidate curves. True or false: the marks land on $y = e^{2x}$ itself, so $\dfrac{d}{dx}(e^{2x}) = e^{2x}$.

Graph and worked example showing that the derivative of e to the x is e to the x, including the chain rule
3

Work through examples

Follow three worked problems, revealing each step as you go.

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

PROBLEM 1 · CHAIN RULE

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

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$u = 3x \Rightarrow u' = 3$
Identify the exponent and differentiate it.
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PROBLEM 2 · PRODUCT RULE

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

Band 4 Apply
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$u = x^2,\; v = e^x \Rightarrow u' = 2x,\; v' = e^x$
Two functions multiplied, product rule required. Differentiate each factor.
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PROBLEM 3 · QUOTIENT RULE + GRADIENT

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

Band 4 Apply
1
$u = e^x,\; v = x + 1 \Rightarrow u' = e^x,\; v' = 1$
Function divided by function, quotient rule. Identify numerator and denominator.
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4

Avoid the common traps

Meet the three errors that cost marks, and check yourself.

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 $x = 1$.

5

Drill the derivatives

Choose your work mode, then work five quick-fire derivatives.

Quick-fire practice · 5 problems

1

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

2

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

3

Differentiate $y = xe^x$.

4

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

5

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 $5$.
6

Revisit and consolidate

Revisit your first answer and confirm what stuck.

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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.

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

7

Practise the exponential rule

Answer the questions on differentiating e to the x.

01
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.

02
Short answer
ApplyBand 42 marks

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

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ApplyBand 43 marks

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

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AnalyseBand 54 marks

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

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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].