The Natural Exponential eˣ

This lesson introduces the number e and the natural exponential function y = eˣ. The constant e is arguably the most important number in A-Level Mathematics after π, and it arises naturally from calculus. Understanding eˣ is essential for differentiation, integration, and modelling.

What Is the Number e?

The number e is an irrational constant approximately equal to:

e ≈ 2.71828...

It is defined as the limit:

e = lim(n→∞) (1 + 1/n)ⁿ

You can see how this works by computing (1 + 1/n)ⁿ for increasing n:

n	(1 + 1/n)ⁿ
1	2
10	2.5937...
100	2.7048...
1000	2.7169...
10000	2.7181...
1000000	2.71828...

As n increases, the value approaches e = 2.71828...

Why Is e Special?

The function y = eˣ is the unique exponential function whose gradient equals its own value at every point:

If y = eˣ, then dy/dx = eˣ

No other exponential function has this property. For y = 2ˣ, the derivative is 2ˣ × ln(2), which includes a constant multiplier. Only when the base is e does that multiplier equal 1.

This self-replicating property makes eˣ fundamental to:

Solving differential equations
Modelling continuous growth and decay
Probability (the normal distribution)
Complex number theory

The Graph of y = eˣ

The graph of y = eˣ shares the same general shape as any exponential growth curve (base > 1):

Property	Value
Domain	All real numbers
Range	y > 0
y-intercept	(0, 1)
Asymptote	y = 0 (the x-axis)
Gradient at (0, 1)	Exactly 1
Shape	Increasing, concave up

The special feature is that the gradient at the y-intercept is exactly 1. At any point on the curve, the gradient equals the y-coordinate.

Transformations of y = eˣ

All standard transformation rules apply:

Function	Transformation	y-intercept	Asymptote
y = eˣ	Standard	(0, 1)	y = 0
y = e⁻ˣ	Reflection in y-axis (decay)	(0, 1)	y = 0
y = 3eˣ	Vertical stretch, scale factor 3	(0, 3)	y = 0
y = eˣ + 5	Translation up by 5	(0, 6)	y = 5
y = e^(x-2)	Translation right by 2	(0, e⁻²)	y = 0
y = -eˣ	Reflection in x-axis	(0, -1)	y = 0

Exam Tip: When the question says "sketch the curve y = eˣ + k", always draw the asymptote y = k as a dashed line and clearly label it.

The Function y = e⁻ˣ

The function y = e⁻ˣ is an exponential decay curve. It is the reflection of y = eˣ in the y-axis.

Key properties:

Passes through (0, 1)
Decreases for all x
Asymptote at y = 0
Gradient at (0, 1) is -1

This function appears frequently in decay models:

Radioactive decay: N = N₀e⁻ᵏᵗ
Newton's cooling: T = T₀e⁻ᵏᵗ + Tₛ
Damped oscillations

Compound Interest and Continuous Growth

The number e arises naturally from compound interest. If you invest £1 at 100% annual interest compounded n times per year:

Annual (n = 1): £2.00
Monthly (n = 12): £2.61
Daily (n = 365): £2.7146
Continuously (n → ∞): £e ≈ £2.7183

The formula for continuous compounding is:

A = Peʳᵗ

where P is the principal, r is the rate, and t is time.

Solving Simple Equations with eˣ

To solve equations involving eˣ, you will typically need logarithms (covered in detail in the next lessons). For now, know that:

eˣ = k has a solution only when k > 0
The solution is x = ln(k), where ln is the natural logarithm
eˣ is never zero and never negative

Example: Solve eˣ = 5

x = ln(5) ≈ 1.609

Example: Solve e²ˣ = 12

2x = ln(12)
x = ln(12)/2 ≈ 1.242

Summary

e ≈ 2.71828 is an irrational constant defined as lim(n→∞) (1 + 1/n)ⁿ.
y = eˣ is the unique function whose derivative equals itself: dy/dx = eˣ.
The graph passes through (0, 1), has asymptote y = 0, and has gradient 1 at the y-intercept.
y = e⁻ˣ is the decay version, reflected in the y-axis.
e arises from continuous compounding and is central to modelling, calculus, and probability.

A-Level Deep Dive: The Natural Exponential e^x

Spec mapping

Edexcel 9MA0 specification section 6 — Exponentials and logarithms covers the function $e^x$ and its graph; know that the gradient of $e^{kx}$ is $ke^{kx}$ ; understand and use the inverse relationship $y = e^x \Leftrightarrow x = \ln y$ (refer to the official specification document for exact wording). The natural exponential sits inside Paper 1 (Pure) section 6 but is genuinely cross-paper: it underwrites section 7 (Differentiation) — $\dfrac{d}{dx}e^x = e^x$ is the foundation of every calculus problem involving growth, decay, or oscillation in a real-world context, section 8 (Integration) — $\int e^x\,dx = e^x + C$ and the more general $\int e^{kx}\,dx = \tfrac{1}{k}e^{kx} + C$ , section 9 (Numerical methods) — Newton–Raphson on $f(x) = e^x - g(x)$ , and section 10 (Differential equations) — separable equations of the form $\dfrac{dy}{dx} = ky$ have solution $y = Ae^{kx}$ . The constant $e$ also reappears in Paper 3 (Statistics and Mechanics) within continuous probability distributions and decay-of-momentum problems. The Edexcel formula booklet supplies $\dfrac{d}{dx}e^{kx} = ke^{kx}$ but does not supply the limit definition or the Taylor series — both must be reasoned with from scratch.

Worked example with full mark scheme

Question (8 marks):

(a) Solve $e^{2x} - 5e^x + 4 = 0$ , giving the exact values of $x$ . (5)

(b) The curve $C$ has equation $y = e^x$ . Find the gradient of $C$ at each of the points whose $x$ -coordinates are the solutions found in part (a). (3)

Solution with mark scheme:

(a) Step 1 — recognise the hidden quadratic and substitute.

The structure "term in $e^{2x}$ + term in $e^x$ + constant" is the signature of a quadratic in $u = e^x$ . Note $e^{2x} = (e^x)^2 = u^2$ .

Let $u = e^x$ . The equation becomes:

$u^2 - 5u + 4 = 0$

M1 — correct substitution $u = e^x$ producing a quadratic in $u$ . Examiners reward the explicit substitution line; candidates who silently leap to a factorisation often lose this method mark because the structural insight is invisible.

Step 2 — factorise the quadratic.

$(u - 1)(u - 4) = 0 \implies u = 1 \text{ or } u = 4$

M1 — correct factorisation (or use of the quadratic formula).

A1 — both roots stated correctly.

Step 3 — back-substitute and apply the inverse relationship.

$u = e^x = 1 \implies x = \ln 1 = 0$ . $u = e^x = 4 \implies x = \ln 4$ .

M1 — correct use of the inverse relationship $e^x = a \implies x = \ln a$ (valid because $a > 0$ in both cases).

A1 — both exact answers $x = 0$ and $x = \ln 4$ stated. Note: $\ln 4 = 2\ln 2$ is an equally acceptable form. Decimal approximations such as $x \approx 1.386$ would lose this A1 because the question asks for exact values.

A presentational note: candidates frequently write $u = e^x = -1$ when, for example, factoring $(u + 1)(u - 4)$ . Here neither root is negative, but examiners look for a defensive sentence such as "since $e^x > 0$ for all real $x$ , both roots are admissible" — this is examiner-rewarded phrasing for AO2.5 reasoning marks on harder variants of the question.

(b) Step 1 — differentiate.

The curve is $y = e^x$ , so $\dfrac{dy}{dx} = e^x$ — the defining property of the natural exponential.

B1 — stated derivative $\dfrac{dy}{dx} = e^x$ .

Step 2 — evaluate at each solution.

At $x = 0$ : $\dfrac{dy}{dx} = e^0 = 1$ .

At $x = \ln 4$ : $\dfrac{dy}{dx} = e^{\ln 4} = 4$ (by the inverse relationship).

M1 — substitution of both $x$ -values into the derivative.

A1 — both gradients $1$ and $4$ correctly evaluated, with exact-form working visible.

Total: 8 marks (M3 A3 B1 + 1 inferred from M1 A1 in part (b)). Split AO1 = 6, AO2 = 2.

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): The curve $C$ has equation $y = 2e^x - 6x$ .

(a) Find $\dfrac{dy}{dx}$ . (2)

(b) Hence find, in exact form, the $x$ -coordinate of the stationary point of $C$ , and determine its nature. (4)

Mark scheme decomposition by AO:

(a)

B1 (AO1.1a) — knowing $\dfrac{d}{dx}e^x = e^x$ , so $\dfrac{d}{dx}(2e^x) = 2e^x$ .
B1 (AO1.1b) — fully correct derivative $\dfrac{dy}{dx} = 2e^x - 6$ .

(b)

M1 (AO1.1b) — setting $\dfrac{dy}{dx} = 0$ , giving $2e^x = 6$ , hence $e^x = 3$ .
A1 (AO1.1b) — exact $x$ -coordinate $x = \ln 3$ .
M1 (AO2.1) — computing the second derivative $\dfrac{d^2y}{dx^2} = 2e^x$ and evaluating at the stationary point: $2e^{\ln 3} = 6 > 0$ .
A1 (AO2.5) — concluding "minimum, since $\dfrac{d^2y}{dx^2} > 0$ at the stationary point", with the reasoning explicit.

Total: 6 marks split AO1 = 4, AO2 = 2. This question targets the simplest synoptic chain: differentiate, set to zero, classify by second derivative. The AO2 marks reward two specific things — the exact-form answer $\ln 3$ (not the decimal $1.0986$ ) and the explicit reasoning sentence "since $\dfrac{d^2y}{dx^2} > 0$ , the point is a minimum". A response that computes the right number but says only "minimum" without justification typically scores 5/6.

Synoptic links

Connects to:

Section 7 — Differentiation: $\dfrac{d}{dx}e^x = e^x$ is the unique fixed point of the differentiation operator on smooth functions (up to scalar multiple). More generally $\dfrac{d}{dx}e^{kx} = ke^{kx}$ , which is the only family of functions whose rate of change is proportional to the function itself — a property that makes $e^x$ irreplaceable in modelling.
Section 8 — Integration: the antiderivative $\int e^x\,dx = e^x + C$ falls out of the differentiation rule by inversion. Combined with substitution and integration by parts, it lets students integrate $xe^x$ , $e^x \sin x$ , and arbitrary polynomial-times-exponential combinations — the workhorse integrals of A-Level applied calculus.
Section 10 — Differential equations: the equation $\dfrac{dy}{dx} = ky$ has general solution $y = Ae^{kx}$ . This is the model for radioactive decay ( $k < 0$ ), unrestricted population growth ( $k > 0$ ), Newton's law of cooling, and continuous compound interest. Recognising the exponential as the solution is a Year-2 A-Level discriminator.
Further Maths section 5 — Complex numbers (Euler's formula): $e^{i\theta} = \cos\theta + i\sin\theta$ extends the natural exponential to imaginary arguments and unifies trigonometry with exponentials. The identity $e^{i\pi} + 1 = 0$ — connecting $e$ , $i$ , $\pi$ , $1$ , $0$ — is often called the most beautiful equation in mathematics.
Paper 3 — Population and decay modelling: examination questions in section 16 (numerical methods on real-world data) and section 6 of the Statistics paper (continuous probability) routinely involve fitting $P(t) = P_0 e^{kt}$ to data, computing decay constants, and solving for half-lives $t_{1/2} = \ln 2 / k$ .

Mark-scheme literacy

Natural-exponential questions on 9MA0 distribute AO marks roughly as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Stating $\dfrac{d}{dx}e^x = e^x$ , applying the substitution $u = e^x$ , evaluating $e^{\ln a} = a$ , factorising the resulting quadratic
AO2 (reasoning / interpretation)	25–35%	Justifying that $e^x > 0$ rejects negative $u$ -roots, choosing exact over decimal form, classifying stationary points with explicit second-derivative reasoning
AO3 (problem-solving)	0–15%	Modelling questions: setting up $N = N_0 e^{kt}$ from a verbal description, interpreting $k$ in context, drawing conclusions about long-term behaviour

Examiner-rewarded phrasing: "by substitution $u = e^x$ , the equation reduces to a quadratic in $u$ "; "since $e^x > 0$ for all real $x$ , we reject $u = -1$ "; "by the inverse relationship $e^x = a \Leftrightarrow x = \ln a$ for $a > 0$ "; "since $\dfrac{d^2y}{dx^2} > 0$ at the stationary point, $C$ has a minimum there". Phrases that lose marks: writing $\ln(e^x) = x$ without comment when the substitution requires it; using $\log$ (base 10) instead of $\ln$ to invert $e^x$ ; giving decimal approximations when "exact" is demanded; classifying a stationary point as "min" or "max" with no derivative-based justification.

A specific Edexcel pattern: questions phrased "show that $x = \ln k$ for some integer $k$ to be found" require the candidate to manipulate to that form exactly. An answer of $x = 2\ln 2$ instead of $x = \ln 4$ , while equivalent, can lose the final A1 because the printed form requires a single $\ln$ of an integer. Read the form constraint precisely.

Grade-band model answers

3-mark question

Question: Evaluate $e^{2\ln 3}$ , giving your answer as an integer.

Grade C response (~210 words):

We have $e^{2\ln 3}$ . Using the index law, $e^{2\ln 3} = (e^{\ln 3})^2$ .

Since $e^{\ln 3} = 3$ (because $e^x$ and $\ln x$ are inverse functions), this gives $3^2 = 9$ .

So $e^{2\ln 3} = 9$ .

Examiner commentary: Full marks (3/3). The candidate identifies the key step — pulling the coefficient $2$ out via $e^{2a} = (e^a)^2$ — and uses the inverse relationship $e^{\ln x} = x$ explicitly. Working is brief but every step is justified. This is the standard Grade C answer for a one-shot index question — efficient and correct. A small presentational tightening would be to write the inverse relationship as a line: $e^{\ln 3} = 3$ because $\ln$ is the inverse of the natural exponential. Many candidates lose marks here by writing $e^{2\ln 3} = 2 \cdot e^{\ln 3} = 2 \cdot 3 = 6$ , which confuses $e^{2a}$ with $2e^a$ — a fundamental mis-application of the index laws. Others write $e^{2\ln 3} = e^{\ln 6}$ by combining $2\ln 3 = \ln 6$ , which is wrong because $2\ln 3 = \ln 9$ (by the power law $n\ln a = \ln a^n$ ). The candidate has avoided both traps.

Grade A response (~290 words):*

We can evaluate $e^{2\ln 3}$ by either of two routes; both rely on the inverse-function relationship between $e^x$ and $\ln x$ , valid because each is the inverse of the other on appropriate domains.

The Natural Exponential e^x