Practice Maths

The Number e and Differentiating Exponentials

Key Terms

For any base a > 0, the limit limh→0 (ah − 1)/h equals a constant that depends on a.
Euler’s number e ≈ 2.71828
is the unique base where limh→0 (eh − 1)/h = 1.
Because of this, d/dx(ex) = ex — the exponential function is its own derivative.
Features of y = ex: domain all reals, range y > 0, y-intercept (0, 1), horizontal asymptote y = 0 as x → −∞, always increasing.
The chain rule extends this: d/dx(ekx) = kekx.
RuleFormula
Derivative of exd/dx(ex) = ex
Derivative of ekxd/dx(ekx) = kekx
Derivative of aekxd/dx(aekx) = akekx
Key limitlimh→0 (eh − 1)/h = 1
Value of ee ≈ 2.71828… (irrational)
x y 1 O (0,1) y = eˣ y = 0 (asymptote)
Hot Tip ex is special because differentiating it gives back itself. No other function (except multiples of ex) has this property. This makes it the natural choice for modelling growth and decay.

Worked Example 1 — Estimating the limit

Question: Complete a table of values for f(h) = (2h − 1)/h as h → 0 using h = 0.1, 0.01, 0.001. What does this suggest about limh→0 (2h − 1)/h?

h = 0.1: (20.1 − 1)/0.1 = (1.07177 − 1)/0.1 = 0.7177

h = 0.01: (20.01 − 1)/0.01 = (1.00695 − 1)/0.01 = 0.6956

h = 0.001: (20.001 − 1)/0.001 ≈ 0.6933

Conclusion: The limit appears to approach approximately 0.693, which equals ln(2). Since this is not 1, the base 2 is not special. For base e, this limit equals exactly 1.

Worked Example 2 — Differentiating exponential functions

Question: Find dy/dx for: (a) y = 3e2x    (b) y = ex + 5e−x

(a) y = 3e2x. Using d/dx(aekx) = akekx with a = 3, k = 2:

dy/dx = 3 × 2 × e2x = 6e2x

(b) y = ex + 5e−x. Differentiate term by term:

dy/dx = ex + 5 × (−1) × e−x = ex − 5e−x

Where Does e Come From?

The number e arises naturally from the question: for what base a does the exponential function y = ax have the simplest possible derivative? To answer this, recall the first-principles definition of the derivative:

d/dx(ax) = limh→0 (ax+h − ax)/h = limh→0 ax(ah − 1)/h = ax × limh→0 (ah − 1)/h

The factor limh→0 (ah − 1)/h is a constant that depends only on the base a. If this constant equals 1, then d/dx(ax) = ax — the function is its own derivative. The unique value of a that makes this happen is called Euler’s number, denoted e.

Numerically, e ≈ 2.71828182845904… It is irrational (cannot be written as a fraction) and transcendental (not a root of any polynomial with rational coefficients). It can also be defined as e = limn→∞ (1 + 1/n)n, which arises naturally in compound interest calculations.

Features of y = ex

The function y = ex has the following key features that you must know for the exam:

  • Domain: all real numbers (x can be any value)
  • Range: y > 0 (always positive, never zero)
  • y-intercept: (0, 1), since e0 = 1
  • No x-intercept (the graph never crosses the x-axis)
  • Horizontal asymptote: y = 0 as x → −∞
  • Always increasing: since dy/dx = ex > 0 for all x
  • Concave up everywhere: d²y/dx² = ex > 0

Transformations apply as expected: y = ex + 3 shifts up 3; y = ex−2 shifts right 2; y = −ex reflects over the x-axis (and now has range y < 0).

Why d/dx(ex) = ex

The formal proof uses the definition of e. Since limh→0 (eh − 1)/h = 1 (by definition of e), we have:

d/dx(ex) = ex × limh→0 (eh − 1)/h = ex × 1 = ex

This is the most remarkable property in calculus: differentiation maps ex to itself. This means ex models any process where the rate of change is proportional to the current quantity — population growth, radioactive decay, bank interest, etc.

For y = ekx, the chain rule gives d/dx(ekx) = k × ekx. Think of it as: the k “comes down” as a multiplier. More generally, d/dx(aekx) = akekx.

Building up from Simpler Derivatives

Exponential terms can be combined with polynomial terms using the sum and constant-multiple rules. For example:

  • y = x² + ex ⇒ dy/dx = 2x + ex
  • y = 4e3x − 2e−x ⇒ dy/dx = 12e3x + 2e−x
  • y = (ex + 1)/3 = (1/3)ex + 1/3 ⇒ dy/dx = (1/3)ex

A common error: students write d/dx(e3x) = e3x and forget the factor of 3. Always apply the chain rule when there is a coefficient in the exponent.

Exam Technique and Common Errors

In QCAA exams, exponential derivatives appear frequently, often embedded in chain rule, product rule or optimisation questions. Key strategies:

  • Always bring coefficients from exponents to the front: d/dx(e5x) = 5e5x, not e5x.
  • Do not confuse e0 = 1 with d/dx(e0) = 0 (the derivative of a constant is zero).
  • For negative exponents: d/dx(e−3x) = −3e−3x. The negative comes from the chain rule, not from “reflecting” the function.
  • When finding stationary points of exponential functions, remember eanything > 0 always. So ex terms never equal zero — solutions come from other factors.
Exam Tip: If asked to “sketch y = ex” or a transformation of it, always mark: (1) the y-intercept, (2) the horizontal asymptote, (3) the direction it approaches as x → ±∞. These three things earn the marks.
Exam Tip: e is not a variable — it is a fixed irrational number like π. Do not treat ex like xe. The derivative of xe is e xe−1 (power rule), while the derivative of ex is ex (completely different).

Mastery Practice

See Answers ➔

  1. Estimate the limit numerically. Fluency

    Complete a table for f(h) = (3h − 1)/h using h = 0.1, 0.01, 0.001, 0.0001.
    1. (a) Calculate f(0.1), f(0.01), f(0.001), f(0.0001) correct to 4 decimal places.
    2. (b) What value does the limit appear to approach?
    3. (c) Explain why this value is not equal to 1, and what it implies about base 3 compared to base e.

  2. Differentiate each function. Fluency

    1. (a) y = e4x
    2. (b) y = e−2x
    3. (c) y = 5e3x
    4. (d) y = −2ex
    5. (e) y = ex/2
    6. (f) y = 4e−x/3

  3. State the key features of each function. Fluency

    1. (a) y = ex + 2: state the y-intercept, asymptote, and range.
    2. (b) y = e−x: state whether this function is increasing or decreasing, and its asymptote.
    3. (c) y = 3ex: state the y-intercept and range.

  4. Differentiate using the sum and constant-multiple rules. Fluency

    1. (a) y = 2ex + 3x2
    2. (b) y = e2x − e−2x
    3. (c) y = (ex + 1)/4
    4. (d) y = 6 − e5x

  5. Tangent line to an exponential curve. Understanding

    Consider the curve y = 2e3x.
    1. (a) Find dy/dx.
    2. (b) Find the gradient of the tangent at x = 0.
    3. (c) Write the equation of the tangent at (0, 2) in the form y = mx + c.
    4. (d) At what x-value is the gradient of the tangent equal to 6e3?

  6. Population growth model. Understanding

    A bacterial population is modelled by P(t) = 500e0.4t, where P is the number of bacteria and t is time in hours.
    1. (a) Find the initial population (at t = 0).
    2. (b) Find P′(t) and interpret it in context.
    3. (c) Find the rate of growth at t = 2 hours. Give your answer correct to the nearest whole number.
    4. (d) Show that P′(t) = 0.4P(t) and explain the significance of this result.

  7. Comparing exponential derivatives. Understanding

    Consider y1 = e2x and y2 = ex.
    1. (a) Find dy1/dx and dy2/dx.
    2. (b) At x = 0, which function has the steeper gradient? Justify.
    3. (c) Find the x-value where dy1/dx = 10. Give your answer exact.

  8. Analysing a transformed exponential. Understanding

    The function f(x) = Aekx passes through (0, 3) and (1, 6).
    1. (a) Use the point (0, 3) to find A.
    2. (b) Use the point (1, 6) to find k. Leave your answer in exact form.
    3. (c) Write the fully defined function f(x).
    4. (d) Find f′(x) and evaluate f′(2) to 2 decimal places.

  9. Drug concentration model. Problem Solving

    The concentration of a drug in the bloodstream is modelled by C(t) = 8e−0.5t mg/L, where t is time in hours after administration.
    1. (a) Find C(0) and interpret its meaning.
    2. (b) Find C′(t) and determine the rate of change at t = 1 and t = 3 (to 2 d.p.).
    3. (c) Explain why C′(t) is always negative for this model and what this means in context.
    4. (d) Find the time when the concentration first falls below 1 mg/L. Give your answer exact and to 2 d.p.

  10. Exploring the limit definition of e. Problem Solving

    The number e can be approximated using e = limn→∞(1 + 1/n)n.
    1. (a) Calculate (1 + 1/n)n for n = 10, 100, 1000. Compare with e ≈ 2.71828.
    2. (b) This formula arises from continuously compounded interest. If $1 is invested at 100% annual interest compounded n times per year, the amount after 1 year is (1 + 1/n)n. What is the limiting amount as n → ∞?
    3. (c) A more general formula is A = Pert for continuous compounding, where P is principal, r is rate, and t is time. If $5000 is invested at 6% p.a. continuously compounded, find A after 10 years (to the nearest dollar) and find dA/dt at t = 10.