A. Rate of Change · Derivatives of Exponential and Sinusoidal FunctionsDiscovering e: The Function That Is Its Own Derivative
A2.6About 70 minutes
Learning goals- I can search numerically for the base a whose exponential function has k = 1.
- I can identify the number e ≈ 2.71828 as the base for which the derivative of aˣ equals aˣ.
- I can state and use the rule d/dx eˣ = eˣ, together with the constant multiple, sum and difference rules.
- I can explain why the slope of y = eˣ at any point equals the y-value at that point, and find tangent lines to y = eˣ.
Words to know
e
The irrational number 2.718281828…, the base for which the exponential function is its own derivative.
natural exponential function
The function f(x) = eˣ. Its derivative is f′(x) = eˣ.
irrational number
A real number that cannot be written as a fraction of integers. Its decimal never ends and never repeats.
tangent line
The line through a point on a curve whose slope is the instantaneous rate of change there.
compound interest
Interest added to the principal at regular intervals, so later interest is earned on earlier interest.
Hunting for the perfect base
In the last lesson you found that every exponential function obeys f′(x) = k aˣ, where k = f′(0). For base 2, k ≈ 0.693 (too small). For base 3, k ≈ 1.099 (too big). So there should be a base between 2 and 3 where k = 1 exactly. For that base, the derivative would equal the function itself: f′(x) = f(x).
Expectation A2.6 asks you to find that base by investigation. Graphing software with a slider for a does it quickly, but a calculator works too. Here is a calculator search using (aʰ − 1) ÷ h with h = 0.000001:
| base a | 2.71 | 2.715 | 2.718 | 2.719 | 2.72 |
|---|
| k ≈ | 0.99695 | 0.99879 | 0.99990 | 1.00026 | 1.00063 |
k passes through 1 between a = 2.718 and a = 2.719.The search pins the base down to about 2.718. More careful searching gives
e = 2.718281828459045…
This number is called e. It is irrational: its decimal never ends and never repeats, just like π. Leonhard Euler made the letter e the standard name for it, in the 1700s. Your calculator has an eˣ key (often the second function of ln). Press eˣ with 1 to see 2.718281828.
The derivative of eˣ
d/dx eˣ = eˣ.
The function f(x) = eˣ is its own derivative. Its slope at any point equals its height at that point. For example, at x = 2 the height is e² ≈ 7.39, and so is the slope.
Three exponential curves pass through (0, 1). The tangent to y = eˣ at (0, 1) has slope exactly 1, so it is the line y = x + 1. The curve y = 2ˣ crosses less steeply and y = 3ˣ more steeply.Seeing slope = height
Take any point (a, eᵃ) on y = eˣ. The tangent has slope eᵃ. Its equation is y − eᵃ = eᵃ(x − a). Where does it meet the x-axis? Set y = 0: −eᵃ = eᵃ(x − a), so x − a = −1, and x = a − 1.
So every tangent to y = eˣ meets the x-axis exactly 1 unit to the left of the point of tangency. That makes a right triangle with run 1 and rise eᵃ: slope = height, every time.
The tangent to y = eˣ at (1, e) meets the x-axis at x = 0, exactly 1 unit to the left. Rise e over run 1 gives slope e, the same as the height of the point.Another road to e: compound interest
e also appears in money. Put $1 in an account paying 100% interest per year (an impossible rate, but it makes the arithmetic clear). If interest is compounded n times a year, after one year you have (1 + 1/n)ⁿ dollars.
- n = 1 (yearly): $2.00
- n = 12 (monthly): about $2.61
- n = 365 (daily): about $2.71
- n = 8760 (hourly): about $2.718
The amounts approach e as n grows: e = lim (n → ∞) (1 + 1/n)ⁿ. The Swiss mathematician Jacob Bernoulli met this limit while studying compound interest in the late 1600s. So e is the natural base both for growth that compounds continuously and for functions that are their own derivatives. That double role is why it appears all through university science, economics and engineering.
Worked exampleExample 1: Combining with earlier rules
Differentiate f(x) = 3eˣ − 2x² + 5, then find f′(1) exactly and to two decimal places.
- Constant multiple rule: d/dx 3eˣ = 3eˣ.
- Power rule: d/dx 2x² = 4x. The derivative of the constant 5 is 0.
- So f′(x) = 3eˣ − 4x.
- f′(1) = 3e − 4 ≈ 3(2.71828) − 4 ≈ 4.15.
f′(x) = 3eˣ − 4x, and f′(1) = 3e − 4 ≈ 4.15.
Worked exampleExample 2: A tangent line
Find the equation of the tangent to y = eˣ at x = 2. Give the exact form and a rounded form.
- Point: (2, e²).
- Slope: y′ = eˣ, so at x = 2 the slope is e².
- Point-slope form: y − e² = e²(x − 2), so y = e²x − 2e² + e² = e²x − e².
- Rounded: e² ≈ 7.389, so y ≈ 7.389x − 7.389.
- Check with the 'one unit to the left' fact: the tangent should meet the x-axis at x = 1. Set y = 0: e²x = e², so x = 1. ✓
y = e²x − e², or y ≈ 7.39x − 7.39.
Worked exampleExample 3: Rewriting before differentiating
Differentiate g(x) = e^(x + 3) without the chain rule.
- Use the exponent law: e^(x + 3) = e³ × eˣ.
- e³ is just a constant (about 20.09).
- Constant multiple rule: g′(x) = e³ × eˣ = e^(x + 3).
g′(x) = e^(x + 3). A horizontal shift of eˣ is still its own derivative.
Hands-on activityFind e with compound interest on a calculator
You need: a scientific calculator and a sheet of paper.
- Make a table with n = 1, 2, 4, 12, 52, 365, 8760 and 525 600 (yearly, half-yearly, quarterly, monthly, weekly, daily, hourly, every minute).
- For each n, compute (1 + 1/n)ⁿ. Use the bracket keys and the power key (xʸ or ^).
- Compare each result with e (press eˣ then 1).
- Now compute (eʰ − 1) ÷ h for h = 0.1, 0.01 and 0.001. What do the answers approach?
- Finally, choose your own base between 2.7 and 2.8 and compute (a^0.000001 − 1) ÷ 0.000001. Adjust a until the answer is as close to 1 as you can get. How many correct digits of e did you find?
What to look for: the compound-interest values should be 2, 2.25, 2.4414, 2.6130, 2.6926, 2.7146, 2.7181 and 2.7183 (rounded), creeping up towards e ≈ 2.71828 but never passing it. In step 4 the answers approach 1 (about 1.0517, 1.0050 and 1.0005). In step 5 most people find 2.718.
Just for funThe digits of e begin 2.7 1828 1828, which makes it look as if it repeats. It doesn't: the next digits are 459045. Being irrational, e never settles into a repeating pattern, so it makes a terrible bank PIN and a great trivia answer.
Your turn
1Differentiate f(x) = 4eˣ + x³.
2Differentiate g(x) = eˣ − 7x + π.
g′(x) = eˣ − 7 (π is a constant, so its derivative is 0).
3Find the slope of y = eˣ at x = −1, exactly and to three decimal places.
y′ = eˣ, so the slope is e⁻¹ = 1/e ≈ 0.368.
4Find the equation of the tangent to y = eˣ at x = 0.
Point (0, 1), slope e⁰ = 1: y = x + 1.
5At what point on y = eˣ is the slope of the tangent equal to 5?
eˣ = 5, so x = ln 5 ≈ 1.609 (the ln key undoes eˣ; the next lesson explains why). The point is (ln 5, 5) ≈ (1.609, 5).
6Differentiate h(x) = e² + e^(x − 1). (Careful: one term is a constant.)
e² is a constant, so its derivative is 0. e^(x − 1) = e⁻¹eˣ, whose derivative is e^(x − 1). So h′(x) = e^(x − 1).
More practice: warm-up
1Give e to five decimal places.
3Evaluate e³ to two decimal places.
5True or false: the slope of y = eˣ at x = 4 equals e⁴.
True. The slope at every point equals the y-value, e⁴ ≈ 54.6.
More practice: on your own
1Find the equation of the tangent to y = 2eˣ at x = 1.
Point (1, 2e); slope y′ = 2eˣ = 2e at x = 1. y − 2e = 2e(x − 1), so y = 2ex (about y = 5.44x). It passes through the origin.
2For f(x) = eˣ − x, find f′(x) and the point where the tangent is horizontal.
f′(x) = eˣ − 1. Set it to 0: eˣ = 1, so x = 0. f(0) = 1 − 0 = 1. The tangent is horizontal at (0, 1).
3Find the second derivative of y = eˣ + x⁴.
y′ = eˣ + 4x³, so y″ = eˣ + 12x².
4Compute (1 + 1/n)ⁿ for n = 1000 and compare with e.
(1.001)¹⁰⁰⁰ ≈ 2.7169, about 0.0014 below e ≈ 2.7183.
5Show that y = 5eˣ satisfies the equation y′ = y.
y′ = 5eˣ, which is the same as y. ✓ (Any constant multiple of eˣ works.)
Challenge
1Find the point on y = eˣ where the tangent passes through the origin.
The tangent at (a, eᵃ) meets the x-axis at a − 1. For it to pass through the origin, which is on the x-axis, a − 1 = 0, so a = 1. The point is (1, e), and the tangent is y = ex.
2For f(x) = eˣ + e^(x + 1), show that f′(x) = (1 + e)eˣ.
e^(x + 1) = e × eˣ, so f(x) = eˣ + e eˣ = (1 + e)eˣ. The constant multiple rule gives f′(x) = (1 + e)eˣ (about 3.718eˣ).
3$1000 is invested at 5% per year compounded continuously, so A = 1000e^(0.05t). It can be shown (in the next lessons) that A′(t) = 50e^(0.05t). Find A(1) and A′(0), and explain why A′(0) is $50 per year.
A(1) = 1000e^0.05 ≈ $1051.27. A′(0) = 50e⁰ = $50 per year: at the start, the balance grows at 5% of $1000 per year, exactly as the interest rate says. The extra $1.27 after one year comes from interest earning interest continuously.
Check your understandingQuiz
1. The number e is defined in this lesson as the base a for which:
2. What is e to four decimal places?
3. If f(x) = 6eˣ + x², then f′(x) =
4. The equation of the tangent to y = eˣ at x = 0 is:
5. What is the slope of y = eˣ at the point where y = 12?
For parents and teachersBig idea (A2.6): among all exponential functions, exactly one has a derivative equal to itself, and its base is the irrational number e ≈ 2.71828. On the graph of y = eˣ, the slope at every point equals the height there.
How to help: have your teen show you the eˣ key on their calculator and compute (1 + 1/n)ⁿ for a large n. Ask: "Why is e not 2 or 3?" A good answer mentions that k = 0.693 for base 2 and 1.099 for base 3, so the base with k = 1 lies in between.
Common difficulties:
- Treating e like a variable. It is a constant, like π, so e² and e³ are just numbers whose derivatives are 0.
- Writing d/dx eˣ = xeˣ⁻¹ by mistake (the power rule does not apply).
- Rounding e too early (use the calculator's eˣ key, not 2.7).
- Forgetting to give exact answers such as e² when asked, and rounded answers only at the end.