A. Rate of Change · Derivatives of Exponential and Sinusoidal FunctionsThe Derivative of an Exponential Function Is Proportional to the Function
A2.5About 70 minutes
Learning goals- I can estimate the instantaneous rate of change of f(x) = aˣ at many points using difference quotients.
- I can show algebraically that the derivative of aˣ is k × aˣ, where k = f′(0).
- I can describe how the graph of f′(x) is related to the graph of f(x) = aˣ (a vertical stretch or compression).
- I can use the value of k to find slopes of tangents and points with a given slope.
Words to know
derivative function f′(x)
The function that gives the instantaneous rate of change (slope of the tangent) of f at every x where it exists.
exponential function
A function f(x) = aˣ with a constant base a > 0, a ≠ 1.
proportional
Two quantities are proportional when one is always a constant multiple of the other: y = kx.
constant of proportionality k
For f(x) = aˣ, the number k with f′(x) = k f(x). It equals f′(0), the slope of the tangent at the y-intercept.
difference quotient
[f(a + h) − f(a)] ÷ h, the slope of a secant. Its limit as h → 0 is f′(a).
Growth that feeds itself
A sourdough starter on a kitchen counter, money in a savings account, the number of people who have seen a video: the more there is, the faster it grows. That is the signature of an exponential function.
In the unit Derivatives and Differentiation Rules you learned to find derivatives of polynomials with the power rule. But the power rule does not apply to 2ˣ. In x², the variable is the base. In 2ˣ, the variable is the exponent. We need a brand-new rule, and in this lesson you will discover what it looks like.
Remember from MHF4U: in the Rates of Change unit of Advanced Functions, you estimated tangent slopes of y = 2ˣ and found that the slope at any point was about 0.693 times the y-value. Now you will prove why that happens and write it as a derivative.
Building a table of tangent slopes
To estimate the instantaneous rate of change of f(x) = 2ˣ at x = a, use a very small h in the difference quotient:
f′(a) ≈ [2^(a + h) − 2ᵃ] ÷ h, with h = 0.0001.
Doing this on a calculator for several values of a gives the table below. Look at the last column.
| x | f(x) = 2ˣ | f′(x) (estimated) | f′(x) ÷ f(x) |
|---|
| −2 | 0.25 | 0.1733 | 0.693 |
| −1 | 0.5 | 0.3466 | 0.693 |
| 0 | 1 | 0.6931 | 0.693 |
| 1 | 2 | 1.3863 | 0.693 |
| 2 | 4 | 2.7726 | 0.693 |
| 3 | 8 | 5.5452 | 0.693 |
What the table shows
The ratio f′(x) ÷ f(x) is constant, about 0.693. So for f(x) = 2ˣ,
f′(x) ≈ 0.693 × 2ˣ.
Three connections between f and f′ (this is what expectation A2.5 asks you to notice):
- f′ is also an exponential function with the same base 2.
- The ratio f′(x) ÷ f(x) is a constant k.
- The graph of f′ is a vertical stretch of the graph of f by the factor k (here k < 1, so it is really a compression toward the x-axis).
Why the ratio is constant: the algebra
Use the definition of the derivative and the exponent law a^(x + h) = aˣ × aʰ:
f′(x) = lim (h → 0) [a^(x + h) − aˣ] ÷ h
= lim (h → 0) [aˣ aʰ − aˣ] ÷ h
= lim (h → 0) aˣ (aʰ − 1) ÷ h
= aˣ × lim (h → 0) (aʰ − 1) ÷ h.
The factor aˣ comes out of the limit because it does not depend on h. The remaining limit does not depend on x at all: it is just a number. Call it k. Then
f′(x) = k aˣ = k f(x).
Put x = 0: f′(0) = k a⁰ = k. So k is the slope of the tangent at the y-intercept (0, 1). Once you know that one slope, you know every slope.
The key fact
For every exponential function f(x) = aˣ:
f′(x) = k × aˣ, where k = f′(0) = lim (h → 0) (aʰ − 1) ÷ h.
The rate of change is proportional to the current value. Double the value and you double the slope.
Different bases, different k
Repeating the estimate (aʰ − 1) ÷ h with h = 0.0001 for several bases gives these values of k, rounded to three decimal places:
| base a | 1.7 | 2 | 2.3 | 3 | 3.5 | 10 |
|---|
| k ≈ | 0.531 | 0.693 | 0.833 | 1.099 | 1.253 | 2.303 |
For a base between 0 and 1, such as f(x) = 0.5ˣ, the function decreases, so its slopes are negative. The same algebra works and k is negative: for 0.5ˣ, k ≈ −0.693. The graph of f′ is then a reflection of f in the x-axis combined with a vertical compression.
Notice that for base 0.5, k is the opposite of the value for base 2. That is not a coincidence, and the lesson The Derivative of aˣ: Where ln a Comes From in this unit will explain it.
Worked exampleExample 1: Estimating k for f(x) = 3ˣ
Estimate k = f′(0) for f(x) = 3ˣ, then write f′(x) and find f′(2).
- Compute (3ʰ − 1) ÷ h for shrinking h: h = 0.1 gives 1.1612; h = 0.01 gives 1.1047; h = 0.001 gives 1.0992; h = 0.0001 gives 1.0987.
- The values settle near k ≈ 1.099.
- So f′(x) ≈ 1.099 × 3ˣ.
- f′(2) ≈ 1.099 × 3² = 1.099 × 9 ≈ 9.89.
f′(x) ≈ 1.099 × 3ˣ, and the tangent at (2, 9) has slope about 9.89.
Worked exampleExample 2: Where is the slope 10?
For f(x) = 5ˣ, k ≈ 1.609. Find f′(2), and find the x-value where the tangent has slope 10.
- f′(x) ≈ 1.609 × 5ˣ.
- f′(2) ≈ 1.609 × 25 ≈ 40.2.
- Set the slope to 10: 1.609 × 5ˣ = 10, so 5ˣ = 10 ÷ 1.609 ≈ 6.215.
- Take the common logarithm of both sides (from MHF4U Exponential and Logarithmic Functions): x = log 6.215 ÷ log 5 ≈ 0.7934 ÷ 0.6990 ≈ 1.135.
f′(2) ≈ 40.2; the slope is 10 at x ≈ 1.135.
Worked exampleExample 3: A growing population
A bacteria culture in a lab has P(t) = 800(1.05)ᵗ cells after t hours. For base 1.05, k ≈ 0.0488. Find and interpret P′(10).
- P′(t) = 800 × k × 1.05ᵗ ≈ 800 × 0.0488 × 1.05ᵗ. (A constant multiple, 800, stays in front, just as in the constant multiple rule.)
- P(10) = 800 × 1.05¹⁰ ≈ 1303.1 cells.
- P′(10) ≈ 0.0488 × 1303.1 ≈ 63.6.
After 10 hours the culture is growing at about 64 cells per hour. Since P′ = kP, the growth rate is always about 4.88% of the current population per hour.
Hands-on activityBuild the derivative of 2ˣ with a calculator and graph paper
You need: a scientific calculator, a sheet of graph paper (or plain paper with a ruled grid), a ruler, a pencil and two different coloured pencils.
- Make a table with x = −2, −1, 0, 1, 2, 3.
- For each x, compute 2ˣ and the estimate (2^(x + 0.001) − 2ˣ) ÷ 0.001. Write both down to four decimal places.
- Divide the second number by the first. Write the ratio in a third column.
- On the grid, plot y = 2ˣ in one colour and the slope values in the other colour. Join each set of points with a smooth curve.
- Pick any x-value on your graph. Measure the height of each curve with your ruler and divide. You should get about 0.69 every time.
- Repeat steps 1 to 3 for 3ˣ. Is the ratio bigger or smaller than 1 now?
What to look for: every ratio for 2ˣ should be about 0.6934 (with h = 0.001 the estimate is a little above the true 0.6931). For 3ˣ the ratio is about 1.099, so the slope curve sits above y = 3ˣ. Your two curves for 2ˣ should never cross, and the slope curve should always be a bit lower.
Just for funAn exponential function walks into a job interview. "Where do you see yourself in five years?" It answers: "Growing at a rate proportional to wherever I am by then."
Your turn
1For f(x) = 2ˣ, use f′(x) ≈ 0.693 × 2ˣ to find f′(4).
2For f(x) = 3ˣ, f′(x) ≈ 1.099 × 3ˣ. What is the slope of the tangent at the y-intercept?
At x = 0: 1.099 × 1 = 1.099. (That is k itself.)
3Estimate k for f(x) = 4ˣ using h = 0.0001.
(4^0.0001 − 1) ÷ 0.0001 ≈ 1.386.
4Is the graph of the derivative of 10ˣ above or below the graph of 10ˣ? Explain.
k ≈ 2.303 > 1, so f′(x) = 2.303 × 10ˣ is above the graph of 10ˣ at every x (a vertical stretch by about 2.3).
5For f(x) = 0.5ˣ, k ≈ −0.693. Find f′(−1) and say what it means.
f′(−1) ≈ −0.693 × 0.5⁻¹ = −0.693 × 2 ≈ −1.386. The function is decreasing there, falling about 1.39 units per unit of x.
6Show that the derivative of f(x) = 7(2ˣ) is about 4.85 × 2ˣ.
The constant multiple rule gives f′(x) = 7 × 0.693 × 2ˣ ≈ 4.85 × 2ˣ.
More practice: warm-up
1True or false: the power rule gives the derivative of 2ˣ as x × 2ˣ⁻¹.
False. The power rule is only for xⁿ, where the base is the variable. In 2ˣ the exponent is the variable.
2For f(x) = aˣ, what is f′(0) called in this lesson?
The constant k (the constant of proportionality).
3If f′(x) = 0.693 f(x) and f(3) = 8, find f′(3).
4Which is larger: k for base 2.3 or k for base 3.5?
k for base 3.5 (about 1.253 compared with 0.833). A bigger base gives a bigger k.
5What kind of function is the derivative of an exponential function?
Another exponential function with the same base (a constant multiple of the original).
More practice: on your own
1For f(x) = 1.7ˣ, k ≈ 0.531. Find the slope of the tangent at x = 3.
f(3) = 1.7³ = 4.913, so f′(3) ≈ 0.531 × 4.913 ≈ 2.61.
2For f(x) = 2ˣ, find the x-value where the tangent has slope 2.
0.693 × 2ˣ = 2, so 2ˣ ≈ 2.886 and x = log 2.886 ÷ log 2 ≈ 1.53.
3An investment grows as A(t) = 2000(1.04)ᵗ dollars after t years. For base 1.04, k ≈ 0.0392. Find A′(5) and interpret it.
A(5) = 2000 × 1.04⁵ ≈ 2433.31. A′(5) ≈ 0.0392 × 2433.31 ≈ 95.4. After 5 years the value is growing at about $95 per year.
4Explain in one or two sentences why f′(x) ÷ f(x) does not depend on x for f(x) = aˣ.
Factoring gives f′(x) = aˣ × lim (aʰ − 1)/h. The limit involves only h and a, never x, so the ratio f′(x)/f(x) is that same number for every x.
5The graph of g′ is the graph of g(x) = aˣ stretched vertically by a factor of 1.25. Use the table of k values to estimate a.
k = 1.25 is closest to the value for a ≈ 3.5 (k ≈ 1.253).
Challenge
1Show that if f(x) = aˣ has constant k, then f(x) = a²ˣ = (a²)ˣ has constant 2k. (Hint: use the chain rule from Derivatives and Differentiation Rules, or compare the difference quotients.)
By the chain rule, d/dx a^(2x) = f′(2x) × 2 = k a^(2x) × 2 = 2k a^(2x). So the base a² has constant 2k. Check: base 2 has k ≈ 0.693 and base 4 has k ≈ 1.386 = 2 × 0.693. ✓
2Use the result above to explain why k for base 0.5 is the opposite of k for base 2.
0.5ˣ = 2^(−x). By the chain rule its derivative is k₂ × 2^(−x) × (−1) = −k₂ × 0.5ˣ. So k for base 0.5 is −k₂ ≈ −0.693.
3The difference quotient (2ʰ − 1)/h with h = 0.1 gives 0.7177, while h = −0.1 gives 0.6697. Explain why their average, 0.6937, is a better estimate of k than either one.
y = 2ˣ is concave up, so a secant to the right is steeper than the tangent and a secant to the left is less steep. Their errors are nearly equal and opposite, so the average (a centred difference) cancels most of the error: 0.6937 is within 0.0006 of k ≈ 0.6931.
Quiz
1. For f(x) = aˣ, which statement is true?
2. For f(x) = 2ˣ, the slope of the tangent at x = 3 is closest to:
3. For f(x) = aˣ, the constant k equals:
4. For which base is the graph of f′(x) above the graph of f(x) = aˣ?
5. If f(x) = aˣ and f′(x) = −0.4 f(x), then:
6. A population is modelled by P(t) = 500(1.1)ᵗ with k ≈ 0.0953 for base 1.1. P′(t) equals about:
Big idea (A2.5): for an exponential function, the rate of change at any moment is a fixed multiple of the current value: f′(x) = k f(x). That is why exponential models describe money, populations and radioactive decay so well.
How to help: ask your teen to explain, without notes, why the power rule does not work for 2ˣ (the variable is in the exponent). Then ask them to pick a value of x, compute (2^(x + 0.001) − 2ˣ) ÷ 0.001 on a calculator, divide by 2ˣ and show you that the answer is about 0.69 every time.
Common difficulties:
- Writing the derivative of 2ˣ as x × 2ˣ⁻¹ (using the power rule where it does not belong).
- Forgetting that k depends on the base but not on x.
- Arithmetic with tiny h: calculators need all the digits, so encourage them to store intermediate values rather than retype rounded ones.
- Mixing up f(x) and f′(x) when reading the two graphs: the derivative graph is the one that gives slopes, not heights of the original.