A. Rate of Change · Derivatives of Exponential and Sinusoidal Functions

The Derivative of aˣ: Where ln a Comes From

A2.8 A2.7About 70 minutes

Learning goals

I can verify numerically that the constant k for f(x) = aˣ equals ln a.
I can derive d/dx aˣ = aˣ ln a by writing aˣ = e^(x ln a).
I can differentiate functions that contain aˣ, together with power and constant terms.
I can interpret the sign of ln a: growth when a > 1 and decay when 0 < a < 1.

Words to know

derivative of aˣ
d/dx aˣ = aˣ ln a, for a > 0, a ≠ 1.

growth factor
The base a in an exponential model; a > 1 means growth and 0 < a < 1 means decay.

decay
Decrease by a constant percentage per time unit, modelled by aˣ with 0 < a < 1.

numerical verification
Checking a rule by calculating with specific numbers, for example comparing a difference quotient with aˣ ln a.

ln a
The natural logarithm of the base. It is the slope of y = aˣ at x = 0.

Solving the mystery of k

In the first lesson of this unit you found that d/dx aˣ = k aˣ, and you estimated k for many bases: about 0.693 for base 2 and about 1.099 for base 3. In the last lesson you noticed that ln 2 ≈ 0.6931 and ln 3 ≈ 1.0986. Is that a coincidence? Expectation A2.8 asks you to check with technology, and then we will prove it.

base a	k from difference quotient (h = 0.0001)	ln a (calculator)
0.5	−0.6931	−0.6931
1.7	0.5306	0.5306
2	0.6932	0.6931
2.3	0.8329	0.8329
3	1.0987	1.0986
3.5	1.2528	1.2528
10	2.3029	2.3026

The two columns agree to three or four decimal places. A smaller h gives even closer agreement.

The values of k from the first lesson, plotted against the base, sit exactly on the curve y = ln a. At a = e the value is 1, as the lesson on e predicted.

The proof

The numbers strongly suggest k = ln a. Here is why it must be true.

From the last lesson, a = e^(ln a). So aˣ = (e^(ln a))ˣ = e^(x ln a).
Now differentiate e^(x ln a) with the chain rule from Derivatives and Differentiation Rules. The outer function is eᵘ (its derivative is eᵘ) and the inner function is u = x ln a (its derivative is the constant ln a).
So d/dx e^(x ln a) = e^(x ln a) × ln a.
Rewrite e^(x ln a) as aˣ again.

Four steps take you from aˣ to its derivative: change to base e, use the chain rule, then change back.

The derivative of an exponential function

d/dx aˣ = aˣ ln a (for a > 0, a ≠ 1).

Check with a = e: d/dx eˣ = eˣ ln e = eˣ × 1 = eˣ. ✓ The rule for eˣ is the special case where ln a = 1. That is exactly why e is the 'natural' base.

Growth or decay: the sign of ln a

If a > 1, then ln a > 0, so the derivative aˣ ln a is positive: the function is increasing (growth).
If 0 < a < 1, then ln a < 0, so the derivative is negative: the function is decreasing (decay).

For example, d/dx 0.5ˣ = 0.5ˣ ln 0.5 ≈ −0.693 × 0.5ˣ. Since ln 0.5 = ln(1/2) = −ln 2, this is the opposite of the constant for base 2. That explains the pattern you saw in the first lesson.

For a decay function such as 0.5ˣ, the derivative curve is below the x-axis: every slope is negative. At x = −2 the height is 4 and the slope is 4 ln 0.5 ≈ −2.77.

Worked example

Example 1: Basic derivatives

Differentiate (a) f(x) = 5ˣ (b) g(x) = 3(10ˣ) − x⁵ (c) h(x) = 2ˣ + x² + e²

(a) f′(x) = 5ˣ ln 5.
(b) Constant multiple rule and power rule: g′(x) = 3(10ˣ ln 10) − 5x⁴ = 3 ln 10 × 10ˣ − 5x⁴.
(c) e² is a constant, so its derivative is 0: h′(x) = 2ˣ ln 2 + 2x.

(a) 5ˣ ln 5 (b) 3 ln 10 × 10ˣ − 5x⁴ (c) 2ˣ ln 2 + 2x

Worked example

Example 2: A tangent to y = 2ˣ

Find the equation of the tangent to y = 2ˣ at x = 3.

Point: (3, 2³) = (3, 8).
Slope: y′ = 2ˣ ln 2, so m = 8 ln 2 ≈ 5.545.
y − 8 = 8 ln 2 (x − 3).
Expand: y = (8 ln 2)x − 24 ln 2 + 8 ≈ 5.545x − 8.636.

y = (8 ln 2)x + 8 − 24 ln 2, or about y = 5.545x − 8.636.

Worked example

Example 3: A decay rate

After a dose, the amount of a medication in a patient's bloodstream is A(t) = 50(0.9)ᵗ milligrams, t hours later. Find A′(3) and interpret it.

A′(t) = 50(0.9)ᵗ ln 0.9.
A(3) = 50(0.9)³ = 50 × 0.729 = 36.45 mg.
A′(3) = 36.45 × ln 0.9 ≈ 36.45 × (−0.10536) ≈ −3.84.

After 3 hours the amount is decreasing at about 3.84 mg per hour. (The negative sign means decay.)

Worked example

Example 4: Where is the slope 1?

Find the point on y = 2ˣ where the tangent has slope 1.

Set y′ = 1: 2ˣ ln 2 = 1, so 2ˣ = 1 ÷ ln 2 ≈ 1.4427.
Take ln of both sides: x ln 2 = ln(1 ÷ ln 2), so x = ln(1/ln 2) ÷ ln 2 ≈ 0.3665 ÷ 0.6931 ≈ 0.529.
y = 2^0.529 ≈ 1.443 (it equals 1/ln 2).

(0.529, 1.443), approximately.

Hands-on activity

Verify the rule with three bases of your choice

You need: a scientific calculator, a pencil and paper. A free graphing tool (such as an online graphing calculator) is optional.

Choose three bases: one between 0 and 1, one between 1 and 2, and one bigger than 5. (For example, 0.8, 1.25 and 7.)
For each base a, choose x = 1 and compute the difference quotient (a^(1.0001) − a) ÷ 0.0001.
Compute aˣ ln a at x = 1, which is a ln a. Compare with step 2.
Divide your step-2 answer by a. Compare with ln a.
Optional, with a graphing tool: graph y = aˣ and y = aˣ ln a for your bases. Add a slider for a and find the value where the two graphs are identical.

What to look for: steps 2 and 3 should agree to about three decimal places. For a = 0.8 you should get about −0.1785, for a = 1.25 about 0.2789 and for a = 7 about 13.62. The ratio in step 4 equals ln a (−0.223, 0.223 and 1.946 for these bases). With the slider, the graphs match only when a ≈ 2.718, which is e.

Just for fun

Bankers have a shortcut called the rule of 72 (or 70): divide 72 by the yearly interest rate in percent to estimate how many years it takes money to double. At 6%, that is about 12 years. The secret ingredient is ln 2 ≈ 0.693, the same number that makes 2ˣ differentiate the way it does.

Your turn

Differentiate y = 4ˣ.

Differentiate y = 7(3ˣ) + 2x³.

Differentiate y = (1/3)ˣ and say whether the function is increasing or decreasing.

Find the exact slope of y = 10ˣ at x = 2.

Verify numerically that the slope of y = 3ˣ at x = 2 is 9 ln 3. Use h = 0.0001.

For y = eˣ, use the rule d/dx aˣ = aˣ ln a to show the derivative is eˣ.

More practice: warm-up

What is d/dx 2ˣ?

What is the value of ln a when a = 1? Why is a = 1 excluded as a base?

Is d/dx 0.8ˣ positive or negative?

Find the slope of y = 5ˣ at x = 0.

Write the slope of y = 2ˣ at x = 1 exactly.

More practice: on your own

A savings account grows as A(t) = 1000(1.06)ᵗ dollars after t years. Find A′(10) and interpret it.

Find the equation of the tangent to y = 3ˣ at x = 0.

For f(x) = 2ˣ − 2x, find f′(x) and the x-value where the tangent is horizontal, to three decimal places.

Find f″(x) for f(x) = 5ˣ.

A town's population is P(t) = 500(1.03)ᵗ thousand people, t years after 2026. How fast is it growing in 2046?

Challenge

Find a formula for the nth derivative of f(x) = aˣ and use it to find the fourth derivative of 2ˣ at x = 0.

For which base a does the tangent to y = aˣ at x = 0 have slope 2?

Show that the graphs of f(x) = aˣ and its derivative intersect only when a = e. (Assume a > 0, a ≠ 1.)

Check your understanding

Quiz

d/dx 6ˣ =

The constant k for f(x) = aˣ (where f′(x) = k f(x)) is:

Which function has a negative derivative everywhere?

The slope of y = 2ˣ at x = 3 is:

To prove d/dx aˣ = aˣ ln a, the first step is to rewrite aˣ as:

If A(t) = 200(1.05)ᵗ, then A′(t) =

For parents and teachers

Big idea (A2.8, building on A2.7): the derivative of any exponential function is d/dx aˣ = aˣ ln a. The mysterious constant from the first lesson is just ln a. When a = e, ln e = 1, which is why eˣ is its own derivative.

How to help: pick a base, say 4, and have your teen compute the slope of y = 4ˣ at x = 1 two ways: with the difference quotient (4^1.0001 − 4) ÷ 0.0001 and with 4 ln 4. Both should be about 5.545.

Common difficulties:

Writing d/dx aˣ = aˣ ÷ ln a, or x aˣ⁻¹. Remind them: multiply by ln a.
Forgetting that ln a is a constant (so it stays as a factor and has no derivative of its own).
Missing that the derivative is negative for bases between 0 and 1.
Confusing ln a with log a (common log). Only the natural logarithm works in this rule.