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

The Chain Rule with Exponential Functions

A3.5 A2.6 A2.8About 70 minutes

Learning goals

I can differentiate e^(g(x)) and a^(g(x)) using the chain rule.
I can differentiate powers of exponential expressions, such as (eˣ + 1)⁴.
I can explain why d/dt (A e^(kt)) = k A e^(kt) means the rate of change is proportional to the amount.
I can use these derivatives to find slopes, tangent lines and rates in growth and decay problems.

Words to know

chain rule
If y = f(g(x)), then y′ = f′(g(x)) × g′(x): the derivative of the outer function, evaluated at the inner function, times the derivative of the inner function.

inner function
The function inside another, such as x² in e^(x²).

outer function
The function applied last, such as eᵘ in e^(x²).

relative growth rate
The rate of change divided by the amount, P′ ÷ P. For P = A e^(kt) it is the constant k.

half-life
The time it takes for half of a decaying quantity to remain.

When the exponent is more than x

Real models rarely look like plain eˣ. A cooling cup of hot chocolate follows something like 20 + 70e^(−0.05t). Radioactive carbon decays as (0.5)^(t/5730). The exponent is a function of its own. To differentiate these, you combine this unit's new rules with the chain rule from Derivatives and Differentiation Rules.

Remember the chain rule: if y = f(g(x)), then y′ = f′(g(x)) × g′(x). In Leibniz notation, with u = g(x): dy/dx = (dy/du)(du/dx).

Chain rule versions for exponentials

d/dx e^(g(x)) = e^(g(x)) × g′(x)
d/dx a^(g(x)) = a^(g(x)) × ln a × g′(x)

In words: copy the exponential function exactly, then multiply by the derivative of the exponent (and by ln a if the base is not e).

For y = e^(x²), the inner function is u = x² and the outer function is eᵘ. Multiplying their derivatives gives 2x e^(x²).

The simplest case: e^(kx)

For y = e^(kx), the inner function is kx with derivative k. So

d/dx e^(kx) = k e^(kx).

The constant in the exponent comes down in front. That is why exponential models in science are usually written with base e: for P(t) = A e^(kt), you get P′(t) = k A e^(kt) = k P(t). The rate of change is k times the amount. A positive k means growth and a negative k means decay. The constant k is called the relative growth rate: for example, k = 0.03 means the quantity grows at 3% of its current size per unit of time.

e^(0.5x), eˣ and e^(2x) all pass through (0, 1), where their slopes are 0.5, 1 and 2: the number in front of x in the exponent.

Worked example

Example 1: Constants in the exponent

Differentiate (a) y = e^(3x) (b) y = 4e^(−0.5x) (c) y = e^(5 − 2x)

(a) Inner 3x has derivative 3: y′ = 3e^(3x).
(b) Inner −0.5x has derivative −0.5: y′ = 4 × (−0.5)e^(−0.5x) = −2e^(−0.5x).
(c) Inner 5 − 2x has derivative −2: y′ = −2e^(5 − 2x).

(a) 3e^(3x) (b) −2e^(−0.5x) (c) −2e^(5 − 2x)

Worked example

Example 2: A polynomial in the exponent

Differentiate f(x) = e^(x² − 2x) and find the point where the tangent is horizontal.

Inner u = x² − 2x, so u′ = 2x − 2.
f′(x) = (2x − 2)e^(x² − 2x).
Set f′(x) = 0. An exponential is never zero, so 2x − 2 = 0 and x = 1.
f(1) = e^(1 − 2) = e⁻¹ ≈ 0.368.

f′(x) = (2x − 2)e^(x² − 2x); horizontal tangent at (1, 1/e) ≈ (1, 0.368).

Worked example

Example 3: Another base

Differentiate y = 5^(2x + 1) and find the slope at x = 1.

Use d/dx a^(g(x)) = a^(g(x)) ln a × g′(x) with g(x) = 2x + 1, g′(x) = 2.
y′ = 2 ln 5 × 5^(2x + 1).
At x = 1: 2 ln 5 × 5³ = 250 ln 5 ≈ 402.4.

y′ = 2 ln 5 × 5^(2x + 1); slope 250 ln 5 ≈ 402.4 at x = 1.

Worked example

Example 4: Radiocarbon decay

Carbon-14 has a half-life of about 5730 years. The percent of the original carbon-14 left in a sample after t years is A(t) = 100(0.5)^(t/5730). Find the rate of decay after 1000 years.

Inner exponent t/5730 has derivative 1/5730.
A′(t) = 100(0.5)^(t/5730) × ln 0.5 × (1/5730).
A(1000) = 100(0.5)^(1000/5730) ≈ 88.61%.
A′(1000) ≈ 88.61 × (−0.6931) ÷ 5730 ≈ −0.01072.

After 1000 years, the amount is decreasing by about 0.0107 percentage points per year, roughly 1% of the original every century.

Worked example

Example 5: A power of an exponential expression

Differentiate g(x) = (eˣ + 1)⁴ and evaluate g′(0).

Outer function u⁴, inner u = eˣ + 1, with u′ = eˣ.
g′(x) = 4(eˣ + 1)³ × eˣ.
g′(0) = 4(1 + 1)³ × 1 = 32.

g′(x) = 4eˣ(eˣ + 1)³; g′(0) = 32.

Hands-on activity

Coin-flip decay

You need: 50 coins (or 50 two-sided counters, or 50 small pieces of paper marked on one side), a cup, a table, and paper and pencil.

Put all 50 coins in the cup, shake, and spill them on the table. Remove every coin that lands heads up. Record how many are left. This is one "half-life".
Put the remaining coins back in the cup and repeat. Keep going until no coins remain or you have done 6 rounds.
Plot round number n against coins left. Compare with the model N(n) = 50(0.5)ⁿ.
Find N′(n) = 50(0.5)ⁿ ln 0.5. Compute N′(1) and N′(3). On your graph, draw a tangent line at n = 1 and n = 3 by eye and estimate its slope. Do they agree with N′?
Explain why the coins disappear fastest at the start.

What to look for: your counts should be close to 50, 25, 12.5, 6.25, 3.1, 1.6, 0.8, though chance will make them wobble. N′(1) = 25 ln 0.5 ≈ −17.3 coins per round and N′(3) = 6.25 ln 0.5 ≈ −4.3 coins per round. The rate is proportional to how many coins are left, so it is largest when there are the most coins: the same reason a radioactive sample decays fastest when it is fresh.

Just for fun

Riddle: Why can't an exponential function ever reach zero?

Because every time it gets closer, it slows down by exactly the same percentage. It is the ultimate procrastinator.

Your turn

Differentiate y = e^(7x).

Differentiate y = 3e^(−2x) + x².

Differentiate y = e^(x³).

Differentiate y = 2^(3x).

Differentiate y = e^(√x). (Write √x as x^(1/2).)

Differentiate y = (e^(2x) − 3)⁵.

Find the slope of y = e^(3x) at x = 2.

More practice: warm-up

d/dx e^(4x) = ?

d/dx e^(−x) = ?

What is the inner function in e^(x² + 1)?

d/dx 10e^(0.1x) = ?

True or false: d/dx e^(x²) = e^(x²).

More practice: on your own

A cup of hot chocolate cools according to T(t) = 20 + 70e^(−0.05t), in °C, t minutes after it is poured. Find T′(10) and interpret it.

Find the equation of the tangent to y = e^(2x) at x = 1.

For f(x) = e^(−x²), find f′(x) and the x-value where the tangent is horizontal.

$1000 grows as A(t) = 1000e^(0.04t). Find A′(10) and the ratio A′(10) ÷ A(10).

Differentiate y = 3^(x²) and evaluate y′(1) exactly.

Challenge

Show that y = 4e^(−3x) satisfies y′ + 3y = 0.

Show that P(t) = A e^(kt) and Q(t) = A bᵗ describe the same model when k = ln b.

Differentiate y = e^(e^x) and find y′(0) exactly.

Check your understanding

Quiz

d/dx e^(5x) =

d/dx e^(x²) =

d/dx 2^(3x) =

If P(t) = 500e^(0.02t), then P′(t) ÷ P(t) equals:

A quantity follows Q(t) = 80e^(−0.3t). Which statement is true?

Where does y = e^(x² − 4x) have a horizontal tangent?

For parents and teachers

Big idea (A3.5, with A2.6 and A2.8): to differentiate an exponential with a function in the exponent, copy the exponential, then multiply by the derivative of the exponent (and by ln a when the base is not e). For P = A e^(kt), this gives P′ = kP: the rate is proportional to the amount.

How to help: ask your teen to point to the 'inner' and 'outer' parts of e^(x²) or 5^(2x + 1) before they differentiate. Saying "copy, then multiply by the derivative of the exponent" out loud helps.

Common difficulties:

Bringing the exponent down like the power rule: writing d/dx e^(x²) = x² e^(x² − 1).
Forgetting the chain-rule factor (for example, writing d/dx e^(3x) = e^(3x)).
Forgetting ln a for bases other than e.
Trying to solve e^(g(x)) = 0. An exponential is never zero, so only the other factor can be zero.