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

Products and Quotients with Exponential Functions

A3.5About 70 minutes

Learning goals

I can use the product rule to differentiate products such as x eˣ and x² e^(−x).
I can differentiate quotients with exponential functions, using the quotient rule or by rewriting as a product.
I can factor derivatives that contain a common exponential factor, and use the fact that eˣ is never zero to find horizontal tangents.
I can interpret derivatives of product models, such as a drug concentration that rises and then falls.

Words to know

product rule
If h(x) = f(x)g(x), then h′(x) = f′(x)g(x) + f(x)g′(x).

quotient rule
If h(x) = f(x)/g(x), then h′(x) = [f′(x)g(x) − f(x)g′(x)] ÷ [g(x)]², where g(x) ≠ 0.

common factor
A factor that appears in every term, such as eˣ in eˣ + x eˣ, which can be factored out: eˣ(1 + x).

horizontal tangent
A tangent with slope 0, where f′(x) = 0.

concentration
The amount of a substance per unit of volume, for example milligrams per litre (mg/L).

When exponentials team up

After a dose of medicine, the amount in the bloodstream first rises as it is absorbed, then falls as the body clears it. A common model multiplies a rising factor (t) by a decaying factor (e^(−0.4t)). To find the rate of change of such a product, you need the product rule from Derivatives and Differentiation Rules, now with exponential functions inside.

Remember: (fg)′ = f′g + fg′, and (f/g)′ = (f′g − fg′)/g². A quotient can also be written as a product with a negative exponent, the approach the Ontario curriculum emphasizes.

Worked example

Example 1: y = x eˣ

Differentiate y = x eˣ and find the point where the tangent is horizontal.

Let f = x (f′ = 1) and g = eˣ (g′ = eˣ).
Product rule: y′ = 1 × eˣ + x × eˣ = eˣ + x eˣ.
Factor out the common eˣ: y′ = eˣ(1 + x).
Set y′ = 0. Since eˣ > 0 for every x, only 1 + x = 0 works: x = −1.
y(−1) = −1 × e⁻¹ = −1/e ≈ −0.368.

y′ = eˣ(1 + x); horizontal tangent at (−1, −1/e).

The graph of y = x eˣ falls until x = −1, has a horizontal tangent at its lowest point (−1, −1/e), then rises. The sign of y′ = eˣ(1 + x) matches what the graph does.

Factor out the exponential

Derivatives of products with eˣ or e^(kx) almost always have the exponential in every term. Factor it out. Since an exponential is never zero (and always positive), the sign of the derivative and its zeros come only from the other factor.

Worked example

Example 2: y = x² e^(−x)

Differentiate y = x² e^(−x). Find all points with horizontal tangents.

f = x², f′ = 2x. g = e^(−x), g′ = −e^(−x) (chain rule).
y′ = 2x e^(−x) + x²(−e^(−x)) = 2x e^(−x) − x² e^(−x).
Factor: y′ = x e^(−x)(2 − x).
y′ = 0 when x = 0 or x = 2.
y(0) = 0 and y(2) = 4e⁻² ≈ 0.541.

y′ = x e^(−x)(2 − x); horizontal tangents at (0, 0) and (2, 4/e²) ≈ (2, 0.541).

Worked example

Example 3: A quotient, two ways

Differentiate y = eˣ/x (x ≠ 0).

Quotient rule: y′ = [eˣ × x − eˣ × 1] ÷ x² = eˣ(x − 1)/x².
As a product: y = eˣ x⁻¹. y′ = eˣ x⁻¹ + eˣ(−x⁻²) = eˣ/x − eˣ/x².
Common denominator: eˣ/x − eˣ/x² = (x eˣ − eˣ)/x² = eˣ(x − 1)/x². Same answer. ✓
Horizontal tangent where x − 1 = 0: at (1, e).

y′ = eˣ(x − 1)/x², with a horizontal tangent at (1, e).

Worked example

Example 4: Drug concentration

The concentration of a medication in the blood is C(t) = 5t e^(−0.4t) mg/L, t hours after a dose. Find C′(t), C′(1) and C′(5), and find when the concentration is greatest.

Product rule: C′(t) = 5e^(−0.4t) + 5t(−0.4e^(−0.4t)) = 5e^(−0.4t)(1 − 0.4t).
C′(1) = 5e^(−0.4)(0.6) ≈ 2.01 mg/L per hour (rising).
C′(5) = 5e^(−2)(−1) ≈ −0.68 mg/L per hour (falling).
C′(t) = 0 when 1 − 0.4t = 0, so t = 2.5 h. The concentration rises before this and falls after, so it is greatest at t = 2.5.
C(2.5) = 12.5e⁻¹ ≈ 4.60 mg/L.

The concentration peaks at about 4.60 mg/L, 2.5 hours after the dose. (Finding maximums like this is studied fully in Applications of Derivatives.)

C(t) = 5t e^(−0.4t) rises, levels off at t = 2.5 h and then falls. The tangents at t = 1, 2.5 and 5 have slopes about 2.01, 0 and −0.68.

Worked example

Example 5: Another base in a product

Differentiate y = x³ 2ˣ and find the slope at x = 1.

y′ = 3x² × 2ˣ + x³ × 2ˣ ln 2 = x² 2ˣ(3 + x ln 2).
At x = 1: 1 × 2 × (3 + ln 2) = 6 + 2 ln 2 ≈ 7.39.

y′ = x² 2ˣ(3 + x ln 2); slope 6 + 2 ln 2 ≈ 7.39 at x = 1.

Hands-on activity

Tangent lines by hand on a medication curve

You need: graph paper, a ruler, a pencil and a scientific calculator.

Make a table of C(t) = 5t e^(−0.4t) for t = 0, 0.5, 1, 1.5, …, 8. Plot the points and join them with a smooth curve.
At t = 1, lay the ruler so it just touches the curve without cutting across it. Draw the tangent. Pick two points on your line and compute its slope.
Repeat at t = 2.5 and t = 5.
Compare your three slopes with C′(1) ≈ 2.01, C′(2.5) = 0 and C′(5) ≈ −0.68.
Use C′(t) = 5e^(−0.4t)(1 − 0.4t) to find C′(0). Why is it the steepest rise of all?

What to look for: hand-drawn slopes usually land within about 0.3 of the exact values. The tangent at t = 2.5 should be level. C′(0) = 5 mg/L per hour: at the very start the concentration climbs fastest, because almost none of the medication has been cleared yet. (This is only a model; real dosing decisions are made by doctors and pharmacists.)

Just for fun

The product rule is the fairest rule in calculus: in (fg)′ = f′g + fg′, each factor gets exactly one turn being differentiated while the other one waits patiently.

Your turn

Differentiate y = x² eˣ.

Differentiate y = (3x − 1)e^(2x).

Differentiate y = e^(−x)/x² by writing it as a product.

Differentiate y = x/eˣ using the quotient rule.

Find the horizontal tangent of y = x e^(−x).

Find the slope of y = x eˣ at x = 2.

More practice: warm-up

State the product rule.

Differentiate y = 5x eˣ.

Can eˣ(x − 3) equal 0 when x = 1? Explain.

Factor: 4x³e^(2x) + 2x⁴e^(2x).

Differentiate y = eˣ(x + 1).

More practice: on your own

Find the equation of the tangent to y = x e^(2x) at x = 0.

Find all points where y = x² e^(−2x) has a horizontal tangent.

Differentiate y = (x² + 1)/eˣ and simplify.

A worker's productivity t hours into a shift is modelled by P(t) = 30t e^(−0.25t) units per hour. Find P′(2) and say whether productivity is rising or falling.

Differentiate y = eˣ/(eˣ + 1).

Challenge

For y = x eˣ, find y″ and y‴. Then guess a formula for the nth derivative.

Find a and b if y = (ax + b)eˣ has y(0) = 2 and y′(0) = 5.

Show that y = eˣ/(eˣ + 1) is increasing for every x, and find where its slope is greatest. (Use the answer to the last 'on your own' question.)

Check your understanding

Quiz

d/dx (x eˣ) =

Why can a factor like e^(−x) be ignored when solving f′(x) = 0?

Where does y = x² e^(−x) have horizontal tangents?

d/dx (eˣ/x) =

C(t) = 5t e^(−0.4t). When is C′(t) = 0?

d/dx (x³ 2ˣ) =

For parents and teachers

Big idea (A3.5): the product and quotient rules work with exponential functions exactly as with polynomials. Factoring out the exponential (which is never zero) makes it easy to find where the derivative is zero and where it is positive or negative.

How to help: ask your teen to say the product rule aloud while writing ("derivative of the first times the second, plus the first times derivative of the second"). Then ask them to factor the answer. Check that they never set eˣ = 0.

Common difficulties:

Differentiating each factor separately and multiplying: (x eˣ)′ ≠ 1 × eˣ.
Forgetting the chain-rule factor inside, as in d/dx e^(−x) = −e^(−x).
Sign errors in the quotient rule (the numerator is f′g − fg′, in that order).
Leaving answers unfactored, which hides the zeros. The curriculum's focus is applying the rules, not heavy simplification, but factoring out the exponential is worth the habit.