A. Rate of Change · Derivatives and Differentiation Rules

The Quotient Rule and Rational Functions

A3.5About 70 minutes

Learning goals

I can rewrite a quotient as a product and differentiate it with the product and chain rules.
I can derive and use the quotient rule.
I can differentiate rational functions and use the derivative to find tangents and horizontal tangents.
I can interpret the derivative of a rational model in context.

Words to know

quotient rule
If q = f/g, then q′ = (f′g − fg′)/g², where g ≠ 0.

rational function
A function that is a quotient of two polynomials, such as (2x + 1)/(x − 3).

reciprocal
For g(x) ≠ 0, the function 1/g(x) = [g(x)]⁻¹.

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

Dividing functions

In MHF4U, unit Rational Functions and Equations, you graphed functions such as y = (2x + 1)/(x − 3), with vertical and horizontal asymptotes. How steep are these graphs? In the lesson on rational exponents, you could split a fraction like (x² + 3)/√x because its denominator had only one term. With a denominator like x − 3, splitting does not work.

There are two good approaches, and both give the same answer:

Rewrite the quotient as a product with a negative exponent, f(x) · [g(x)]⁻¹, and use the product and chain rules. This is the approach the Ontario curriculum emphasizes.
Use the quotient rule, a formula built from approach 1. It is used in most university calculus courses.

Deriving the quotient rule

Let q(x) = f(x)/g(x) = f(x) · [g(x)]⁻¹.

Product rule: q′ = f′ · g⁻¹ + f · (g⁻¹)′.

Chain rule: (g⁻¹)′ = −g⁻² · g′.

So q′ = f′/g − fg′/g². Over the common denominator g²:

q′(x) = [f′(x)g(x) − f(x)g′(x)] ÷ [g(x)]²

The quotient rule

If q(x) = f(x)/g(x) and g(x) ≠ 0, then

q′(x) = [f′(x)g(x) − f(x)g′(x)] / [g(x)]²

A memory aid: "low d-high minus high d-low, over the square of what's below". The order matters because of the minus sign: start with the derivative of the top.

Worked example

Example 1: Both methods

Differentiate y = (2x + 1)/(x − 3) (a) as a product and (b) with the quotient rule.

(a) y = (2x + 1)(x − 3)⁻¹. Product and chain rules: y′ = 2(x − 3)⁻¹ + (2x + 1)(−1)(x − 3)⁻²(1).
Common denominator (x − 3)²: y′ = [2(x − 3) − (2x + 1)]/(x − 3)² = (2x − 6 − 2x − 1)/(x − 3)² = −7/(x − 3)².
(b) Top f = 2x + 1, f′ = 2. Bottom g = x − 3, g′ = 1.
y′ = [2(x − 3) − (2x + 1)(1)]/(x − 3)² = −7/(x − 3)².

Both give y′ = −7/(x − 3)². Since this is negative for every x ≠ 3, the function is decreasing on each side of its asymptote.

y = (2x + 1)/(x − 3) has asymptotes x = 3 and y = 2 (dashed), as in MHF4U. Its derivative −7/(x − 3)² is negative for every x ≠ 3, so every tangent slopes down. Points equally far from x = 3 have equal slopes.

Worked example

Example 2: Horizontal tangents

Differentiate y = x²/(x² + 1) and find any horizontal tangents.

f = x², f′ = 2x; g = x² + 1, g′ = 2x.
y′ = [2x(x² + 1) − x²(2x)]/(x² + 1)² = (2x³ + 2x − 2x³)/(x² + 1)² = 2x/(x² + 1)².
y′ = 0 when the numerator is 0: 2x = 0, x = 0. (The denominator is never 0.)
y(0) = 0.

y′ = 2x/(x² + 1)², with one horizontal tangent at (0, 0), a local minimum (y′ changes from − to +).

Worked example

Example 3: A tangent line

Find the equation of the tangent to y = (x² − 1)/(x + 2) at x = 1.

y′ = [2x(x + 2) − (x² − 1)(1)]/(x + 2)² = (2x² + 4x − x² + 1)/(x + 2)² = (x² + 4x + 1)/(x + 2)².
At x = 1: y′ = 6/9 = 2/3. y(1) = 0/3 = 0.
Tangent: y − 0 = (2/3)(x − 1).

y = (2/3)x − 2/3 (or 2x − 3y − 2 = 0).

Worked example

Example 4: Medicine in the bloodstream

After a dose of a medicine, its concentration in a patient's blood is modelled by c(t) = 5t/(t² + 4) mg/L, where t is hours after the dose. (a) Find c′(t). (b) Find c′(1) and c′(4) and interpret them. (c) When is the concentration highest?

(a) c′(t) = [5(t² + 4) − 5t(2t)]/(t² + 4)² = (5t² + 20 − 10t²)/(t² + 4)² = (20 − 5t²)/(t² + 4)².
(b) c′(1) = 15/25 = 0.6 mg/L per hour (rising). c′(4) = (20 − 80)/400 = −0.15 mg/L per hour (falling).
(c) c′(t) = 0 when 20 − 5t² = 0, t² = 4, t = 2 (t ≥ 0). c′ > 0 before and < 0 after, so this is a maximum: c(2) = 10/8 = 1.25 mg/L.

(a) c′(t) = (20 − 5t²)/(t² + 4)² (b) 0.6 mg/L/h rising at 1 h; 0.15 mg/L/h falling at 4 h (c) at t = 2 h, 1.25 mg/L.

The concentration model c(t) = 5t/(t² + 4) mg/L. The quotient rule gives c′(t) = (20 − 5t²)/(t² + 4)²: positive before t = 2 h, zero at the peak, and negative afterwards as the body clears the medicine.

Choose the quickest tool

If the numerator is a constant, such as y = 3/(x² + 1), the quotient rule works but the chain rule is faster: y = 3(x² + 1)⁻¹, so y′ = −3(x² + 1)⁻²(2x) = −6x/(x² + 1)². If the denominator is a single power of x, divide first.

Just for fun

Some teachers like to joke that the quotient rule hardly deserves its own name, since it is "just" the product rule and chain rule wearing a trench coat. Either way, "low d-high minus high d-low" has rescued many tests.

Hands-on activity

Two routes, one answer

You need: paper, a pencil and a calculator. Work with a family member or friend if you can.

Choose y = (3x − 2)/(x + 1).
Person A differentiates it with the quotient rule. Person B rewrites it as (3x − 2)(x + 1)⁻¹ and uses the product and chain rules.
Compare simplified answers.
Check numerically: compute [y(1.001) − y(0.999)] ÷ 0.002 and compare with your formula at x = 1.
Bonus: rewrite y as 3 − 5/(x + 1) (do the division) and differentiate that. Which route was quickest?

What to look for: every route gives y′ = 5/(x + 1)², so y′(1) = 5/4 = 1.25. The centred estimate is about 1.25000. The bonus route (y′ = 5(x + 1)⁻²) is usually the quickest.

Your turn

Differentiate y = x/(x + 1).

Differentiate f(x) = (3x − 4)/(2x + 5).

Differentiate y = (x² + 1)/(x − 1).

Differentiate g(x) = 4/(x² − 9) using the chain rule.

Find the slope of y = x/(x² + 1) at x = 2.

Find where y = x/(x² + 1) has horizontal tangents.

Explain why the order of terms in the quotient rule matters but in the product rule it does not.

More practice: warm-up

State the quotient rule.

Rewrite (x + 2)/(x − 5) as a product.

Differentiate y = 1/x using the quotient rule.

What is the denominator of the derivative of f/g?

Differentiate y = (x² + x)/x for x ≠ 0 by simplifying first.

More practice: on your own

Find the equation of the tangent to y = (2x)/(x − 1) at x = 2.

Show that y = (x − 1)/(x + 1) is increasing on each interval of its domain.

Differentiate y = √x/(x + 1).

For c(t) = 5t/(t² + 4), find the average rate of change from t = 0 to t = 2 and compare with c′(0).

The average cost per item of making x items is A(x) = (2000 + 15x)/x. Find A′(100) and interpret it.

Challenge

Find the points on y = (x² + 3)/(x − 1) where the tangent is horizontal.

Find a and b if y = (ax + b)/(x + 2) passes through (0, 1) and has slope 1 at x = 0.

Use the quotient rule to show that d/dx (xⁿ) = nxⁿ⁻¹ when n is a negative integer, n = −m. (Write x⁻ᵐ = 1/xᵐ.)

Check your understanding

Quiz

Which is the quotient rule for (f/g)′?

What is the derivative of y = x/(x − 2)?

Which rewriting turns (x + 1)/(x² + 3) into a product?

For c(t) = 5t/(t² + 4), at what time is c′(t) = 0 (t ≥ 0)?

y′ = −7/(x − 3)² for y = (2x + 1)/(x − 3). What does this tell you about the graph?

What is the quickest way to differentiate y = 6/(x² + 2)?

For parents and teachers

Big idea (A3.5): a quotient f/g can be differentiated by rewriting it as f · g⁻¹ and using the product and chain rules, or with the quotient rule (f′g − fg′)/g². Both give the same result.

How to help: Have your teen differentiate one quotient both ways and compare. Ask them to say which part is the "top" and which is the "bottom" before starting. For rational functions from MHF4U, ask what the sign of the derivative says about the graph.

Common difficulties:

Reversing the order in the numerator (fg′ − f′g), which flips the sign.
Forgetting to square the denominator.
Expanding (x − 3)² in the denominator; it is usually best left factored.
Splitting a fraction over a sum, such as 6/(x² + 2) = 6/x² + 3, which is false.