The Quotient Rule: Study Guide (Calculus AB 2.9)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so fractions are written on one line. Brackets matter:

[g(x)f′(x) − f(x)g′(x)] / [g(x)]² means the whole numerator in square brackets divided by the square of g(x).

On paper, write it as a stacked fraction with a long bar. f′(x) is read “f prime of x” and means the derivative of f. Angles are in radians throughout.

Why you need a new rule

In Topic 2.8 you saw that the derivative of a product is not the product of the derivatives. The same warning applies to division. Try a quotient whose derivative you already know:

q(x) = x²/x, which equals x for every x ≠ 0, so q′(x) = 1.

If you divide the derivatives instead, you get 2x/1 = 2x. That is wrong for every x except x = 1/2. So (f/g)′ is not f′/g′. You need a rule that uses f, g and both derivatives.

The rule

Let f and g be differentiable at x, and suppose g(x) ≠ 0. Then the quotient q(x) = f(x)/g(x) is differentiable at x, and

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

In words: bottom times derivative of top, minus top times derivative of bottom, all over bottom squared. Some students remember it as “low d-high minus high d-low, over low squared”.

Three features to notice:

Feature	What it means for you
A minus sign in the numerator	Order matters. Start with the bottom function g, multiplied by f′. Swapping the terms gives the negative of the correct answer.
The denominator is [g(x)]²	Square the original bottom function. Do not differentiate it.
The condition g(x) ≠ 0	The quotient, and its derivative, exist only where the bottom is nonzero.

Where the rule comes from

You can build the quotient rule from the product rule. Suppose q = f/g is differentiable. Then

f(x) = q(x) · g(x)

Differentiate both sides with the product rule:

f′(x) = q′(x)g(x) + q(x)g′(x)

Solve for q′(x), then replace q(x) by f(x)/g(x):

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

Multiply top and bottom by g(x):

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

That is the quotient rule. (This argument assumes q is differentiable. A proof from the limit definition shows that it is, whenever f and g are differentiable and g(x) ≠ 0. You do not need that proof for this course.)

A special case: the reciprocal rule

If the top is the constant 1, then f′ = 0 and the rule shrinks to

d/dx [1/g(x)] = −g′(x) / [g(x)]²

For example, d/dx [1/(x² + 1)] = −2x/(x² + 1)².

When not to use the quotient rule

The quotient rule always works, but it is not always the best tool. Check the fraction first.

What you see	Better approach	Example
A constant on the bottom	Treat it as a constant multiple	(x² + 3)/5 = (1/5)(x² + 3), so the derivative is 2x/5
A constant on the top and a power of x on the bottom	Rewrite with a negative exponent	7/x³ = 7x⁻³, so the derivative is −21x⁻⁴ = −21/x⁴
A single-term bottom that divides each top term	Split and simplify, then use the power rule	See below
Anything else, such as (2x + 5)/(x² + 3) or (sin x)/x	Use the quotient rule	See Worked example 1

Simplify first, an example. Let k(x) = (x³ − 4√x)/x for x > 0. Divide each term by x:

k(x) = x² − 4x^(−1/2)

Now the power rule gives

k′(x) = 2x + 2x^(−3/2)

So k′(4) = 8 + 2/8 = 8 + 1/4 = 33/4. The quotient rule gives the same answer, but with more algebra to simplify.

Trig, exponential and log quotients usually call for the quotient rule. For example,

d/dx [(sin x)/x] = [x cos x − sin x] / x², for x ≠ 0.

Worked example 1: a derivative and a tangent line

Question. Let h(x) = (2x + 5)/(x² + 3). Find h′(x). Then find the equation of the line tangent to the graph of h at x = 1.

1. Name the parts. Top: f(x) = 2x + 5, so f′(x) = 2. Bottom: g(x) = x² + 3, so g′(x) = 2x. The bottom is never 0, so h is differentiable for every x.
2. Write the rule with the bottom first. h′(x) = [(x² + 3)(2) − (2x + 5)(2x)] / (x² + 3)²
3. Expand the numerator carefully. (x² + 3)(2) = 2x² + 6. (2x + 5)(2x) = 4x² + 10x. Subtract the whole second product: 2x² + 6 − 4x² − 10x = −2x² − 10x + 6.
4. State the derivative. Leave the denominator factored: h′(x) = (−2x² − 10x + 6) / (x² + 3)²
5. Evaluate at x = 1. h′(1) = (−2 − 10 + 6)/(1 + 3)² = −6/16 = −3/8.
6. Find the point. h(1) = (2 + 5)/(1 + 3) = 7/4.
7. Write the tangent line in point-slope form: y = 7/4 − (3/8)(x − 1)

Answer. h′(x) = (−2x² − 10x + 6)/(x² + 3)², and the tangent line is y = 7/4 − (3/8)(x − 1), which is the same as y = 17/8 − (3/8)x.

Check. A symmetric difference quotient with a small step, [h(1.001) − h(0.999)]/0.002, gives −0.375 to six decimal places, and −3/8 = −0.375.

Interpretation. At x = 1 the graph is falling. h′(x) = 0 where −2x² − 10x + 6 = 0, that is x² + 5x − 3 = 0, so x = (−5 ± √37)/2. The value x = (−5 + √37)/2 ≈ 0.54 gives the high point in Figure 1.

The graph agrees with the algebra: the curve rises while h′(x) > 0, levels off where h′(x) = 0 and falls where h′(x) < 0.

Worked example 2: using values from a table

Many exam questions give values of functions instead of formulas. You then use the rule with numbers.

Question. Functions f and g are differentiable, with

x	f(x)	f′(x)	g(x)	g′(x)
3	4	−2	2	5

(a) Let p(x) = f(x)/g(x). Find p′(3). (b) Let r(x) = g(x)/f(x). Find r′(3).

(a)

1. Check the condition. g(3) = 2, which is not 0, so the quotient rule applies at x = 3.
2. Write the rule, then substitute. p′(3) = [g(3)f′(3) − f(3)g′(3)] / [g(3)]².
3. Calculate. p′(3) = [(2)(−2) − (4)(5)] / 2² = (−4 − 20)/4 = −24/4 = −6.

(b)

1. The top and bottom have swapped. Now the top is g and the bottom is f. Check: f(3) = 4, which is not 0.
2. Write the rule with the new bottom first. r′(3) = [f(3)g′(3) − g(3)f′(3)] / [f(3)]².
3. Calculate. r′(3) = [(4)(5) − (2)(−2)] / 4² = (20 + 4)/16 = 24/16 = 3/2.

What the wrong methods give.

Error	Result for p′(3)
Dividing the derivatives: f′(3)/g′(3)	−2/5
Swapping the order in the numerator	+6
Forgetting to square the bottom	−12

Interpretation. p is decreasing at x = 3 (p′(3) < 0), while r is increasing there (r′(3) > 0). That makes sense: r = 1/p, so when p goes down, r goes up.

Check. The reciprocal rule links the answers. r = 1/p, so r′(3) = −p′(3)/[p(3)]². Here p(3) = 4/2 = 2, so r′(3) = −(−6)/4 = 3/2. The two methods agree.

Common misconceptions

• “(f/g)′ = f′/g′.” Dividing the derivatives is wrong. The example x²/x shows this: the true derivative is 1, but f′/g′ gives 2x.
• Swapping the order in the numerator. g f′ − f g′ and f g′ − g f′ differ by a minus sign. Always start with the bottom function times the derivative of the top.
• Differentiating the denominator. The bottom of the answer is [g(x)]², the original bottom squared. It is not g′(x), and not [g′(x)]².
• Losing a bracket. In [(x² + 3)(2) − (2x + 5)(2x)], the minus sign applies to the whole second product. Write the products in brackets, then expand.
• Expanding the denominator. Leave (x² + 3)² as it is. Expanding wastes time and hides useful factors.
• Using the quotient rule when the bottom is a constant. It is not wrong, but (x² + 3)/5 is just a constant multiple. Rewriting is quicker.
• Ignoring where the bottom is zero. The derivative of f/g does not exist where g(x) = 0, because the function itself is undefined there.
• “Simplifying” a quotient by cancelling terms. In (x² + 3)/(x + 3) you cannot cancel the 3s. Only whole factors cancel, as in Topic 1.6.

Where this leads

The quotient rule completes your set of rules for combining functions: sum, difference, constant multiple, product and quotient. Next, in Topic 2.10, you rewrite tan x as (sin x)/(cos x), and similar identities, and use the quotient rule to find the derivatives of tangent, cotangent, secant and cosecant. In Unit 3 you will combine the quotient rule with the chain rule. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap. If the product rule still feels shaky, review Topic 2.8 first.

Try the practice questions now, then use the revision notes and the checklist to consolidate.