Quotient Rule

The quotient rule finds the derivative of one function divided by another. For \frac{f(x)}{g(x)}, the derivative is \frac{f'g - fg'}{g^2}: the bottom times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared. The order in the numerator matters.

Before this lesson: Product Rule, Power Rule and Zero and Negative Exponents

The idea behind the quotient rule

A fraction changes when its top changes and when its bottom changes. If the top grows, the fraction grows. If the bottom grows, the fraction shrinks. That is why the numerator of the rule has a minus sign: the change in the bottom works against the change in the top.

Dividing by the bottom squared scales the result. A change in a small denominator has a big effect, and a change in a large denominator has a small effect.

The quotient rule formula

\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{\left[g(x)\right]^2}, \qquad g(x) \ne 0

A memory trick: “low d-high minus high d-low, over the square of what’s below.” Here “high” is the numerator f, “low” is the denominator g, and “d” means “the derivative of.” Unlike the product rule, you can’t swap the two terms, because subtraction is not commutative.

1. Name the top f(x) and the bottom g(x).
2. Find f′(x) and g′(x).
3. Substitute into \frac{f'g - fg'}{g^2}.
4. Simplify the numerator. Usually leave the denominator squared, not expanded.

Quotient rule examples

Example 1: Two linear functions

Differentiate y = \frac{3x + 2}{x - 1}.

f(x) = 3x + 2, \; g(x) = x - 1

f'(x) = 3, \; g'(x) = 1

y' = \frac{3(x - 1) - (3x + 2)(1)}{(x - 1)^2}

y' = \frac{3x - 3 - 3x - 2}{(x - 1)^2}

y' = \frac{-5}{(x - 1)^2}

Answer: y' = -\frac{5}{(x - 1)^2}. The derivative is negative for every x ≠ 1, so the graph falls on both sides of its vertical asymptote.

Example 2: Quadratics on top and bottom

Differentiate y = \frac{x^2}{x^2 + 4}.

f(x) = x^2, \; g(x) = x^2 + 4

f'(x) = 2x, \; g'(x) = 2x

y' = \frac{2x(x^2 + 4) - x^2(2x)}{(x^2 + 4)^2}

y' = \frac{2x^3 + 8x - 2x^3}{(x^2 + 4)^2}

y' = \frac{8x}{(x^2 + 4)^2}

Answer: y' = \frac{8x}{(x^2 + 4)^2}.

Example 3: Tangent line to a quotient

Find the tangent line to f(x) = \frac{x^2 + 1}{x + 2} at x = 0.

f(0) = \frac{0 + 1}{0 + 2} = \frac{1}{2}

f'(x) = \frac{2x(x + 2) - (x^2 + 1)(1)}{(x + 2)^2}

f'(x) = \frac{2x^2 + 4x - x^2 - 1}{(x + 2)^2}

f'(x) = \frac{x^2 + 4x - 1}{(x + 2)^2}

f'(0) = \frac{-1}{4}

y - \frac{1}{2} = -\frac{1}{4}(x - 0)

y = -\frac{1}{4}x + \frac{1}{2}

Answer: the tangent line is y = -\frac{1}{4}x + \frac{1}{2}.

Example 4: Using a table of values

Suppose f(3) = 4, f′(3) = 2, g(3) = −2 and g′(3) = 5. Find the derivative of q(x) = \frac{f(x)}{g(x)} at x = 3.

q'(3) = \frac{f'(3)g(3) - f(3)g'(3)}{[g(3)]^2}

q'(3) = \frac{(2)(-2) - (4)(5)}{(-2)^2}

q'(3) = \frac{-4 - 20}{4}

q'(3) = \frac{-24}{4} = -6

Answer: q'(3) = -6.

Example 5: Word problem with a medicine

The concentration of a medicine in the blood t hours after a dose is C(t) = \frac{5t}{t^2 + 9} milligrams per liter. Find how fast the concentration is changing at t = 1 and at t = 5, and find when it is highest.

C'(t) = \frac{5(t^2 + 9) - 5t(2t)}{(t^2 + 9)^2}

C'(t) = \frac{5t^2 + 45 - 10t^2}{(t^2 + 9)^2}

C'(t) = \frac{45 - 5t^2}{(t^2 + 9)^2}

C'(1) = \frac{45 - 5}{10^2} = \frac{40}{100} = 0.4

C'(5) = \frac{45 - 125}{34^2} = \frac{-80}{1156} \approx -0.069

45 - 5t^2 = 0

t^2 = 9

t = 3

C(3) = \frac{15}{18} = \frac{5}{6} \approx 0.83

Answer: at 1 hour the concentration rises by about 0.4 mg/L per hour. At 5 hours it falls by about 0.069 mg/L per hour. It peaks at 3 hours, at about 0.83 mg/L.

Common mistakes

Related methods

Simplify before you differentiate

Sometimes you can divide first. For \frac{x^2 - 1}{x}, split it into x - x^{-1}, so the derivative is 1 + x^{-2}. That matches the quotient rule’s answer, \frac{x^2 + 1}{x^2}, after you split that fraction too.

Rewrite as a product

Any quotient is a product: \frac{f}{g} = f \cdot g^{-1}. Later, the chain rule lets you differentiate g^{-1} directly, and then the product rule gives the same result as the quotient rule. Use whichever is easier.

Where the formula comes from

Let q = \frac{f}{g}, so f = qg. The product rule gives f' = q'g + qg'. Solve for q′: q' = \frac{f' - qg'}{g}. Replace q with \frac{f}{g} and multiply the top and bottom by g to get \frac{f'g - fg'}{g^2}.

Practice problems

1. Differentiate y = \frac{x}{x + 1}.
2. Differentiate y = \frac{2x - 5}{3x + 1}.
3. Differentiate y = \frac{x^2 - 1}{x} with the quotient rule.
4. Differentiate y = \frac{5}{x^2 + 1}.
5. Differentiate y = \frac{x^2 + 3x}{x - 2}.
6. Find f′(2) for f(x) = \frac{x^3}{x + 1}.
7. If f(0) = 6, f′(0) = −3, g(0) = 2 and g′(0) = 1, find the derivative of f/g at x = 0.
8. Find the tangent line to y = \frac{x + 3}{x - 1} at x = 3.
9. Find every x where y = \frac{x}{x^2 + 16} has a horizontal tangent line.
10. Making x phone cases costs $2,000 in setup plus $15 per case, so the average cost per case is A(x) = \frac{2000 + 15x}{x} dollars. Find A′(100) and say what it means.

y' = \frac{1(x + 1) - x(1)}{(x + 1)^2}

y' = \frac{x + 1 - x}{(x + 1)^2}

y' = \frac{1}{(x + 1)^2}

y' = \frac{2(3x + 1) - (2x - 5)(3)}{(3x + 1)^2}

y' = \frac{6x + 2 - 6x + 15}{(3x + 1)^2}

y' = \frac{17}{(3x + 1)^2}

y' = \frac{2x \cdot x - (x^2 - 1)(1)}{x^2}

y' = \frac{2x^2 - x^2 + 1}{x^2}

y' = \frac{x^2 + 1}{x^2}

y' = \frac{0(x^2 + 1) - 5(2x)}{(x^2 + 1)^2}

y' = \frac{-10x}{(x^2 + 1)^2}

y' = \frac{(2x + 3)(x - 2) - (x^2 + 3x)(1)}{(x - 2)^2}

y' = \frac{2x^2 - x - 6 - x^2 - 3x}{(x - 2)^2}

y' = \frac{x^2 - 4x - 6}{(x - 2)^2}

f'(x) = \frac{3x^2(x + 1) - x^3(1)}{(x + 1)^2}

f'(x) = \frac{2x^3 + 3x^2}{(x + 1)^2}

f'(2) = \frac{16 + 12}{9} = \frac{28}{9}

\left(\frac{f}{g}\right)'(0) = \frac{(-3)(2) - (6)(1)}{2^2}

\left(\frac{f}{g}\right)'(0) = \frac{-6 - 6}{4} = -3

y(3) = \frac{6}{2} = 3

y' = \frac{(x - 1) - (x + 3)}{(x - 1)^2} = \frac{-4}{(x - 1)^2}

y'(3) = \frac{-4}{4} = -1

y - 3 = -1(x - 3)

y = -x + 6

y' = \frac{(x^2 + 16) - x(2x)}{(x^2 + 16)^2}

y' = \frac{16 - x^2}{(x^2 + 16)^2}

16 - x^2 = 0

x^2 = 16

x = \pm 4

A'(x) = \frac{15x - (2000 + 15x)(1)}{x^2}

A'(x) = \frac{-2000}{x^2}

A'(100) = \frac{-2000}{10000} = -0.2

Quotient rule FAQ

What is the quotient rule formula?

\frac{d}{dx}\left[\frac{f}{g}\right] = \frac{f'g - fg'}{g^2}. Take the bottom times the derivative of the top, subtract the top times the derivative of the bottom, and divide by the bottom squared.

Does the order matter in the quotient rule?

Yes. Switching the two terms in the numerator changes the sign of the answer. Always start with f′g, the derivative of the top times the bottom.

Should I expand the denominator?

Usually not. Leaving it as g², like (x - 1)^2, keeps the answer simpler and makes it easy to see where the derivative is undefined.

Can I use the product rule instead of the quotient rule?

Yes. Write \frac{f}{g} as f \cdot g^{-1} and use the product rule with the chain rule. Both methods give the same derivative.

Summary

• The quotient rule: \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}.
• Order matters: derivative of the top times the bottom comes first.
• Distribute the minus sign to every term, then combine like terms.
• Leave the denominator squared.
• If the bottom is a constant, or the fraction splits easily, simplify first and use the power rule.

Previous lesson: Product Rule

Next lesson: Chain Rule

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