§ 01What Is the Quotient Rule?
Whenever you need to differentiate a fraction where both the numerator and denominator contain the variable x, the quotient rule is your tool.
Functions like y = sin x ÷ x², y = (x² + 1) ÷ (x − 3), or y = eˣ ÷ cos x all share the same structure: a function of x sitting on top, divided by another function of x on the bottom. The denominator is not a plain constant, it contains x, and that changes everything.
Just as you cannot differentiate a product by simply multiplying the individual derivatives, you cannot differentiate a quotient by dividing the individual derivatives. The correct method accounts for how both the numerator and denominator are changing simultaneously, and the result, the quotient rule, is a formula with a very specific structure and a very specific sign that must not be reversed.
In words: differentiate the top, multiply by the bottom, then subtract the top times the derivative of the bottom, all divided by the bottom squared.
A classic memory rhyme used in many classrooms: "lo d-hi minus hi d-lo, over lo squared": where "hi" is the numerator (the high part of the fraction) and "lo" is the denominator (the low part).
§ 02Proof from First Principles
The quotient rule can be derived directly from the limit definition of the derivative, or more elegantly by treating the quotient as a product and applying the product rule. We show both approaches.
Method 1 — From the Limit Definition
We want to differentiate h(x) = u(x) ÷ v(x). Write the limit definition:
h′(x) = lim[Δx→0] [u(x+Δx)/v(x+Δx) − u(x)/v(x)] ÷ ΔxCombine the two fractions in the numerator over a common denominator of v(x+Δx)·v(x):
= lim[Δx→0] [u(x+Δx)·v(x) − u(x)·v(x+Δx)] ÷ [Δx · v(x+Δx) · v(x)]Add and subtract u(x)·v(x) in the numerator, this is the same algebraic trick used in the product rule proof. It introduces zero but breaks the expression into two recognisable difference-quotient shapes:
numerator = u(x+Δx)·v(x) − u(x)·v(x) − u(x)·v(x+Δx) + u(x)·v(x) = v(x)[u(x+Δx) − u(x)] − u(x)[v(x+Δx) − v(x)]Divide through by Δx and take the limit as Δx → 0. Each bracketed difference quotient converges to the corresponding derivative, and v(x+Δx) → v(x) since v is continuous:
[u(x+Δx) − u(x)] ÷ Δx → u′(x) [v(x+Δx) − v(x)] ÷ Δx → v′(x) v(x+Δx) → v(x)Assemble the final result. The denominator becomes v(x) · v(x) = [v(x)]²:
h′(x) = [v(x)·u′(x) − u(x)·v′(x)] ÷ [v(x)]² = (u′v − uv′) ÷ v² ✓ This confirms the formula: the minus sign comes from the algebra of the common denominator step. It is not arbitrary.Method 2 — From the Product Rule (Quicker)
There is a slicker derivation. Write y = u ÷ v as y = u · v⁻¹. Now differentiate using the product rule, treating v⁻¹ as the second factor and using the chain rule to differentiate it:
Rewrite the quotient as a product with a negative exponent:
y = u · v⁻¹Apply the product rule: d/dx[u · v⁻¹] = u′ · v⁻¹ + u · d/dx[v⁻¹]
Use the chain rule to find d/dx[v⁻¹]. The outer function is "something to the −1", and the inner function is v:
d/dx[v⁻¹] = −1 · v⁻² · v′ = −v′ ÷ v² Chain rule: outer derivative (−1·v⁻²) times inner derivative (v′).Substitute back and combine over a common denominator of v²:
dy/dx = u′/v − uv′/v² = u′v/v² − uv′/v² = (u′v − uv′) ÷ v² ✓ Same result. This approach is faster and reinforces that the quotient rule is really a special case of the product rule.Method 2 also shows something useful in practice: if the denominator is a simple power of x (like x² or x³) you can often avoid the quotient rule entirely by rewriting as a product with a negative power and applying the product rule directly. This is frequently cleaner.
§ 03How to Apply the Quotient Rule — Step by Step
A clear four-step routine keeps the sign correct and the algebra organised every time.
- Step 1: Identify u (numerator) and v (denominator).
- Step 2: Differentiate to find u′ and v′.
- Step 3: Apply the formula: dy/dx = (u′v − uv′) ÷ v².
- Step 4: Expand the numerator, collect like terms, and simplify. Cancel common factors if possible.
Step 1: u = sin x, v = x²
Step 2: u′ = cos x, v′ = 2x
Step 3: dy/dx = (u′v − uv′) ÷ v²
= (cos x · x² − sin x · 2x) ÷ x⁴
Step 4: Factor x from numerator:
= x(x cos x − 2 sin x) ÷ x⁴
= (x cos x − 2 sin x) ÷ x³
When to Use Quotient Rule vs. Rewriting as a Product
You have a genuine choice whenever the denominator is a power of x or a simple exponential. For example:
- y = (x² + 1) ÷ x³ can be split into y = x⁻¹ + x⁻² and differentiated term by term, no rule needed.
- y = sin x ÷ eˣ can be written y = sin x · e⁻ˣ and differentiated with the product rule.
- y = (x² + 3) ÷ (x − 1) is genuinely a case where the quotient rule (or long division followed by the power rule) makes more sense than rewriting.
Neither approach is wrong. Choose whichever leads to less algebra. If both seem equal, the quotient rule is usually faster.
Trig Functions as Quotients
Four of the six standard trigonometric derivatives are found directly by applying the quotient rule to their fractional definitions. This is worth understanding because it shows where those results come from, not just what they are.
u = sin x, u′ = cos x
v = cos x, v′ = −sin x
d/dx[tan x] = (cos x · cos x − sin x · (−sin x)) ÷ cos²x
= (cos²x + sin²x) ÷ cos²x
= 1 ÷ cos²x (using the identity sin²x + cos²x = 1)
= sec²x ✓
The same process applied to cot x = cos x ÷ sin x gives d/dx[cot x] = −cosec²x. The derivatives of sec x and cosec x also follow from the quotient rule applied to their definitions as 1 ÷ cos x and 1 ÷ sin x respectively. See the Trig Derivatives page for all four full derivations.
§ 04Common Mistakes
The quotient rule has four classic error patterns. Recognise them now and you will not fall into them under exam pressure.
The formula is (u′v − uv′) ÷ v². Many students write (uv′ − u′v) ÷ v² instead: reversing the order of the two terms in the numerator. This changes the sign of every term in the numerator and produces a completely wrong answer.
Write the memory phrase "lo d-hi minus hi d-lo" every time until the order is instinctive.
The denominator of the quotient rule is always v squared: the original denominator raised to the power 2. Students sometimes write just v in the denominator, especially when they are rushed.
When the numerator or denominator is itself a composite function, for example, y = sin(3x) ÷ x, you must use the chain rule to differentiate it. A student who writes u′ = cos(3x) without multiplying by the inner derivative 3 has made a chain rule error inside the quotient rule.
Before writing u′ or v′, always ask: is this a composite function? If yes, apply the chain rule to find it.
After applying the formula, students often make sign errors when expanding and collecting terms in the numerator. The most common culprit is distributing a negative sign incorrectly. For example, if u = x² + 1 and v = x − 3, then uv′ = (x² + 1)(1) = x² + 1. When this gets subtracted, it becomes −(x² + 1) = −x² − 1, not −x² + 1.
Write out the subtraction explicitly with brackets: (u′v) − (uv′), keeping the parentheses until you have fully expanded.
§ 05Worked Exercises — 12 Problems
Work each problem yourself before revealing the solution. Every step is fully explained, the algebra, the reasoning, and the simplification. Nothing is skipped.
Group A — Polynomial ÷ Polynomial (Exercises 1 – 3)
- The function is a fraction with x in both numerator and denominator, the denominator is not a constant. We need the quotient rule. If the denominator were a plain number like 5, we could just pull it out as a constant factor and use the power rule. But here the denominator changes with x.
- Label the numerator and denominator: u = x + 4 v = x − 2
- Differentiate each separately. For linear functions like these, the derivative is just the coefficient of x: u′ = d/dx[x + 4] = 1 v′ = d/dx[x − 2] = 1 The constants +4 and −2 both disappear when differentiated.
- Apply the quotient rule: dy/dx = (u′v − uv′) ÷ v². Substitute all four values: dy/dx = (1 · (x − 2) − (x + 4) · 1) ÷ (x − 2)²
- Expand the numerator carefully: numerator = (x − 2) − (x + 4) Distribute the minus: −(x + 4) = −x − 4. So: x − 2 − x − 4. = x − 2 − x − 4 = −6
- The x terms cancel, leaving a constant numerator: dy/dx = −6 ÷ (x − 2)² Interesting result: the derivative of any function of the form (x + a)÷(x + b) is always a constant over (x + b)². This is a useful pattern to recognise. Answer: dy/dx = −6 ÷ (x − 2)²
- Fraction with x in both parts, quotient rule needed. Label: u = x² + 5x v = 2x + 1
- Differentiate each: u′ = d/dx[x² + 5x] = 2x + 5 v′ = d/dx[2x + 1] = 2
- Apply the quotient rule: dy/dx = (u′v − uv′) ÷ v²: dy/dx = ((2x + 5)(2x + 1) − (x² + 5x) · 2) ÷ (2x + 1)²
- Expand (2x + 5)(2x + 1) using FOIL: (2x)(2x) + (2x)(1) + (5)(2x) + (5)(1) = 4x² + 2x + 10x + 5 = 4x² + 12x + 5
- Expand (x² + 5x) · 2: = 2x² + 10x
- Subtract the second expansion from the first (keeping brackets to stay safe): numerator = (4x² + 12x + 5) − (2x² + 10x) = 4x² + 12x + 5 − 2x² − 10x = 2x² + 2x + 5
- Write the final answer: Answer: dy/dx = (2x² + 2x + 5) ÷ (2x + 1)²
- Label the parts: u = x³ v = x² + 1
- Differentiate: u′ = 3x² v′ = 2x
- Apply the quotient rule: dy/dx = (3x² · (x² + 1) − x³ · 2x) ÷ (x² + 1)²
- Expand the numerator. First term: 3x²(x² + 1) = 3x⁴ + 3x². Second term: x³ · 2x = 2x⁴: numerator = (3x⁴ + 3x²) − 2x⁴ = 3x⁴ + 3x² − 2x⁴ = x⁴ + 3x²
- Factor x² from the numerator: x⁴ + 3x² = x²(x² + 3)
- Write the final answer: Answer: dy/dx = x²(x² + 3) ÷ (x² + 1)²
Group B — Trigonometric Quotients (Exercises 4 – 6)
- Label the parts: u = sin x v = x² + 1
- Differentiate each: u′ = cos x v′ = 2x
- Apply the quotient rule: dy/dx = (u′v − uv′) ÷ v²: dy/dx = (cos x · (x² + 1) − sin x · 2x) ÷ (x² + 1)²
- Expand the numerator. There are no like terms to combine here, but we can write it clearly: numerator = (x² + 1) cos x − 2x sin x We cannot simplify further: sin x and cos x are different functions and cannot be combined algebraically.
- Write the final answer: Answer: dy/dx = ((x² + 1) cos x − 2x sin x) ÷ (x² + 1)²
- Write the function using its definition as a quotient: y = cos x ÷ sin x
- Label the parts: u = cos x v = sin x
- Differentiate each. Be careful: the derivative of cos x is negative sin x: u′ = −sin x v′ = cos x
- Apply the quotient rule: dy/dx = (u′v − uv′) ÷ v²: dy/dx = (−sin x · sin x − cos x · cos x) ÷ sin²x
- Expand the numerator: −sin x · sin x = −sin²x −cos x · cos x = −cos²x numerator = −sin²x − cos²x
- Factor out −1 from the numerator: = −(sin²x + cos²x)
- Apply the Pythagorean identity: sin²x + cos²x = 1: = −1
- Write the full result and recognise the trig form: dy/dx = −1 ÷ sin²x = −cosec²x Recall cosec x = 1/sin x, so 1/sin²x = cosec²x. This is the standard result d/dx[cot x] = −cosec²x, now fully derived. Answer: d/dx[cot x] = −cosec²x
- Label the parts: u = x² − 1 v = cos x
- Differentiate each: u′ = 2x v′ = −sin x The derivative of cos x is −sin x. The minus sign must be carried through correctly.
- Apply the quotient rule: dy/dx = (u′v − uv′) ÷ v²: dy/dx = (2x · cos x − (x² − 1) · (−sin x)) ÷ cos²x
- Carefully distribute the double negative in the second term. Subtracting a negative gives a positive: −(x² − 1)(−sin x) = +(x² − 1) sin x Two negatives multiplied: minus times minus-sin x gives plus sin x times the bracket.
- Write the expanded numerator: numerator = 2x cos x + (x² − 1) sin x
- The denominator is cos²x. We can optionally rewrite using sec x = 1/cos x, but leaving it as cos²x is perfectly acceptable: Answer: dy/dx = (2x cos x + (x² − 1) sin x) ÷ cos²x
Group C — Exponential & Logarithm Quotients (Exercises 7 – 9)
- Label the parts: u = eˣ v = x + 3
- Differentiate each: u′ = eˣ (eˣ is its own derivative) v′ = 1
- Apply the quotient rule: dy/dx = (eˣ · (x + 3) − eˣ · 1) ÷ (x + 3)²
- Both terms in the numerator contain eˣ. Factorise it out: numerator = eˣ[(x + 3) − 1] = eˣ(x + 2)
- Write the final answer: Answer: dy/dx = eˣ(x + 2) ÷ (x + 3)²
- Label the parts: u = x² + 2x v = eˣ
- Differentiate each: u′ = 2x + 2 v′ = eˣ (eˣ is its own derivative)
- Apply the quotient rule: dy/dx = ((2x + 2) · eˣ − (x² + 2x) · eˣ) ÷ (eˣ)²
- Note that (eˣ)² = e²ˣ. Both terms in the numerator share a factor of eˣ: factorise it: numerator = eˣ[(2x + 2) − (x² + 2x)]
- Expand inside the bracket, distributing the minus sign carefully: = eˣ[2x + 2 − x² − 2x] = eˣ[2 − x²] The 2x and −2x cancel.
- The denominator is e²ˣ. Cancel eˣ from numerator and denominator (eˣ ÷ e²ˣ = e⁻ˣ = 1/eˣ): dy/dx = eˣ(2 − x²) ÷ e²ˣ = (2 − x²) ÷ eˣ Alternatively write as (2 − x²)e⁻ˣ. Both forms are equivalent. Answer: dy/dx = (2 − x²) ÷ eˣ
- Label the parts: u = ln x v = x
- Differentiate each: u′ = 1/x (standard result: d/dx[ln x] = 1/x) v′ = 1
- Apply the quotient rule: dy/dx = ((1/x) · x − ln x · 1) ÷ x²
- Simplify the first term in the numerator: (1/x) · x = 1: numerator = 1 − ln x
- Write the final answer: dy/dx = (1 − ln x) ÷ x² This is also a well-known result: worth remembering. You will encounter it again when finding the maximum of the function y = ln x / x, which occurs at x = e. Answer: dy/dx = (1 − ln x) ÷ x²
Group D — Quotient Rule with Chain Rule Inside (Exercises 10 – 12)
- Label the parts: u = sin(2x) v = x²
- Differentiate v, straightforward power rule: v′ = 2x
- Differentiate u, sin(2x) is a composite function (2x is inside the sine), so the chain rule is needed here: Outer function: sin(something) → derivative is cos(something). Inner function: 2x → derivative is 2. u′ = cos(2x) · 2 = 2 cos(2x)
- Apply the quotient rule: dy/dx = (u′v − uv′) ÷ v²: dy/dx = (2 cos(2x) · x² − sin(2x) · 2x) ÷ x⁴
- Factor 2x from both terms in the numerator: numerator = 2x[x cos(2x) − sin(2x)]
- Cancel the 2x factor with part of the x⁴ denominator: 2x ÷ x⁴ = 2 ÷ x³: dy/dx = 2[x cos(2x) − sin(2x)] ÷ x³ Answer: dy/dx = 2(x cos(2x) − sin(2x)) ÷ x³
- Label the parts: u = x v = e^(x²)
- Differentiate u: u′ = 1
- Differentiate v: e^(x²) is a composite function (x² is inside the exponential), so the chain rule applies: Outer function: e^(something) → derivative is e^(something). Inner function: x² → derivative is 2x. v′ = e^(x²) · 2x = 2x e^(x²)
- Apply the quotient rule: dy/dx = (1 · e^(x²) − x · 2x e^(x²)) ÷ (e^(x²))²
- Note that (e^(x²))² = e^(2x²). Factorise e^(x²) from the numerator: numerator = e^(x²)(1 − 2x²)
- Cancel e^(x²) from numerator and denominator (e^(x²) ÷ e^(2x²) = e^(−x²)): dy/dx = e^(x²)(1 − 2x²) ÷ e^(2x²) = (1 − 2x²) ÷ e^(x²) Answer: dy/dx = (1 − 2x²) ÷ e^(x²)
- Both the numerator and denominator are composite functions. We will need the chain rule for each, then assemble using the quotient rule. Label: u = e^(3x) v = sin(2x)
- Differentiate u using the chain rule. The outer function is e^(something); the inner function is 3x: Outer derivative: e^(3x). Inner derivative: 3. u′ = e^(3x) · 3 = 3e^(3x)
- Differentiate v using the chain rule. The outer function is sin(something); the inner function is 2x: Outer derivative: cos(2x). Inner derivative: 2. v′ = cos(2x) · 2 = 2 cos(2x)
- Apply the quotient rule: dy/dx = (u′v − uv′) ÷ v²: dy/dx = (3e^(3x) · sin(2x) − e^(3x) · 2 cos(2x)) ÷ sin²(2x)
- Factorise e^(3x) from both terms in the numerator: numerator = e^(3x)[3 sin(2x) − 2 cos(2x)]
- Write the final answer: dy/dx = e^(3x)(3 sin(2x) − 2 cos(2x)) ÷ sin²(2x) This cannot be simplified further. Both terms inside the bracket are different trig functions and cannot be combined. The answer is fully simplified. Answer: dy/dx = e^(3x)(3 sin(2x) − 2 cos(2x)) ÷ sin²(2x)
§ 0610-Question Quiz
Select your answer for each question, then click Submit Quiz to see your score with a full explanation for every question.
Quotient Rule — Test Yourself
The quotient rule for y = u ÷ v states that dy/dx equals:
Find d/dx[x ÷ (x + 1)].
What is d/dx[tan x] derived from the quotient rule?
Why is the denominator of the quotient rule v² and not v?
Find d/dx[eˣ ÷ x].
Find d/dx[ln x ÷ x²].
You want to differentiate y = (3x + 1) ÷ x². Which is the simplest correct approach?
When finding d/dx[sin(3x) ÷ x], what extra rule is needed for the numerator?
Find d/dx[x² ÷ sin x] at x = π/2.
A student writes d/dx[(x+1)÷(x−1)] = (1·(x−1) + (x+1)·1) ÷ (x−1)². What is their mistake?