01 · Check the subtraction order
Differentiate 1/x for x ≠ 0 using the quotient rule.
Hint
Here u′ = 0 and v′ = 1.
Worked solution
y′ = (x·0 − 1·1)/x² = −1/x². The reverse subtraction would incorrectly give a positive derivative.
Understand · explore · practise
Quotient rule
Learn the quotient rule, including the subtraction order, squared denominator and original domain. Compare the two contributions to a quotient’s gradient, then practise with worked solutions.
Before you startProduct rule; chain rule; simplifying fractions
01 / Track numerator and denominator changes
A growing denominator contributes a minus sign.
If y = u/v, then y′ = (vu′ − uv′)/v², where v ≠ 0.
Both factors must be differentiable at the input, and the original denominator must be non-zero.
The same rule can be written y′ = u′/v − uv′/v². The first term measures the effect of changing the numerator. The second measures the effect of changing the denominator. Use the explorer to compare them.
Two effects on a quotientExplore
x = 0; u = 1; v = 2; y = 0.5.
Numerator contribution 0; denominator contribution −0.25; total gradient −0.25.
The denominator x + 2 grows at rate 1. With the positive numerators in this window, that denominator change contributes negatively to the quotient’s rate. The numerator change may outweigh it.
The graph shows −1 ≤ x ≤ 2, away from the excluded input x = −2. Bars show the two instantaneous rate contributions on a shared scale; gold is their sum and the tangent’s slope.
02 / Derive the rule from a product
Write the denominator as a negative power.
y = u v⁻¹Worked example
y′ = u′v⁻¹ + u(−v⁻²v′)
Use the product rule outside and the chain rule inside v⁻¹.
y′ = u′/v − uv′/v²
The denominator derivative creates the minus sign.
y′ = (vu′ − uv′)/v²
Put both terms over the common denominator v².
The order vu′ − uv′ matters: reversing it changes the sign of the derivative. Multiplying by the denominator and differentiating vy = u leads to the same rule.
Compare the two contributions to the quotient gradient
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Use a clear numerator and denominator
Keep brackets around the whole subtraction.
Differentiate (3x + 1)/(x − 2)Worked example
u = 3x + 1, u′ = 3; v = x − 2, v′ = 1
The original domain excludes x = 2.
y′ = [3(x − 2) − (3x + 1)]/(x − 2)²
Subtract the complete second bracket.
y′ = −7/(x − 2)², x ≠ 2
It is negative on each interval of the domain.
02 · Linear over linear
Differentiate (2x − 5)/(x + 3).
Hint
Expand 2(x + 3) − (2x − 5).
Worked solution
y′ = 11/(x + 3)², with x ≠ −3.
03 · A constant numerator
Differentiate 7/(3x + 2).
Hint
The denominator derivative is 3.
Worked solution
y′ = −21/(3x + 2)² for x ≠ −2/3. The chain rule on 7(3x + 2)⁻¹ is an equally good method.
04 / Look for a simpler form first
Cancelling a factor keeps the original missing point.
Differentiate (x² − 9)/(x − 3)Worked example
y = x + 3, but only for x ≠ 3
Cancel the factor x − 3 on the original domain.
y′ = 1 for x ≠ 3
The simplification does not define the original quotient at 3.
At x = 3 the original graph has a hole
There is no derivative of the original function at a point where it is undefined.
04 · Split a fraction
Differentiate (x + 1)/x² using a simpler expression.
Hint
Write x⁻¹ + x⁻².
Worked solution
y′ = −x⁻² − 2x⁻³ = −(x + 2)/x³, for x ≠ 0.
05 · A removable hole
Differentiate (x² − 4)/(x − 2) and explain what happens at x = 2.
Hint
The quotient equals x + 2 only when x ≠ 2.
Worked solution
The derivative is 1 for x ≠ 2. At x = 2 the original quotient is undefined; its simplified extension would be a different function there.
05 / Differentiate a powered denominator fully
A bracket in v can need the chain rule.
Differentiate (x + 1)/(2x − 3)²Worked example
u′ = 1; v′ = 4(2x − 3)
The denominator is the entire squared bracket.
y′ = [(2x − 3)² − 4(x + 1)(2x − 3)]/(2x − 3)⁴
Square all of v in the denominator of the quotient rule.
y′ = (−2x − 7)/(2x − 3)³, x ≠ 3/2
Cancel one common factor and preserve the exclusion.
06 · A squared denominator
Differentiate x/(x + 2)².
Hint
The denominator derivative is 2(x + 2).
Worked solution
y′ = [(x + 2)² − 2x(x + 2)]/(x + 2)⁴ = (2 − x)/(x + 2)³, for x ≠ −2.
06 / Use complete exponential and log derivatives
The quotient rule sits outside the other derivative rules.
Differentiate e²ˣ/(x + 1)Worked example
u′ = 2e²ˣ; v′ = 1
Include the exponent factor 2.
y′ = [(x + 1)2e²ˣ − e²ˣ]/(x + 1)²
Keep the subtraction order.
y′ = (2x + 1)e²ˣ/(x + 1)², x ≠ −1
Factor out e²ˣ.
07 · A logarithmic quotient
Differentiate (ln x)/x and state the domain.
Hint
Use x(1/x) − ln x in the numerator.
Worked solution
y′ = (1 − ln x)/x² for x > 0.
08 · Exponential over a square
Differentiate eˣ/x².
Hint
Factor the numerator before cancelling.
Worked solution
y′ = [x²eˣ − 2xeˣ]/x⁴ = (x − 2)eˣ/x³, for x ≠ 0.
07 / Simplify with an identity when useful
Keep the original forbidden angles.
Differentiate sin x/(1 + cos x)Worked example
u′ = cos x; v′ = −sin x
Angles are in radians.
y′ = [(1 + cos x)cos x + sin²x]/(1 + cos x)²
The two minus signs give a plus.
y′ = (1 + cos x)/(1 + cos x)² = 1/(1 + cos x)
Use sin²x + cos²x = 1.
Exclude x = (2k + 1)π, k an integer
The original denominator vanishes there.
09 · A trig numerator
Differentiate sin(3x)/x on its original domain.
Hint
Differentiate sin(3x) as 3cos(3x).
Worked solution
y′ = [3x cos(3x) − sin(3x)]/x², for x ≠ 0. The original quotient has no value at zero even though a continuous extension can be defined.
10 · Reverse the factors
Differentiate x/sin x.
Hint
The original domain excludes every integer multiple of π.
Worked solution
y′ = [sin x − x cos x]/sin²x, where x ≠ kπ for integer k.
08 / Choose rules for each part
A numerator can itself be a product.
Differentiate [(x + 1)eˣ]/(x² + 1)Worked example
u′ = (x + 2)eˣ; v′ = 2x
Use the product rule for u.
y′ = eˣ[(x + 2)(x² + 1) − 2x(x + 1)]/(x² + 1)²
Now apply the quotient rule.
y′ = eˣ(x³ − x + 2)/(x² + 1)²
The denominator is positive for all real x.
11 · A root numerator
Differentiate √(x + 1)/(x + 2), giving the original domain and where the derivative formula applies.
Hint
u′ = 1/[2√(x + 1)].
Worked solution
The original domain is x ≥ −1. For x > −1, y′ = −x/[2√(x + 1)(x + 2)²]. There is no finite derivative at the endpoint x = −1.
09 / Find a tangent or normal
Substitute into the derivative and original quotient separately.
Tangent to y = (x² + 1)/(x + 2) at x = 0Worked example
The point is (0, 1/2)
Evaluate the original quotient.
y′ = [2x(x + 2) − (x² + 1)]/(x + 2)²
Apply the quotient rule.
y′(0) = −1/4
Evaluate at the requested input.
y = 1/2 − x/4
Use the contact point and slope.
12 · A normal
Find the normal to y = (x² + 1)/(x + 1) at x = 1.
Hint
The point is (1, 1) and the tangent gradient is 1/2.
Worked solution
The normal slope is −2. Its equation is y − 1 = −2(x − 1), or y = 3 − 2x.
10 / Solve the numerator, then check the domain
A zero denominator is never a stationary-point solution.
Stationary points of y = (x² + 4)/(x + 1)Worked example
y′ = (x² + 2x − 4)/(x + 1)²
The domain excludes x = −1.
(x + 1)² = 5, so x = −1 ± √5
Both roots are in the domain.
y = −2 ± 2√5, with matching signs
Write y = (x + 1) − 2 + 5/(x + 1).
The left point is a maximum; the right is a minimum
The squared denominator is positive; read the numerator’s signs on each side.
13 · Logarithmic maximum
Find and classify the stationary point of y = (ln x)/x.
Hint
The domain is x > 0 and y′ = (1 − ln x)/x².
Worked solution
x = e and y = 1/e. The derivative changes from positive to negative, so this is a maximum.
14 · Check the denominator
For y = x/(x² + 1), find all stationary points and classify them.
Hint
The denominator never vanishes; y′ = (1 − x²)/(x² + 1)².
Worked solution
The stationary points are (−1, −1/2), a minimum, and (1, 1/2), a maximum. The derivative is negative outside [−1, 1] and positive between the two roots.
11 / Check the order and preserve the exclusions
The denominator is squared; the subtraction is not interchangeable.
- Try simplifying first, while recording the original domain.
- Identify the whole numerator u and whole denominator v.
- Use vu′ − uv′ above v².
- Apply chain and product rules inside u′ or v′ when needed.
- For stationary points, solve the derivative numerator and reject forbidden inputs.
Section 1 of 11 · Track numerator and denominator changes