Differentiate quotients and preserve the original denominator exclusions
Keep the subtraction order and squared denominator, then evaluate gradients or rates on the correct connected domain interval.
Section 2.4 · Differentiation
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.4 and 3.4 cover exponential, natural logarithmic and trigonometric derivatives, composites, products and quotients, and first derivatives of parametric and implicit curves, including tangents and normals. Paper 3 additionally requires the derivative of inverse tangent; use its separate lesson for that method. Derivatives of inverse sine and inverse cosine are not required in section 3.4. Trigonometric differentiation uses radians. Keep original domains and check zero derivative denominators separately. Written practice is self-checked.
Before you begin
- Differentiate polynomial, log and radian trigonometric terms.
- Find original denominator restrictions and use point-gradient line equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the quotient-rule numerator for u/v?
What you will learn
- Apply (u/v)′ = (u′v − uv′)/v² and simplify without changing the original domain.
- Use quotient derivatives for curve lines or a supplied rate model and distinguish separate domain intervals.
Keep the numerator order and the denominator square
For y = u/v where v ≠ 0, the quotient rule is y′ = (u′v − uv′)/v². The numerator begins with the top derivative times the unchanged bottom, then subtracts the unchanged top times the bottom derivative. Reversing those two terms changes the sign. Deriving it as the product uv^(−1) also gives u′/v − uv′/v², the same rule on the original denominator domain.
For y = (2x − 4)/(3x + 2), the numerator is 2(3x + 2) − 3(2x − 4) = 16. Thus y′ = 16/(3x + 2)² for x ≠ −2/3. It is positive on each interval x < −2/3 and x > −2/3. This does not make the disconnected domain one unbroken increasing curve: crossing the excluded pole is not an interval on which the derivative argument applies.
The original expression determines allowed evaluation points
For y = sin x/x with radian x and x ≠ 0, the derivative is (x cos x − sin x)/x². A possible limiting extension at zero would be a separately defined function; the original quotient remains undefined there. For y = ln x/(x + 1), require x > 0, which also ensures the denominator is non-zero. Its derivative is [(x + 1)/x − ln x]/(x + 1)². At x = 1 the curve point is (1, 0) and its gradient is 1/2.
A supplied distance model q(t) = 10t/(t + 2) metres for t ≥ 0 seconds has derivative q′(t) = 20/(t + 2)² m/s. At t = 3, its distance is 6 metres and instantaneous speed is 4/5 m/s. The model's time domain is separate from the algebraic quotient exclusion t = −2; the non-negative domain already avoids that excluded input. Do not confuse a distance value with its speed.
Pause and explain
Since 16/(3x + 2)² is positive, what can be concluded about (2x − 4)/(3x + 2)?
Worked example
Differentiate y = (2x − 4)/(3x + 2) and state where the result applies.
Show the worked solution
- The original denominator requires x ≠ −2/3; use u′ = 2 and v′ = 3.
- Form the quotient numerator 2(3x + 2) − 3(2x − 4) = 16.
- Divide by (3x + 2)² to obtain y′ = 16/(3x + 2)² on the original domain.
Answer y′ = 16/(3x + 2)², with x ≠ −2/3; increasing on each separate domain interval.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Differentiate y = sin x/x in radians and state the original excluded input.
Give me a hint
Use u = sin x and v = x, retaining the quotient subtraction order.
Compare my reasoning
- The original function requires x ≠ 0, and the derivatives are u′ = cos x and v′ = 1.
- The quotient numerator is x cos x − sin x.
- Hence y′ = (x cos x − sin x)/x² for x ≠ 0; the original quotient is not defined at zero.
y′ = (x cos x − sin x)/x², with x ≠ 0.
Look for these in your work
- Apply the radian sine derivative and correct numerator order.
- Retain the original exclusion after forming the derivative.
Differentiate y = ln x/(x + 1) for x > 0 and find tangent and normal equations at x = 1.
Give me a hint
Use the quotient rule, then find the original point and evaluate the derivative there.
Compare my reasoning
- The derivative is [(x + 1)/x − ln x]/(x + 1)² on x > 0.
- At x = 1 the point is (1, 0) and the gradient is (2 − 0)/4 = 1/2.
- The tangent is y = (x − 1)/2 and the normal is y = −2(x − 1).
y′ = [(x + 1)/x − ln x]/(x + 1)²; tangent y = (x − 1)/2; normal y = −2(x − 1).
Look for these in your work
- Use the log derivative and original positive domain.
- Build both lines from the original point with perpendicular gradients.
A supplied distance model is q(t) = 10t/(t + 2) metres for t ≥ 0 seconds. Find its distance and instantaneous speed at t = 3.
Give me a hint
Use the original quotient for distance and its derivative for speed.
Compare my reasoning
- The quotient rule gives q′(t) = [10(t + 2) − 10t]/(t + 2)² = 20/(t + 2)².
- The original distance at t = 3 is 30/5 = 6 metres.
- The speed is 20/25 = 4/5 m/s; the evaluation time belongs to the supplied non-negative domain.
Distance 6 m; instantaneous speed 4/5 m/s at t = 3 s.
Look for these in your work
- Simplify both quotient-rule contributions before evaluating.
- Keep the quantity, derivative, units and model domain distinct.
Common mistakes
- Reversing the quotient numerator order reverses its sign.
- The denominator is squared in the derivative, but its original zeros stay excluded.
- A positive derivative on two separate intervals does not justify comparing across an excluded pole.
What can you explain now?
Differentiate y = x²/(x + 1), keeping its excluded input.
Compare with the explanation
y′ = x(x + 2)/(x + 1)², with x ≠ −1.
The numerator is 2x(x + 1) − x² = x² + 2x; the original zero denominator remains excluded.
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Make it stick
Write the quotient rule and verify its sign using a simple linear-over-linear example. Explain why an original denominator exclusion survives differentiation and why separated domain intervals matter.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.