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O-Level A-Maths 2021 Paper 2 Q3 – Differentiation Using Quotient Rule

Source: O-Level Additional Mathematics 2021 Paper 2 Question 3

Introduction

In O-Level A-Maths, quotient rule questions often test whether students can set up u and v correctly, differentiate carefully, and simplify neatly.
In this question, we differentiate a function in the form \(y=\frac{u}{v}\) and show the final simplified derivative exactly as done on the teacher’s board.

 

The Question

Given that:

\(y=\frac{x}{(2x+1)^{\frac12}}\)

show that:

\(\frac{dy}{dx}=\frac{x+1}{(2x+1)^{\frac32}}\)

 

Step-by-step working (Teacher’s method)

Setup:

\(y=\frac{u}{v},\quad u=x,\quad v=(2x+1)^{\frac12}\)

Step 1: Differentiate \(u\)

\(\frac{du}{dx}=1\)

Step 2: Differentiate \(v\)

\(\frac{dv}{dx}=\frac{1}{2}(2x+1)^{-\frac12}\cdot 2=(2x+1)^{-\frac12}\)

Step 3: Apply quotient rule

Quotient rule: \(\displaystyle \frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\)

\(\frac{dy}{dx}=\frac{(2x+1)^{\frac12}(1)-x(2x+1)^{-\frac12}}{\left((2x+1)^{\frac12}\right)^2}\)

Step 4: Simplify

Since: \(\left((2x+1)^{\frac12}\right)^2=2x+1\)

\(\frac{dy}{dx}=\frac{(2x+1)^{\frac12}-\frac{x}{(2x+1)^{\frac12}}}{2x+1}\)

Combine numerator (same denominator):

\((2x+1)^{\frac12}-\frac{x}{(2x+1)^{\frac12}}=\frac{2x+1-x}{(2x+1)^{\frac12}}=\frac{x+1}{(2x+1)^{\frac12}}\)

Step 5: Final simplification

\(\frac{dy}{dx}=\frac{\frac{x+1}{(2x+1)^{\frac12}}}{2x+1}=\frac{x+1}{(2x+1)^{\frac12}(2x+1)}=\frac{x+1}{(2x+1)^{\frac32}}\)

✅ Hence shown

\(\displaystyle \frac{dy}{dx}=\frac{x+1}{(2x+1)^{\frac32}}\)

 

Key concepts

ConceptWhy it matters
Quotient ruleCorrect structure for \(y=\frac{u}{v}\)
Chain ruleNeeded to differentiate \((2x+1)^{\frac12}\)
Index lawsCombine powers: \((2x+1)^{\frac12}(2x+1)=(2x+1)^{\frac32}\)
Careful algebraPrevents sign and fraction mistakes

 

Tips for students

  • Label first: Write \(u\) and \(v\) clearly before differentiating.
  • Use chain rule: Power rule and derivative of the inside for \(v=(2x+1)^{\frac12}\).
  • Simplify in indices: Express terms like \((2x+1)^{\frac32}\) to keep working clean.
  • Keep layout neat: Stick to the quotient rule structure to avoid missing terms.

 

For parents

Many students lose marks in differentiation not because they “don’t know calculus,” but due to small algebra slips while simplifying (O-Level A-Maths).
At MasterMaths, we train students to (1) set up the method correctly, (2) write clean working, and (3) simplify confidently—exactly what exam markers want to see.

Want your child to master O-Level A-Maths differentiation and score confidently in exams?
Book a Free Trial Lesson with MasterMaths today!
Frequently Asked Questions
\(y = \frac{u}{v}\)

Use the quotient rule whenever the function is written as a fraction \(\tfrac{u}{v}\). It prevents common errors that happen when students try to rewrite and apply the product rule incorrectly.

\(\frac{dv}{dx}\) for \(v = (2x+1)^{\tfrac12}\)

The most frequent mistake is forgetting the chain rule in \(\tfrac{dv}{dx}\), especially missing the “\(\cdot 2\)” from differentiating the inside term \(2x+1\).

\((2x+1)^{\tfrac12} \cdot (2x+1) = (2x+1)^{\tfrac12+1} = (2x+1)^{\tfrac32}\)

Combine powers using index laws: multiplying \((2x+1)^{\tfrac12}\) by \((2x+1)\) adds the exponents, giving \((2x+1)^{\tfrac32}\).