Edexcel Paper 2 Specimen — Question 3 4 marks

Exam BoardEdexcel
ModulePaper 2 (Paper 2)
SessionSpecimen
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeShow derivative equals given algebraic form
DifficultyModerate -0.8 This is a straightforward product rule application with a chain rule component. Students must differentiate x(2x+1)^4, factor out (2x+1)^3, and identify constants by inspection. It's routine calculus with clear structure and no problem-solving insight required—easier than average A-level questions.
Spec1.07q Product and quotient rules: differentiation

3. Given \(y = x ( 2 x + 1 ) ^ { 4 }\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x + 1 ) ^ { n } ( A x + B )$$ where \(n , A\) and \(B\) are constants to be found.

Question 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Attempts product and chain rule on \(y = x(2x+1)^4\)M1 Apply product rule to reach \(\frac{dy}{dx} = (2x+1)^4 + Bx(2x+1)^3\)
\(\frac{dy}{dx} = (2x+1)^4 + 8x(2x+1)^3\)A1 Correct unsimplified derivative
Takes out common factor: \(\frac{dy}{dx} = (2x+1)^3\{(2x+1)+8x\}\)M1 Takes out common factor of \((2x+1)^3\)
\(\frac{dy}{dx} = (2x+1)^3(10x+1) \Rightarrow n=3, A=10, B=1\)A1 Form of answer is given; identify correct values
## Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts product and chain rule on $y = x(2x+1)^4$ | M1 | Apply product rule to reach $\frac{dy}{dx} = (2x+1)^4 + Bx(2x+1)^3$ |
| $\frac{dy}{dx} = (2x+1)^4 + 8x(2x+1)^3$ | A1 | Correct unsimplified derivative |
| Takes out common factor: $\frac{dy}{dx} = (2x+1)^3\{(2x+1)+8x\}$ | M1 | Takes out common factor of $(2x+1)^3$ |
| $\frac{dy}{dx} = (2x+1)^3(10x+1) \Rightarrow n=3, A=10, B=1$ | A1 | Form of answer is given; identify correct values |

---
3. Given $y = x ( 2 x + 1 ) ^ { 4 }$, show that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x + 1 ) ^ { n } ( A x + B )$$

where $n , A$ and $B$ are constants to be found.\\

\hfill \mbox{\textit{Edexcel Paper 2  Q3 [4]}}