| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of quotient |
| Difficulty | Moderate -0.3 This is a straightforward application of standard differentiation rules (product rule for part a, quotient rule for part b) with no algebraic complications. Both parts are routine textbook exercises requiring direct application of learned techniques, making it slightly easier than the average A-level question which typically involves more steps or conceptual challenge. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d}{dx}(\ln(3x)) \rightarrow \frac{B}{x}\) for any constant \(B\) | M1 | Differentiates the \(\ln(3x)\) term to \(\frac{B}{x}\). Note that \(\frac{1}{3x}\) is fine for this mark. |
| Applies product rule \(vu'+uv'\): \(\ln(3x) \times 2x + x\) | M1, A1 | Applies product rule to \(x^2\ln(3x)\). If rule is quoted it must be correct. There must be some attempt to differentiate both terms. If rule is not quoted (or implied by their working) only accept answers of the form \(\ln(3x) \times Ax + x^2 \times \frac{B}{x}\) where A and B are non-zero constants |
| One term correct and simplified, either \(2x\ln(3x)\) or \(x\) | A1 | \(\ln 3x^{2x}\) and \(\ln(3x)2x\) are acceptable forms |
| Both terms correct and simplified: \(2x\ln(3x) + x\) | A1 | \(\ln(3x) \times 2x + x\), \(x(2\ln 3x+1)\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Applies quotient rule \(\frac{vu'-uv'}{v^2}\) | M1 | If formula is not quoted (or implied by their working) only accept answers of the form \(\frac{x^3 \times \pm A\cos(4x) - \sin(4x) \times Bx^2}{(x^3)^2 \text{ or } x^6 \text{ or } x^5 \text{ or } x^9}\) with \(B>0\) |
| Correct first term on numerator: \(x^3 \times 4\cos(4x)\) | A1 | |
| Correct second term on numerator: \(-\sin(4x) \times 3x^2\) | A1 | |
| Correct denominator \(x^6\); the \((x^3)^2\) needs to be simplified | A1 | |
| Fully correct simplified expression: \(\frac{4x\cos(4x)-3\sin(4x)}{x^4}\) | A1 | \(\frac{\cos(4x)4x-\sin(4x)3}{x^4}\) oe. Accept \(4x^{-3}\cos(4x) - 3x^{-4}\sin(4x)\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Writes \(\frac{\sin(4x)}{x^3}\) as \(\sin(4x) \times x^{-3}\) and applies product rule | M1, A1 | If formula is quoted it must be correct. If not quoted (nor implied by their working) only accept answers of the form \(x^{-3} \times A\cos(4x) + \sin(4x) \times \pm Bx^{-4}\) |
| One term correct, either \(x^{-3} \times 4\cos(4x)\) or \(\sin(4x) \times -3x^{-4}\) | A1 | |
| Both terms correct: \(x^{-3} \times 4\cos(4x) + \sin(4x) \times -3x^{-4}\) | A1 | |
| Fully correct expression: \(4x^{-3}\cos(4x) - 3x^{-4}\sin(4x)\) or \(4\cos(4x)x^{-3} - 3\sin(4x)x^{-4}\) oe | A1 | The negative must have been dealt with for the final mark |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}(\ln(3x)) \rightarrow \frac{B}{x}$ for any constant $B$ | M1 | Differentiates the $\ln(3x)$ term to $\frac{B}{x}$. Note that $\frac{1}{3x}$ is fine for this mark. |
| Applies product rule $vu'+uv'$: $\ln(3x) \times 2x + x$ | M1, A1 | Applies product rule to $x^2\ln(3x)$. If rule is quoted it must be correct. There must be some attempt to differentiate both terms. If rule is **not quoted (or implied by their working)** only accept answers of the form $\ln(3x) \times Ax + x^2 \times \frac{B}{x}$ where A and B are non-zero constants |
| One term correct and simplified, either $2x\ln(3x)$ or $x$ | A1 | $\ln 3x^{2x}$ and $\ln(3x)2x$ are acceptable forms |
| Both terms correct and simplified: $2x\ln(3x) + x$ | A1 | $\ln(3x) \times 2x + x$, $x(2\ln 3x+1)$ oe | **(4 marks)** |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Applies quotient rule $\frac{vu'-uv'}{v^2}$ | M1 | If formula is **not quoted (or implied by their working)** only accept answers of the form $\frac{x^3 \times \pm A\cos(4x) - \sin(4x) \times Bx^2}{(x^3)^2 \text{ or } x^6 \text{ or } x^5 \text{ or } x^9}$ with $B>0$ |
| Correct first term on numerator: $x^3 \times 4\cos(4x)$ | A1 | |
| Correct second term on numerator: $-\sin(4x) \times 3x^2$ | A1 | |
| Correct denominator $x^6$; the $(x^3)^2$ needs to be simplified | A1 | |
| Fully correct simplified expression: $\frac{4x\cos(4x)-3\sin(4x)}{x^4}$ | A1 | $\frac{\cos(4x)4x-\sin(4x)3}{x^4}$ oe. Accept $4x^{-3}\cos(4x) - 3x^{-4}\sin(4x)$ oe | **(5 marks)** |
**Alternative method using product rule:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Writes $\frac{\sin(4x)}{x^3}$ as $\sin(4x) \times x^{-3}$ and applies product rule | M1, A1 | If formula is quoted it must be correct. If **not quoted (nor implied by their working)** only accept answers of the form $x^{-3} \times A\cos(4x) + \sin(4x) \times \pm Bx^{-4}$ |
| One term correct, either $x^{-3} \times 4\cos(4x)$ or $\sin(4x) \times -3x^{-4}$ | A1 | |
| Both terms correct: $x^{-3} \times 4\cos(4x) + \sin(4x) \times -3x^{-4}$ | A1 | |
| Fully correct expression: $4x^{-3}\cos(4x) - 3x^{-4}\sin(4x)$ or $4\cos(4x)x^{-3} - 3\sin(4x)x^{-4}$ oe | A1 | The negative must have been dealt with for the final mark |
**(9 MARKS TOTAL)**Differentiate with respect to $x$, giving your answer in its simplest form,
\begin{enumerate}[label=(\alph*)]
\item $x ^ { 2 } \ln ( 3 x )$
\item $\frac { \sin 4 x } { x ^ { 3 } }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2012 Q1 [9]}}