| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Moderate -0.3 This is a straightforward multi-part differentiation question testing standard techniques: chain rule for (i) and (ii), and quotient rule for (b). Part (b) requires algebraic simplification to reach the given form, but all steps are routine for C3 level with no novel problem-solving required. Slightly easier than average due to being purely procedural. |
| 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 |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(6\sin x\cos x + 2\sec 2x\tan 2x\) or \(3\sin 2x + 2\sec 2x\tan 2x\) | M1A1A1 | M1 for \(6\sin x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(3(x + \ln 2x)^2\left(1 + \frac{1}{x}\right)\) | B1M1A1 | B1 for \(3(x+\ln 2x)^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Differentiating numerator to obtain \(10x - 10\) | B1 | |
| Differentiating denominator to obtain \(2(x-1)\) | B1 | |
| Using quotient rule formula correctly | M1 | |
| \(\frac{dy}{dx} = \frac{(x-1)^2(10x-10)-(5x^2-10x+9)\cdot 2(x-1)}{(x-1)^4}\) | A1 | |
| Simplifying to form \(\frac{2(x-1)[5(x-1)^2-(5x^2-10x+9)]}{(x-1)^4}\) | M1 | |
| \(= -\frac{8}{(x-1)^3}\) * | A1 | c.s.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Using product rule formula correctly | M1 | |
| Obtaining \(10x - 10\) | B1 | |
| Obtaining \(-2(x-1)^{-3}\) | B1 | |
| \(\frac{dy}{dx} = (5x^2-10x+9)\{-2(x-1)^{-3}\} + (10x-10)(x-1)^{-2}\) | A1 cao | |
| Simplifying to form \(\frac{10(x-1)^2 - 2(5x^2-10x+9)}{(x-1)^3}\) | M1 | |
| \(= -\frac{8}{(x-1)^3}\) * | A1 | c.s.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Splitting fraction to give \(5 + \frac{4}{(x-1)^2}\) then differentiating | M1 B1 B1 M1 A1 A1 |
## Question 2:
### Part (a)(i):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $6\sin x\cos x + 2\sec 2x\tan 2x$ or $3\sin 2x + 2\sec 2x\tan 2x$ | M1A1A1 | M1 for $6\sin x$ |
### Part (a)(ii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $3(x + \ln 2x)^2\left(1 + \frac{1}{x}\right)$ | B1M1A1 | B1 for $3(x+\ln 2x)^2$ |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Differentiating numerator to obtain $10x - 10$ | B1 | |
| Differentiating denominator to obtain $2(x-1)$ | B1 | |
| Using quotient rule formula correctly | M1 | |
| $\frac{dy}{dx} = \frac{(x-1)^2(10x-10)-(5x^2-10x+9)\cdot 2(x-1)}{(x-1)^4}$ | A1 | |
| Simplifying to form $\frac{2(x-1)[5(x-1)^2-(5x^2-10x+9)]}{(x-1)^4}$ | M1 | |
| $= -\frac{8}{(x-1)^3}$ * | A1 | c.s.o. |
**Alternative (product rule):**
| Working/Answer | Marks | Guidance |
|---|---|---|
| Using product rule formula correctly | M1 | |
| Obtaining $10x - 10$ | B1 | |
| Obtaining $-2(x-1)^{-3}$ | B1 | |
| $\frac{dy}{dx} = (5x^2-10x+9)\{-2(x-1)^{-3}\} + (10x-10)(x-1)^{-2}$ | A1 cao | |
| Simplifying to form $\frac{10(x-1)^2 - 2(5x^2-10x+9)}{(x-1)^3}$ | M1 | |
| $= -\frac{8}{(x-1)^3}$ * | A1 | c.s.o. |
**Alternative (splitting fraction):**
| Working/Answer | Marks | Guidance |
|---|---|---|
| Splitting fraction to give $5 + \frac{4}{(x-1)^2}$ then differentiating | M1 B1 B1 M1 A1 A1 | |
---2. (a) Differentiate with respect to $x$
\begin{enumerate}[label=(\roman*)]
\item $3 \sin ^ { 2 } x + \sec 2 x$,
\item $\{ x + \ln ( 2 x ) \} ^ { 3 }$.
Given that $y = \frac { 5 x ^ { 2 } - 10 x + 9 } { ( x - 1 ) ^ { 2 } } , \quad x \neq 1$,\\
(b) show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 8 } { ( x - 1 ) ^ { 3 } }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2005 Q2 [12]}}