| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Moderate -0.8 Part (a) is direct application of chain rule with standard trig derivative (2 marks). Part (b) is routine quotient rule application with straightforward algebraic simplification to reach given form (3 marks). Both are standard textbook exercises requiring only procedural knowledge with no problem-solving or insight needed. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 3\sec^2 3x\) | M1 | For \(\sec 3x\) |
| Correct answer | A1 | \(SC/3\sec^2 x\) B1 |
| Alternative: Use of product/Quotient rule | (M1) | Good attempt |
| \(\frac{3\cos^2 3x + 3\sin^2 3x}{\cos^2 3x}\) | (A1) | Correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{(2x+1)3 - 2(3x+1)}{(2x+1)^2} = \frac{6x+3-6x-2}{(2x+1)^2}\) | M1 | Use of quotient rule |
| Numerator correct | A1 | |
| \(= \frac{1}{(2x+1)^2}\) | A1 | AG (no errors) |
| Alternative: \(-2(3x+1)(2x+1)^{-2} + 3(2x+1)^{-1}\) | (M1A1) | |
| \(= \frac{1}{(2x+1)^2}\) | (A1) | |
| Alternative: \(y = \frac{3}{2} - \frac{1}{2}(2x+1)^{-1}\) | M1A1 | |
| \(\frac{dy}{dx} = (2x+1)^{-2}\) | A1 | |
| \(= \frac{1}{(2x+1)^2}\) | AG |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 3\sec^2 3x$ | M1 | For $\sec 3x$ |
| Correct answer | A1 | $SC/3\sec^2 x$ B1 |
| **Alternative:** Use of product/Quotient rule | (M1) | Good attempt |
| $\frac{3\cos^2 3x + 3\sin^2 3x}{\cos^2 3x}$ | (A1) | Correct |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{(2x+1)3 - 2(3x+1)}{(2x+1)^2} = \frac{6x+3-6x-2}{(2x+1)^2}$ | M1 | Use of quotient rule |
| Numerator correct | A1 | |
| $= \frac{1}{(2x+1)^2}$ | A1 | AG (no errors) |
| **Alternative:** $-2(3x+1)(2x+1)^{-2} + 3(2x+1)^{-1}$ | (M1A1) | |
| $= \frac{1}{(2x+1)^2}$ | (A1) | |
| **Alternative:** $y = \frac{3}{2} - \frac{1}{2}(2x+1)^{-1}$ | M1A1 | |
| $\frac{dy}{dx} = (2x+1)^{-2}$ | A1 | |
| $= \frac{1}{(2x+1)^2}$ | AG | |
---1
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ when $y = \tan 3 x$.\\
(2 marks)
\item Given that $y = \frac { 3 x + 1 } { 2 x + 1 }$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { ( 2 x + 1 ) ^ { 2 } }$.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2006 Q1 [5]}}