| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of product |
| Difficulty | Moderate -0.3 This is a straightforward multi-part differentiation question testing standard techniques: chain rule, product rule, implicit differentiation, and tangent equations. All parts are routine applications with no problem-solving required, making it slightly easier than average, though the variety of techniques prevents it from being significantly below average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = (4x^2 + 3x + 2)^{10}\) | ||
| \(\dfrac{dy}{dx} = 10(4x^2 + 3x + 2)^9(8x + 3)\) | M1, A1 | For \(f(x)(\;)^9\) where \(f(x) \neq k\) and is linear |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = x^2\tan x\) | M1 | Product rule |
| \(\dfrac{dy}{dx} = x^2\sec^2 x + 2x\tan x\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = 2y^3 + \ln y\) | ||
| \(\dfrac{dx}{dy} = 6y^2 + \dfrac{1}{y}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At \((2, 1)\): \(\dfrac{dx}{dy} = 6 + 1 = 7\) | M1 | |
| \(\dfrac{dy}{dx} = \dfrac{1}{7}\) | A1\(\checkmark\) | May be implied |
| \((y - 1) = \dfrac{1}{7}(x - 2)\) | A1 | OE |
## Question 6:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = (4x^2 + 3x + 2)^{10}$ | | |
| $\dfrac{dy}{dx} = 10(4x^2 + 3x + 2)^9(8x + 3)$ | M1, A1 | For $f(x)(\;)^9$ where $f(x) \neq k$ and is linear |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = x^2\tan x$ | M1 | Product rule |
| $\dfrac{dy}{dx} = x^2\sec^2 x + 2x\tan x$ | A1 | |
### Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 2y^3 + \ln y$ | | |
| $\dfrac{dx}{dy} = 6y^2 + \dfrac{1}{y}$ | B1 | |
### Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| At $(2, 1)$: $\dfrac{dx}{dy} = 6 + 1 = 7$ | M1 | |
| $\dfrac{dy}{dx} = \dfrac{1}{7}$ | A1$\checkmark$ | May be implied |
| $(y - 1) = \dfrac{1}{7}(x - 2)$ | A1 | OE |
**Total: 8 marks**6
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ when:
\begin{enumerate}[label=(\roman*)]
\item $y = \left( 4 x ^ { 2 } + 3 x + 2 \right) ^ { 10 }$;
\item $y = x ^ { 2 } \tan x$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } x } { \mathrm {~d} y }$ when $x = 2 y ^ { 3 } + \ln y$.
\item Hence find an equation of the tangent to the curve $x = 2 y ^ { 3 } + \ln y$ at the point $( 2,1 )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2007 Q6 [8]}}