| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of composite quotient/product |
| Difficulty | Moderate -0.3 This is a straightforward multi-part differentiation question testing standard application of product rule (part a), quotient rule (part b), and chain rule with logarithms (part c). All three parts are routine textbook exercises requiring direct application of learned techniques with no problem-solving insight needed. Part (b) being a 'show that' provides additional scaffolding. Slightly easier than average due to its mechanical nature. |
| 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 | Mark | Guidance |
| \(\frac{dy}{dx} = m(4x+1)^3\cos 2x + n(4x+1)^2\sin 2x\) | M1 | \(m, n \neq 0\) |
| \(m = 2\) and \(n = 4 \times 3 = 12\) | A1 | isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{(3x^2+4)4x - (2x^2+3)6x}{(3x^2+4)^2}\) | M1 | Or \((2x^2+3)(-1)(3x^2+4)^{-2}6x + (3x^2+4)^{-1}4x\) |
| \(= \frac{-2x}{(3x^2+4)^2}\) | A1 | Accept \(p = -2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2x^2+3} \times \text{their } b(i)\) | M1 | 'their \(b(i)\)' must be in the correct form \(\frac{kx}{(3x^2+4)^2}\); PI |
| \(\frac{dy}{dx} = \frac{(3x^2+4)}{(2x^2+3)} \times \frac{-2x}{(3x^2+4)^2}\) | A1 | isw |
| \(\left(= \frac{-2x}{(2x^2+3)(3x^2+4)}\right)\) |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = \frac{ax}{2x^2+3} - \frac{bx}{3x^2+4}\), \(a>0, b>0\) | M1 |
| \(a=4,\ b=6\) | A1 |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = m(4x+1)^3\cos 2x + n(4x+1)^2\sin 2x$ | M1 | $m, n \neq 0$ |
| $m = 2$ and $n = 4 \times 3 = 12$ | A1 | isw |
**Note:** If expanded, $\frac{dy}{dx} = (ax^2+bx+c)\sin 2x + (dx^3+ex^2+fx+1)2\cos 2x$; $a=192, b=96, c=12, d=64, e=48, f=12$ scores A1
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{(3x^2+4)4x - (2x^2+3)6x}{(3x^2+4)^2}$ | M1 | Or $(2x^2+3)(-1)(3x^2+4)^{-2}6x + (3x^2+4)^{-1}4x$ |
| $= \frac{-2x}{(3x^2+4)^2}$ | A1 | Accept $p = -2$ |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{2x^2+3} \times \text{their } b(i)$ | M1 | 'their $b(i)$' must be in the correct form $\frac{kx}{(3x^2+4)^2}$; PI |
| $\frac{dy}{dx} = \frac{(3x^2+4)}{(2x^2+3)} \times \frac{-2x}{(3x^2+4)^2}$ | A1 | isw |
| $\left(= \frac{-2x}{(2x^2+3)(3x^2+4)}\right)$ | | |
**Or** (using rules of logs): $y = \ln(2x^2+3) - \ln(3x^2+4)$
| $\frac{dy}{dx} = \frac{ax}{2x^2+3} - \frac{bx}{3x^2+4}$, $a>0, b>0$ | M1 | |
| $a=4,\ b=6$ | A1 | |
---1
\begin{enumerate}[label=(\alph*)]
\item Given that $y = ( 4 x + 1 ) ^ { 3 } \sin 2 x$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Given that $y = \frac { 2 x ^ { 2 } + 3 } { 3 x ^ { 2 } + 4 }$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { p x } { \left( 3 x ^ { 2 } + 4 \right) ^ { 2 } }$, where $p$ is a constant.
\item Given that $y = \ln \left( \frac { 2 x ^ { 2 } + 3 } { 3 x ^ { 2 } + 4 } \right)$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2016 Q1 [6]}}