| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find dy/dx at a point |
| Difficulty | Standard +0.3 This is a standard implicit differentiation question requiring the product rule and chain rule. Part (a) is routine substitution after differentiation. Part (b) requires setting dy/dx=0 and algebraic manipulation, which is slightly more challenging than typical textbook exercises but still follows a predictable method for C4 level. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(x^2\frac{dy}{dx} + 2xy\) | M1, A1 | Product rule used. Allow 1 error |
| \(+3y^2\frac{dy}{dx}\) | B1 | Chain rule |
| \(= 2\) | B1 | RHS and equation with no spurious \(\frac{dy}{dx}\) unless recovered |
| At \((2,1)\): \(4\frac{dy}{dx} + 4 + 3\frac{dy}{dx} = 2\) | M1 | Substitute \((2,1)\) |
| \(\frac{dy}{dx} = -\frac{2}{7}\) | A1 | CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{dy}{dx} = 0 \Rightarrow\) | M1 | Derivative \(= 0\) used |
| \(xy = 1\) | A1 | OE |
| \(x^2 \times \frac{1}{x} + \frac{1}{x^3} = 2x + 1\) | m1 | Use \(xy = k\) to eliminate \(y\) on LHS |
| \(\frac{1}{x^3} = x + 1\) | A1 | Answer given; CSO |
# Question 6:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $x^2\frac{dy}{dx} + 2xy$ | M1, A1 | Product rule used. Allow 1 error |
| $+3y^2\frac{dy}{dx}$ | B1 | Chain rule |
| $= 2$ | B1 | RHS and equation with no spurious $\frac{dy}{dx}$ unless recovered |
| At $(2,1)$: $4\frac{dy}{dx} + 4 + 3\frac{dy}{dx} = 2$ | M1 | Substitute $(2,1)$ |
| $\frac{dy}{dx} = -\frac{2}{7}$ | A1 | CSO | **Total: 6** |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{dy}{dx} = 0 \Rightarrow$ | M1 | Derivative $= 0$ used |
| $xy = 1$ | A1 | OE |
| $x^2 \times \frac{1}{x} + \frac{1}{x^3} = 2x + 1$ | m1 | Use $xy = k$ to eliminate $y$ on LHS |
| $\frac{1}{x^3} = x + 1$ | A1 | Answer given; CSO | **Total: 4** |
---6 A curve is defined by the equation $x ^ { 2 } y + y ^ { 3 } = 2 x + 1$.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the curve at the point $( 2,1 )$.
\item Show that the $x$-coordinate of any stationary point on this curve satisfies the equation
$$\frac { 1 } { x ^ { 3 } } = x + 1$$
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2009 Q6 [10]}}