| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find dy/dx at a point |
| Difficulty | Moderate -0.3 This is a straightforward implicit differentiation question requiring standard application of the chain rule and substitution. Part (a) is routine differentiation, and part (b) requires finding x-coordinate then substituting—slightly easier than average due to minimal algebraic complexity and being a standard two-part C4 question. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2y - 3 = 3x^2\frac{dx}{dy}\) | M1 | Differentiates implicitly to include \(\pm kx^2\frac{dx}{dy}\); ignore \(\left(\frac{dx}{dy} = \right)\) |
| Correct equation | A1 | |
| \(2y - 3 = 3x^2 \cdot \frac{1}{\left(\frac{dy}{dx}\right)}\) | dM1 | Applies \(\frac{dx}{dy} = \frac{1}{\left(\frac{dy}{dx}\right)}\) |
| \(\frac{dy}{dx} = \frac{3x^2}{2y-3}\) | A1 oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = (y^2 - 3y - 8)^{\frac{1}{3}}\) | — | |
| \(\frac{dx}{dy} = \frac{1}{3}(y^2 - 3y - 8)^{-\frac{2}{3}}(2y-3)\) | M1 | Differentiates in the form \(\frac{1}{3}(f(y))^{-\frac{2}{3}}(f'(y))\) |
| Correct differentiation | A1 | |
| \(\frac{dy}{dx} = \frac{3(y^2 - 3y - 8)^{\frac{2}{3}}}{2y-3}\) | dM1 | Applies \(\frac{dy}{dx} = \frac{1}{\left(\frac{dx}{dy}\right)}\) |
| \(\frac{dy}{dx} = \frac{3(x^3)^{\frac{2}{3}}}{2y-3} \Rightarrow \frac{dy}{dx} = \frac{3x^2}{2y-3}\) | A1 oe | \(\frac{3(x^3)^{\frac{2}{3}}}{2y-3}\) or \(\frac{3x^2}{2y-3}\) |
# Question 1 (Appendix):
## Part (a) — Aliter Way 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| $2y - 3 = 3x^2\frac{dx}{dy}$ | M1 | Differentiates implicitly to include $\pm kx^2\frac{dx}{dy}$; ignore $\left(\frac{dx}{dy} = \right)$ |
| Correct equation | A1 | |
| $2y - 3 = 3x^2 \cdot \frac{1}{\left(\frac{dy}{dx}\right)}$ | dM1 | Applies $\frac{dx}{dy} = \frac{1}{\left(\frac{dy}{dx}\right)}$ |
| $\frac{dy}{dx} = \frac{3x^2}{2y-3}$ | A1 oe | |
## Part (a) — Aliter Way 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = (y^2 - 3y - 8)^{\frac{1}{3}}$ | — | |
| $\frac{dx}{dy} = \frac{1}{3}(y^2 - 3y - 8)^{-\frac{2}{3}}(2y-3)$ | M1 | Differentiates in the form $\frac{1}{3}(f(y))^{-\frac{2}{3}}(f'(y))$ |
| Correct differentiation | A1 | |
| $\frac{dy}{dx} = \frac{3(y^2 - 3y - 8)^{\frac{2}{3}}}{2y-3}$ | dM1 | Applies $\frac{dy}{dx} = \frac{1}{\left(\frac{dx}{dy}\right)}$ |
| $\frac{dy}{dx} = \frac{3(x^3)^{\frac{2}{3}}}{2y-3} \Rightarrow \frac{dy}{dx} = \frac{3x^2}{2y-3}$ | A1 oe | $\frac{3(x^3)^{\frac{2}{3}}}{2y-3}$ or $\frac{3x^2}{2y-3}$ |
---A curve $C$ has the equation $y ^ { 2 } - 3 y = x ^ { 3 } + 8$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
\item Hence find the gradient of $C$ at the point where $y = 3$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2009 Q1 [7]}}