| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Standard +0.3 This is a standard implicit differentiation question from C4. Part (a) requires applying the product rule and chain rule systematically to differentiate implicitly, then rearranging for dy/dx - a routine technique at this level. Part (b) is straightforward substitution into the derivative and using point-slope form. While implicit differentiation is conceptually more advanced than basic calculus, this is a textbook exercise with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| \(2x - 3y - 3x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0\) | M1 A2 |
| \(\frac{dy}{dx} = \frac{2x-3y}{3x+2y}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{grad} = 5\) | M1 | |
| \(\therefore y + 2 = 5(x - 2)\) | M1 A1 | \([y = 5x - 12]\) |
**(a)**
$2x - 3y - 3x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0$ | M1 A2 |
$\frac{dy}{dx} = \frac{2x-3y}{3x+2y}$ | M1 A1 |
**(b)**
$\text{grad} = 5$ | M1 |
$\therefore y + 2 = 5(x - 2)$ | M1 A1 | $[y = 5x - 12]$ | (8)A curve has the equation
$$x^2 - 3xy - y^2 = 12.$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\frac{dy}{dx}$ in terms of $x$ and $y$. [5]
\item Find an equation for the tangent to the curve at the point $(2, -2)$. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q5 [8]}}