| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Show dy/dx equals given expression |
| Difficulty | Moderate -0.3 This is a straightforward implicit differentiation question with verification. Part (i) requires standard application of the chain rule (differentiating e^{2y} gives 2e^{2y}·dy/dx), then rearranging. Part (ii) involves taking natural logs to make y the subject, then differentiating explicitly—both are routine C3 techniques with no problem-solving insight required. Slightly easier than average due to the simple structure and verification nature. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^y = 1 + \sin x \Rightarrow 2e^y \frac{dy}{dx} = \cos x \Rightarrow \frac{dy}{dx} = \frac{\cos x}{2e^y}\) | M1, B1, A1 | [3] |
| Their \(2e^y \times \frac{dy}{dx} = 2e^y\) | M1 | |
| o.e. cao | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2y = \ln(1 + \sin x) \Rightarrow y = \frac{1}{2}\ln(1 + \sin x) \Rightarrow \frac{dy}{dx} = \frac{1}{2} \cdot \frac{\cos x}{1 + \sin x}\) | B1, M1 | |
| \(= \frac{\cos x}{2e^y}\) as before | B1, E1 | [4] |
| Chain rule (can be within 'correct' quotient rule with \(\frac{dy}{dx} = 0\)) or \(1/u\) or \(1/(1 + \sin x)\) soi www |
**Part (i)**
| $e^y = 1 + \sin x \Rightarrow 2e^y \frac{dy}{dx} = \cos x \Rightarrow \frac{dy}{dx} = \frac{\cos x}{2e^y}$ | M1, B1, A1 | [3] |
| Their $2e^y \times \frac{dy}{dx} = 2e^y$ | M1 | |
| o.e. cao | A1 | |
**Part (ii)**
| $2y = \ln(1 + \sin x) \Rightarrow y = \frac{1}{2}\ln(1 + \sin x) \Rightarrow \frac{dy}{dx} = \frac{1}{2} \cdot \frac{\cos x}{1 + \sin x}$ | B1, M1 | |
| $= \frac{\cos x}{2e^y}$ as before | B1, E1 | [4] |
| Chain rule (can be within 'correct' quotient rule with $\frac{dy}{dx} = 0$) or $1/u$ or $1/(1 + \sin x)$ soi www | | |
---5 The equation of a curve is given by $\mathrm { e } ^ { 2 y } = 1 + \sin x$.\\
(i) By differentiating implicitly, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.\\
(ii) Find an expression for $y$ in terms of $x$, and differentiate it to verify the result in part (i).
\hfill \mbox{\textit{OCR MEI C3 2009 Q5 [7]}}