| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| 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 standard techniques. Part (i) requires applying the chain rule to e^(2y) and differentiating sin x, then rearranging. Part (ii) involves taking logarithms to make y the subject and differentiating explicitly—both are routine C3 procedures with no novel problem-solving required. Slightly easier than average due to the simple structure and verification nature. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(e^{2y} = 1 + \sin x \Rightarrow 2e^{2y}\frac{dy}{dx} = \cos x\) | M1 | Their \(2e^{2y} \times \frac{dy}{dx}\) |
| B1 | \(2e^{2y}\) | |
| \(\Rightarrow \frac{dy}{dx} = \frac{\cos x}{2e^{2y}}\) | A1 [3] | o.e. cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2y = \ln(1 + \sin x) \Rightarrow y = \frac{1}{2}\ln(1 + \sin x)\) | B1 | |
| \(\Rightarrow \frac{dy}{dx} = \frac{1}{2} \cdot \frac{\cos x}{1 + \sin x}\) | M1 | chain rule (can be within 'correct' quotient rule with \(\frac{dy}{dx} = 0\)) |
| B1 | \(\frac{1}{u}\) or \(\frac{1}{(1+\sin x)}\) soi | |
| \(= \frac{\cos x}{2e^{2y}}\) as before | E1 [4] | www |
# Question 4:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $e^{2y} = 1 + \sin x \Rightarrow 2e^{2y}\frac{dy}{dx} = \cos x$ | M1 | Their $2e^{2y} \times \frac{dy}{dx}$ |
| | B1 | $2e^{2y}$ |
| $\Rightarrow \frac{dy}{dx} = \frac{\cos x}{2e^{2y}}$ | A1 [3] | o.e. cao |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2y = \ln(1 + \sin x) \Rightarrow y = \frac{1}{2}\ln(1 + \sin x)$ | B1 | |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{2} \cdot \frac{\cos x}{1 + \sin x}$ | M1 | chain rule (can be within 'correct' quotient rule with $\frac{dy}{dx} = 0$) |
| | B1 | $\frac{1}{u}$ or $\frac{1}{(1+\sin x)}$ soi |
| $= \frac{\cos x}{2e^{2y}}$ as before | E1 [4] | www |
---4 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 Q4 [7]}}