| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - direct evaluation |
| Difficulty | Standard +0.3 This is a straightforward chain rule differentiation followed by direct substitution. While it involves transcendental functions (ln and trig), the technique is standard C3 material requiring dy/dx = 2sin(2x)/(1-cos(2x)) then evaluating at x=π/6. The algebra is routine and no problem-solving insight is needed, making it slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
# Question 2
**M1** $1/(1 - \cos 2x)$ seen or implied
**M1** $\frac{d}{dx}(1 - \cos 2x) = \pm 2\sin 2x$
**A1cao** $\frac{dy}{dx} = \frac{2\sin 2x}{1 - \cos 2x}$
**M1** substituting $\pi/6$ or $30°$ into their derivative (must be in at least two places)
**A1cao** $2\sqrt{3}$
**Guidance:** isw after correct answer seen
**Total: [5]**2 Find the exact gradient of the curve $y = \ln ( 1 - \cos 2 x )$ at the point with $x$-coordinate $\frac { 1 } { 6 } \pi$.
\hfill \mbox{\textit{OCR MEI C3 Q2 [5]}}