| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Differentiation of reciprocal functions |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation problem requiring knowledge of derivatives of sec and tan, followed by algebraic manipulation using trig identities. While it involves reciprocal trig functions, the technique is standard C3 material with no novel insight required—slightly easier than average due to being a 'show that' question with a clear target. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dy} = 2\sec y \times \sec y \tan y + \sec^2 y = \sec^2 y(2\tan y + 1) = \frac{2\tan y + 1}{\cos^2 y}\) | M1 A1 | |
| \(\frac{dy}{dx} = 1 + \frac{dx}{dy} = \frac{\cos^2 y}{2\tan y + 1}\) | M1 A1 | (4 marks) |
$\frac{dx}{dy} = 2\sec y \times \sec y \tan y + \sec^2 y = \sec^2 y(2\tan y + 1) = \frac{2\tan y + 1}{\cos^2 y}$ | M1 A1 |
$\frac{dy}{dx} = 1 + \frac{dx}{dy} = \frac{\cos^2 y}{2\tan y + 1}$ | M1 A1 | (4 marks)\begin{enumerate}
\item Given that
\end{enumerate}
$$x = \sec ^ { 2 } y + \tan y ,$$
show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \cos ^ { 2 } y } { 2 \tan y + 1 } .$$
\hfill \mbox{\textit{Edexcel C3 Q1 [4]}}