| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Implicit differentiation |
| Difficulty | Standard +0.3 Part (a) is a standard implicit differentiation exercise requiring the chain rule and knowledge that d(sec u)/du = sec u tan u, plus a trigonometric identity to simplify. Part (b) is routine differentiation of a composite function followed by finding a tangent line equation. Both are textbook-style questions with straightforward application of techniques, making this slightly easier than average for C3. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| \(\frac{dx}{dy} = \frac{1}{2}\sec^2\frac{y}{2}\tan\frac{y}{2}\) | M1 |
| \(0 \leq y < \pi \therefore \tan\frac{y}{2} \geq 0 \therefore \frac{dx}{dy} = \frac{1}{2}\sec^2\frac{y}{2}\sqrt{\sec^2\frac{y}{2} - 1} = \frac{1}{2}\sqrt{x^2 - 1}\) | M1 A1 |
| \(\frac{dy}{dx} = 1 + \frac{dx}{dy} = \frac{2}{\sqrt{x^2 - 1}}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{1}{2}(3 + 2\cos x)^{-1} \times (-2\sin x) = -\frac{\sin x}{\sqrt{3} + 2\cos x}\) | M1 A1 | |
| At \(x = \frac{\pi}{3}, y = 2\), grad \(= -\frac{1}{4}\sqrt{3}\) | M1 A1 | |
| \(\therefore y - 2 = -\frac{1}{4}\sqrt{3}(x - \frac{\pi}{3})\) | M1 A1 | |
| \([3\sqrt{3}x + 12y - 24 - \pi\sqrt{3} = 0]\) | M1 A1 | (11 marks) |
**(a)**
$\frac{dx}{dy} = \frac{1}{2}\sec^2\frac{y}{2}\tan\frac{y}{2}$ | M1 |
$0 \leq y < \pi \therefore \tan\frac{y}{2} \geq 0 \therefore \frac{dx}{dy} = \frac{1}{2}\sec^2\frac{y}{2}\sqrt{\sec^2\frac{y}{2} - 1} = \frac{1}{2}\sqrt{x^2 - 1}$ | M1 A1 |
$\frac{dy}{dx} = 1 + \frac{dx}{dy} = \frac{2}{\sqrt{x^2 - 1}}$ | M1 A1 |
**(b)**
$\frac{dy}{dx} = \frac{1}{2}(3 + 2\cos x)^{-1} \times (-2\sin x) = -\frac{\sin x}{\sqrt{3} + 2\cos x}$ | M1 A1 |
At $x = \frac{\pi}{3}, y = 2$, grad $= -\frac{1}{4}\sqrt{3}$ | M1 A1 |
$\therefore y - 2 = -\frac{1}{4}\sqrt{3}(x - \frac{\pi}{3})$ | M1 A1 |
$[3\sqrt{3}x + 12y - 24 - \pi\sqrt{3} = 0]$ | M1 A1 | (11 marks)\begin{enumerate}
\item (a) Given that
\end{enumerate}
$$x = \sec \frac { y } { 2 } , \quad 0 \leq y < \pi ,$$
show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x \sqrt { x ^ { 2 } - 1 } } .$$
(b) Find an equation for the tangent to the curve $y = \sqrt { 3 + 2 \cos x }$ at the point where $x = \frac { \pi } { 3 }$.\\
\hfill \mbox{\textit{Edexcel C3 Q4 [11]}}