| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard techniques: differentiate tan y = x to find dy/dx, then differentiate again using the chain rule and product rule. Part (ii) is simple substitution. While it involves second derivatives and implicit differentiation (FP1 topics), the execution is mechanical with no conceptual challenges or novel insights required. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1+x^2)\frac{dy}{dx} = 1\) or \(\sec^2 y \frac{dy}{dx} = 1 \Rightarrow (1+x^2)\frac{dy}{dx} = 1\) or \(\frac{dy}{dx} = \cos^2 y\) | M1 | Using implicit differentiation |
| \(\Rightarrow 2x\frac{dy}{dx} + (1+x^2)\frac{d^2y}{dx^2} = 0 \Rightarrow \frac{d^2y}{dx^2} = -2x\left(\frac{dy}{dx}\right)^2\) (AG) | M1A1 | M1 for good attempt at product rule |
| Alt method: \(\frac{dy}{dx} = \cos^2 y \Rightarrow \frac{d^2y}{dx^2} = \cos y(-\sin y)\frac{dy}{dx} = -2x\left(\frac{dy}{dx}\right)^2\) | M1A1 | M1 for good attempt at implicit differentiation |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y'(1) = \cos^2\left(\frac{\pi}{4}\right) \Rightarrow y'(1) = \frac{1}{2}\) | B1 | |
| \(\Rightarrow y''(1) = -\frac{1}{2}\) | B1 FT | |
| Total: 2 |
## Question 3:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1+x^2)\frac{dy}{dx} = 1$ or $\sec^2 y \frac{dy}{dx} = 1 \Rightarrow (1+x^2)\frac{dy}{dx} = 1$ or $\frac{dy}{dx} = \cos^2 y$ | M1 | Using implicit differentiation |
| $\Rightarrow 2x\frac{dy}{dx} + (1+x^2)\frac{d^2y}{dx^2} = 0 \Rightarrow \frac{d^2y}{dx^2} = -2x\left(\frac{dy}{dx}\right)^2$ (AG) | M1A1 | M1 for good attempt at product rule |
| Alt method: $\frac{dy}{dx} = \cos^2 y \Rightarrow \frac{d^2y}{dx^2} = \cos y(-\sin y)\frac{dy}{dx} = -2x\left(\frac{dy}{dx}\right)^2$ | M1A1 | M1 for good attempt at implicit differentiation |
| **Total: 3** | | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y'(1) = \cos^2\left(\frac{\pi}{4}\right) \Rightarrow y'(1) = \frac{1}{2}$ | B1 | |
| $\Rightarrow y''(1) = -\frac{1}{2}$ | B1 FT | |
| **Total: 2** | | |
---3 A curve $C$ has equation $\tan y = x$, for $x > 0$.\\
(i) Use implicit differentiation to show that
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - 2 x \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 }$$
(ii) Hence find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at the point $\left( 1 , \frac { 1 } { 4 } \pi \right)$ on $C$.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q3 [5]}}