| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find second derivative d²y/dx² |
| Difficulty | Standard +0.8 This is a Further Maths implicit differentiation question requiring two derivatives. Part (a) is straightforward implicit differentiation with verification. Part (b) requires differentiating the first derivative implicitly again, which involves careful application of the quotient/product rule and substitution of known values—more demanding than standard single-variable calculus but a standard Further Maths technique. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dx}(4y^3) = 12y^2y'\) | B1 | Differentiates \(y^3\) correctly |
| \(\frac{d}{dx}((x+y)^6) = 6(x+y)^5(1+y')\) | B1 | Differentiates \((x+y)^6\) correctly |
| \(12(3^2)y' + 6(-1)(1+y') = 0 \Rightarrow y' = \frac{1}{17}\) | B1 | Substitutes \((-4,3)\), AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(2y^2y'' + 4y(y'')^2\) | B1 | Differentiates \(y^2y'\) |
| \(+(x+y)^5y'' + 5(x+y)^4(1+y')^2 = 0\) | B1 B1 | Differentiates \((x+y)^5(1+y')\) |
| \(17y'' + 12\left(\frac{1}{17}\right)^2 + 5\times(-1)^4\times\left(\frac{18}{17}\right)^2 = 0 \Rightarrow 17y'' + \frac{96}{17} = 0\) | M1 | Substitutes \((-4,3)\) |
| \(y'' = -\frac{96}{289}\) | A1 |
## Question 4(a):
| $\frac{d}{dx}(4y^3) = 12y^2y'$ | B1 | Differentiates $y^3$ correctly |
|---|---|---|
| $\frac{d}{dx}((x+y)^6) = 6(x+y)^5(1+y')$ | B1 | Differentiates $(x+y)^6$ correctly |
| $12(3^2)y' + 6(-1)(1+y') = 0 \Rightarrow y' = \frac{1}{17}$ | B1 | Substitutes $(-4,3)$, AG |
## Question 4(b):
| $2y^2y'' + 4y(y'')^2$ | B1 | Differentiates $y^2y'$ |
|---|---|---|
| $+(x+y)^5y'' + 5(x+y)^4(1+y')^2 = 0$ | B1 B1 | Differentiates $(x+y)^5(1+y')$ |
| $17y'' + 12\left(\frac{1}{17}\right)^2 + 5\times(-1)^4\times\left(\frac{18}{17}\right)^2 = 0 \Rightarrow 17y'' + \frac{96}{17} = 0$ | M1 | Substitutes $(-4,3)$ |
| $y'' = -\frac{96}{289}$ | A1 | |
---4 The curve $C$ has equation
$$4 y ^ { 3 } + ( x + y ) ^ { 6 } = 109 .$$
\begin{enumerate}[label=(\alph*)]
\item Show that, at the point $( - 4,3 )$ on $C , \frac { \mathrm { dy } } { \mathrm { dx } } = \frac { 1 } { 17 }$.
\item Find the value of $\frac { d ^ { 2 } y } { d x ^ { 2 } }$ at the point $( - 4,3 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q4 [8]}}