| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | November |
| 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 standard implicit differentiation question requiring two derivatives. Part (a) is routine application of the chain rule and product rule to verify a given result. Part (b) requires differentiating the first derivative implicitly again, which involves careful algebraic manipulation and substitution of known values. While methodical, it's more demanding than typical A-level questions due to the algebraic complexity of second derivatives in implicit form, placing it moderately above average difficulty. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dx}(y^2) = 2yy'\) | B1 | Differentiates \(y^2\) correctly |
| \(\frac{d}{dx}\left((xy+1)^2\right) = 2(xy+1)(xy'+y)\) | B1 | Differentiates \((xy+1)^2\) correctly |
| \((1)y' + ((1)+1)(y'+1) = 0 \Rightarrow y' = -\frac{2}{3}\) | B1 | Substitutes \((1,1)\); AG. \(y' = -\frac{xy^2+y}{y+x^2y+x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(yy'' + (y')^2\) | B1 | Differentiates \(yy'\) |
| \(+(xy+1)(xy''+y'+y') + (xy'+y)(xy'+y) = 0\) | B1 B1 | Differentiates \((xy+1)(xy'+y)\) |
| \(y'' + \left(-\frac{2}{3}\right)^2 + 2\left(y'' - \frac{4}{3}\right) + \left(-\frac{2}{3}+1\right)\left(\frac{1}{3}\right) = 0\) | M1 | Substitutes \((1,1)\) and \(y' = -\frac{2}{3}\) |
| \(y'' = \frac{19}{27}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y'' = -\frac{(y+x^2y+x)(2xyy'+y^2+y') - (xy^2+y)(y'+2xy+x^2y'+1)}{(y+x^2y+x)^2}\) | M1 A1 A1 | Differentiate \(y' = -\frac{xy^2+y}{y+x^2y+x}\) using quotient rule; A1 for \(xy^2\) correctly; A1 for everything correct |
| \(y'' = \frac{19}{27}\) | M1 A1 | Substitutes \((1,1)\) and \(y'=-\frac{2}{3}\) |
## Question 5(a):
| $\frac{d}{dx}(y^2) = 2yy'$ | B1 | Differentiates $y^2$ correctly |
|---|---|---|
| $\frac{d}{dx}\left((xy+1)^2\right) = 2(xy+1)(xy'+y)$ | B1 | Differentiates $(xy+1)^2$ correctly |
| $(1)y' + ((1)+1)(y'+1) = 0 \Rightarrow y' = -\frac{2}{3}$ | B1 | Substitutes $(1,1)$; AG. $y' = -\frac{xy^2+y}{y+x^2y+x}$ |
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## Question 5(b):
| $yy'' + (y')^2$ | B1 | Differentiates $yy'$ |
|---|---|---|
| $+(xy+1)(xy''+y'+y') + (xy'+y)(xy'+y) = 0$ | B1 B1 | Differentiates $(xy+1)(xy'+y)$ |
| $y'' + \left(-\frac{2}{3}\right)^2 + 2\left(y'' - \frac{4}{3}\right) + \left(-\frac{2}{3}+1\right)\left(\frac{1}{3}\right) = 0$ | M1 | Substitutes $(1,1)$ and $y' = -\frac{2}{3}$ |
| $y'' = \frac{19}{27}$ | A1 | |
**Alternative:**
| $y'' = -\frac{(y+x^2y+x)(2xyy'+y^2+y') - (xy^2+y)(y'+2xy+x^2y'+1)}{(y+x^2y+x)^2}$ | M1 A1 A1 | Differentiate $y' = -\frac{xy^2+y}{y+x^2y+x}$ using quotient rule; A1 for $xy^2$ correctly; A1 for everything correct |
|---|---|---|
| $y'' = \frac{19}{27}$ | M1 A1 | Substitutes $(1,1)$ and $y'=-\frac{2}{3}$ |
---5 The curve $C$ has equation
$$y ^ { 2 } + ( x y + 1 ) ^ { 2 } = 5$$
\begin{enumerate}[label=(\alph*)]
\item Show that, at the point $( 1,1 )$ on $C , \frac { \mathrm { dy } } { \mathrm { dx } } = - \frac { 2 } { 3 }$.
\item Find the value of $\frac { d ^ { 2 } y } { d x ^ { 2 } }$ at the point $( 1,1 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q5 [8]}}