Standard +0.3 This is a straightforward implicit differentiation question requiring product rule application, algebraic manipulation to find dy/dx, then solving dy/dx=0. While it involves multiple steps, the techniques are standard C4 material with no conceptual surprises, making it slightly easier than average.
3 The equation of a curve is \(x y ^ { 2 } = x ^ { 2 } + 1\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\), and hence find the coordinates of the stationary points on the curve.
\(\frac{dy}{dx} = \frac{2x - y^2}{2xy}\) or \(\frac{1 - x^{-2}}{2y}\)
A1
Award B1 for \((\pm)\frac{1}{2}(x + x^{-1})^{-1/2}(1 - x^{-2})\)
Stationary point \(\rightarrow\) (their) \(\frac{dy}{dx} = 0\) soi
M1
\(x^2 = 1\) or \(y^2 = 2\) or \(y^4 = 4\)
A1
Ignore any other values
\((1, \sqrt{2})\), \((1, -\sqrt{2})\)
A1, A1
Accept \(1.41\) or \(4^{1/4}\) for \(\sqrt{2}\); SR: award A1 only if extra coordinates presented with both correct answers
[7]
# Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| For attempt at product rule on $xy^2$ | M1 | Or changing equation to $y^2 = x + x^{-1}$ |
| $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ | B1 | soi in the differentiating process |
| $\frac{dy}{dx} = \frac{2x - y^2}{2xy}$ or $\frac{1 - x^{-2}}{2y}$ | A1 | Award **B1** for $(\pm)\frac{1}{2}(x + x^{-1})^{-1/2}(1 - x^{-2})$ |
| Stationary point $\rightarrow$ (their) $\frac{dy}{dx} = 0$ soi | M1 | |
| $x^2 = 1$ or $y^2 = 2$ or $y^4 = 4$ | A1 | Ignore any other values |
| $(1, \sqrt{2})$, $(1, -\sqrt{2})$ | A1, A1 | Accept $1.41$ or $4^{1/4}$ for $\sqrt{2}$; SR: award A1 only if extra coordinates presented with both correct answers |
| **[7]** | | |
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3 The equation of a curve is $x y ^ { 2 } = x ^ { 2 } + 1$. Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$, and hence find the coordinates of the stationary points on the curve.
\hfill \mbox{\textit{OCR C4 2013 Q3 [7]}}