C4 June 2011 Q5
5. Find the gradient of the curve with equation\[\ln y = 2x\ln x, \qquad x \gt 0,\ y \gt 0\]at the point on the curve where \(x = 2\). Give your answer as an exact value. (7)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{y}\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\) | B1 |
| \(\ldots = 2\ln x + 2x\left(\dfrac{1}{x}\right)\) | M1 A1 |
| At \(x = 2\), \(\qquad \ln y = 2(2)\ln 2\) | M1 |
| leading to \(\qquad y = 16\) Accept \(y = \mathrm{e}^{4\ln 2}\) | A1 |
| At \((2, 16)\) \(\qquad \dfrac{1}{16}\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\ln 2 + 2\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 16(2 + 2\ln 2)\) | A1 |
| (7) | |
| (7 marks) |
Notes
In the printed scheme a bracket joins the M1 for \(2\ln x + 2x\left(\frac{1}{x}\right)\) to the final M1, and another joins the M1 for \(\ln y = 2(2)\ln 2\) to the final M1: the final M1 is dependent on these.
Alternative
| Scheme | Marks |
|---|---|
| \(y = \mathrm{e}^{2x\ln x}\) | B1 |
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}(2x\ln x) = 2\ln x + 2x\left(\dfrac{1}{x}\right)\) | M1 A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \left(2\ln x + 2x\left(\dfrac{1}{x}\right)\right)\mathrm{e}^{2x\ln x}\) | M1 A1 |
| At \(x = 2\), \(\qquad \dfrac{\mathrm{d}y}{\mathrm{d}x} = (2\ln 2 + 2)\mathrm{e}^{4\ln 2}\) | M1 |
| \(= 16(2 + 2\ln 2)\) | A1 |
| (7) |
In the printed scheme a bracket joins these method marks: each later M mark is dependent on the M mark before it.