Question 5 - A-Level Maths

Edexcel C4 2011 June — Question 5 7 marks

Exam Board	Edexcel
Module	C4 (Core Mathematics 4)
Year	2011
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Implicit equations and differentiation
Type	Find dy/dx at a point
Difficulty	Moderate -0.3 This is a straightforward implicit differentiation question requiring the product rule and chain rule. The steps are routine: differentiate both sides (1/y · dy/dx = 2ln(x) + 2x/x), rearrange for dy/dx, then substitute x=2. While it requires multiple techniques, it's a standard textbook exercise with no novel insight needed, making it slightly easier than average.
Spec	1.06d Natural logarithm: ln(x) function and properties 1.07s Parametric and implicit differentiation

Find the gradient of the curve with equation

$$\ln y = 2 x \ln x , \quad x > 0 , y > 0$$ at the point on the curve where $x = 2$. Give your answer as an exact value.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
$\frac{1}{y}\frac{dy}{dx} = \ldots$	B1
$\ldots = 2\ln x + 2x\left(\frac{1}{x}\right)$	M1 A1
At $x = 2$: $\ln y = 2(2)\ln 2$	M1
leading to $y = 16$ Accept $y = e^{4\ln 2}$	A1
At $(2, 16)$: $\frac{1}{16}\frac{dy}{dx} = 2\ln 2 + 2$	M1
$\frac{dy}{dx} = 16(2 + 2\ln 2)$	A1	(7) [7]

Alternative:

Answer	Marks	Guidance
$y = e^{2x\ln x}$	B1
$\frac{d}{dx}(2x\ln x) = 2\ln x + 2x\left(\frac{1}{x}\right)$	M1 A1
$\frac{dy}{dx} = \left(2\ln x + 2x\left(\frac{1}{x}\right)\right)e^{2x\ln x}$	M1 A1
At $x = 2$: $\frac{dy}{dx} = (2\ln 2 + 2)e^{4\ln 2}$	M1
$= 16(2 + 2\ln 2)$	A1	(7) [7]

$\frac{1}{y}\frac{dy}{dx} = \ldots$ | B1 |

$\ldots = 2\ln x + 2x\left(\frac{1}{x}\right)$ | M1 A1 |

At $x = 2$: $\ln y = 2(2)\ln 2$ | M1 |

leading to $y = 16$ Accept $y = e^{4\ln 2}$ | A1 |

At $(2, 16)$: $\frac{1}{16}\frac{dy}{dx} = 2\ln 2 + 2$ | M1 |

$\frac{dy}{dx} = 16(2 + 2\ln 2)$ | A1 | (7) [7] |

Alternative:

$y = e^{2x\ln x}$ | B1 |

$\frac{d}{dx}(2x\ln x) = 2\ln x + 2x\left(\frac{1}{x}\right)$ | M1 A1 |

$\frac{dy}{dx} = \left(2\ln x + 2x\left(\frac{1}{x}\right)\right)e^{2x\ln x}$ | M1 A1 |

At $x = 2$: $\frac{dy}{dx} = (2\ln 2 + 2)e^{4\ln 2}$ | M1 |

$= 16(2 + 2\ln 2)$ | A1 | (7) [7] |

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Show LaTeX source

\begin{enumerate}
  \item Find the gradient of the curve with equation
\end{enumerate}

$$\ln y = 2 x \ln x , \quad x > 0 , y > 0$$

at the point on the curve where $x = 2$. Give your answer as an exact value.\\

\hfill \mbox{\textit{Edexcel C4 2011 Q5 [7]}}

This paper (7 questions)

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Q1 4 Q2 6 Q3 6 Q4 15 Q5 7 Q6 14 Q7 15

Answer	Marks	Guidance
\(\frac{1}{y}\frac{dy}{dx} = \ldots\)	B1
\(\ldots = 2\ln x + 2x\left(\frac{1}{x}\right)\)	M1 A1
At \(x = 2\): \(\ln y = 2(2)\ln 2\)	M1
leading to \(y = 16\) Accept \(y = e^{4\ln 2}\)	A1
At \((2, 16)\): \(\frac{1}{16}\frac{dy}{dx} = 2\ln 2 + 2\)	M1
\(\frac{dy}{dx} = 16(2 + 2\ln 2)\)	A1	(7) [7]

Answer	Marks	Guidance
\(y = e^{2x\ln x}\)	B1
\(\frac{d}{dx}(2x\ln x) = 2\ln x + 2x\left(\frac{1}{x}\right)\)	M1 A1
\(\frac{dy}{dx} = \left(2\ln x + 2x\left(\frac{1}{x}\right)\right)e^{2x\ln x}\)	M1 A1
At \(x = 2\): \(\frac{dy}{dx} = (2\ln 2 + 2)e^{4\ln 2}\)	M1
\(= 16(2 + 2\ln 2)\)	A1	(7) [7]