题目

Problem

The equation of a curve is $x^2\ln2y-y\ln(2+x^2)=\ln6$ .

Find the exact value of the gradient of the curve at the point $(2,3)$ . Give your answer in simplified form.

(5)

题目中文翻译

一条曲线的方程为 $x^2\ln2y-y\ln(2+x^2)=\ln6$ 。

求该曲线在点 $(2,3)$ 处的斜率的精确值。答案须化简。

(5)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

解法一

思路

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两个乘积都要用乘积法则，含 $y$ 的对数还要用链式法则。代入点后再解斜率，最终消去分式内的分式。

答题过程

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The product and chain rules give

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\big[x^2\ln(2y)\big] =&\,2x\ln(2y)+\frac{x^2}{y}\frac{\mathrm{d}y}{\mathrm{d}x},\\[4mm] \frac{\mathrm{d}}{\mathrm{d}x}\big[y\ln(2+x^2)\big] =&\,\ln(2+x^2)\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{2pt}+\frac{2xy}{2+x^2}. \end{align*}

Since the derivative of $\ln6$ is zero,

\begin{align*} &\,2x\ln(2y)+\frac{x^2}{y}\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{2pt}-\ln(2+x^2)\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{4pt}-\frac{2xy}{2+x^2}=0. \end{align*}

At $(2,3)$ ,

4\ln6+\frac43\frac{\mathrm{d}y}{\mathrm{d}x} -\ln6\frac{\mathrm{d}y}{\mathrm{d}x}-2=0.

Thus

\bigg(\frac43-\ln6\bigg)\frac{\mathrm{d}y}{\mathrm{d}x} =2-4\ln6.

Multiplying numerator and denominator by $3$ , the gradient is

\boxed{\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{6-12\ln6}{4-3\ln6}}.

CIE 9709 2025 November Paper 35 Q4

题目

解答

解法一

思路

答题过程