一条曲线的方程为

(a) 证明在点 $(0,-2)$ 处，$\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{2}{3}$。

(b) 求在点 $(0,-2)$ 处 $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ 的值。

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。

对 $(x+1)y$ 用乘积法则，再与 $y^2$ 的导数合并。代入给定点求 $y'$。

The product and chain rules give $\frac{\mathrm{d}}{\mathrm{d}x}[(x+1)y]=y+(x+1)y'$ and $\frac{\mathrm{d}}{\mathrm{d}x}(y^2)=2yy'$. Combining them,

At $(0,-2)$ this becomes $-2-3y'=0$, so $y'=-2/3$.

对 (a) 的等式再求一次导数。括号求导得到 $1+2y'$，再代入已求的斜率。

Differentiating $y+(x+1+2y)y'=0$ gives $y'+(1+2y')y'+(x+1+2y)y''=0$. Combining terms,

At $(0,-2)$ with $y'=-2/3$, this is $-3y''-\frac43+\frac89=0$. Therefore $\boxed{y''=-4/27}$.

也可先从 (a) 解出 $y'$，再用商法则。

From (a), $y'=-y/(x+1+2y)$. With $H=x+1+2y$,

At the given point, $H=-3$, $y=-2$ and $y'=-2/3$. Thus $H'=1+2y'=-1/3$ and