曲线 $C$ 的方程为

(a) 证明在曲线 $C$ 上的点 $(-1,-2)$ 处，$\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{8}{11}$。

(b) 求点 $(-1,-2)$ 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。

隐式求导。对数项用 $\frac{\mathrm{d}}{\mathrm{d}x}\ln(xy)=1/x+y'/y$，这在题给点附近的 $xy>0$ 区域成立。

Near the given point, $xy>0$. By the chain and product rules, $(\ln(xy))'=(y+xy')/(xy)=1/x+y'/y$. Differentiating gives

At $(-1,-2)$ this is $5(1-y')-1-\frac12y'=0$. Hence $y'=\frac8{11}$.

对 (a) 再求导。注意 $x-y$ 的导数和乘在外面的 $1-y'$ 会相乘，形成 $(1-y')^2$。

Differentiating gives

At the given point, with $y'=8/11$,

Thus $\boxed{y''=\frac{86}{1331}}$.

也可先解出 $y'$，再用商法则。分别计算分子和分母在给定点的值及导数。

From (a), $y'=N/H$, where $N=5(x-y)^4+1/x$ and $H=5(x-y)^4-1/y$. At the point $N=4$ and $H=11/2$. Their derivatives there are

The quotient rule gives