曲线 $C$ 的方程为

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

(b) 求曲线在点 $(-1,1)$ 处的 $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ 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。

对方程两边求导，将含 $y'$ 的项合并，再代入给定点。

Differentiating implicitly by the product rule gives

Collecting the terms in $y'$ gives

At $(x,y)=(-1,1)$ this becomes $y'-11=0$, so $y'=11$.

对 (a) 的隐式等式再求导，保留 $y'$ 的乘积项。最后同时代入 $x,y,y'$。

The derivative of $3xy^2-4x^3$ is $3y^2+6xyy'-12x^2$. Applying the product rule to the equation in (a) and collecting terms gives

At the given point with $y'=11$, this gives

Therefore $\boxed{y''=900}$.

也可先从 (a) 解出 $y'$。分别计算分子和分母在给定点的值及导数，再代入商法则。

Write $y'=N/D$, where $N=12x^2y-y^3$ and $D=3xy^2-4x^3$. At $(-1,1)$, $N=11$ and $D=1$. With $y'=11$ there,

Thus $y''=(N'D-ND')/D^2=75-11(-75)=900$.