一条曲线的方程为 $y=\dfrac{\mathrm{e}^{\sin x}}{\cos^2x}$，其中 $0\le x\le2\pi$。

求 $\frac{\mathrm{d}y}{\mathrm{d}x}$，并由此求曲线的驻点的 $x$ 坐标。

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。

用商法则后，分子提出恒正的指数项及可约余弦因子。驻点条件可化成正弦的二次方程，须排除分母为零处。

The curve is defined where $\cos x\ne0$, excluding $x=\pi/2,3\pi/2$.

By the quotient rule,

Since $e^{\sin x}>0$, the stationary points require

Using $\cos^2x=1-\sin^2x$,

so $\sin x=1\pm\sqrt2$. Only $1-\sqrt2$ is in $[-1,1]$.

Let $\beta=\sin^{-1}(\sqrt2-1)$. On $[0,2\pi]$, the solutions are

Both have $\cos x\ne0$. Thus

因为函数值恒正，也可对数求导。余弦可能为负，因此对数要写 $\ln|\cos x|$。

Since $y=e^{\sin x}/\cos^2x>0$ on its domain,

Differentiating,

Hence

At a stationary point, $\cos x+2\tan x=0$. Multiplying by $\cos x\ne0$ gives $\cos^2x+2\sin x=0$.

Therefore $\sin^2x-2\sin x-1=0$, with the only valid sine value $\sin x=1-\sqrt2$. The two solutions on $[0,2\pi]$ are