[剑桥/9709/2007]
方程为 $y = \text{e}^{-x} \sin x$ 的曲线在区间 $0 \le x \le \pi$ 内有一个驻点。

1. Differentiate $y$ with respect to $x$ using the product rule.

Let $y = \text{e}^{-x} \sin x$.
Using the product rule:
$\frac{dy}{dx} = \text{e}^{-x} \frac{d}{dx}(\sin x) + \sin x \frac{d}{dx}(\text{e}^{-x})$
$\frac{dy}{dx} = \text{e}^{-x} \cos x - \text{e}^{-x} \sin x$
$\frac{dy}{dx} = \text{e}^{-x}(\cos x - \sin x)$

2. Equate the derivative to zero to find the condition for a stationary point.

For a stationary point, $\frac{dy}{dx} = 0$:
$\text{e}^{-x}(\cos x - \sin x) = 0$

3. Solve the trigonometric equation for $x$ in the given domain.

Since $\text{e}^{-x} > 0$ for all real $x$, we have:
$\cos x - \sin x = 0$
$\sin x = \cos x$
Dividing by $\cos x$ (as $\cos x = 0$ is not a solution here):
$\tan x = 1$
Given the domain $0 \le x \le \pi$, the only solution is:
$x = \frac{\pi}{4}$

1. State the first derivative and the stationary point from part (i)

From part (i), we have the first derivative:
$\frac{\text{d}y}{\text{d}x} = \text{e}^{-x} \cos x - \text{e}^{-x} \sin x = \text{e}^{-x}(\cos x - \sin x)$
Equating to $0$ for the stationary point gives:
$\cos x = \sin x \implies \tan x = 1 \implies x = \frac{\pi}{4}$

2. Find the second derivative to apply the second derivative test

Differentiating $\frac{\text{d}y}{\text{d}x}$ with respect to $x$ using the product rule:
$\frac{\text{d}^2 y}{\text{d}x^2} = \frac{\text{d}}{\text{d}x}[\text{e}^{-x}(\cos x - \sin x)]$
$\frac{\text{d}^2 y}{\text{d}x^2} = -\text{e}^{-x}(\cos x - \sin x) + \text{e}^{-x}(-\sin x - \cos x)$
$\frac{\text{d}^2 y}{\text{d}x^2} = -\text{e}^{-x}\cos x + \text{e}^{-x}\sin x - \text{e}^{-x}\sin x - \text{e}^{-x}\cos x$
$\frac{\text{d}^2 y}{\text{d}x^2} = -2\text{e}^{-x} \cos x$

3. Evaluate the second derivative at the stationary point and conclude

Substitute $x = \frac{\pi}{4}$ into the second derivative:
$\frac{\text{d}^2 y}{\text{d}x^2} = -2\text{e}^{-\frac{\pi}{4}} \cos \left(\frac{\pi}{4}\right)$
$\frac{\text{d}^2 y}{\text{d}x^2} = -2\text{e}^{-\frac{\pi}{4}} \left(\frac{\sqrt{2}}{2}\right) = -\sqrt{2}\text{e}^{-\frac{\pi}{4}}$
Since $\text{e}^{-\frac{\pi}{4}} > 0$, it follows that $\frac{\text{d}^2 y}{\text{d}x^2} < 0$.
Therefore, the stationary point is a maximum point.