第二章随堂测验答案解析
=========================

1. 设函数 :math:`f(x) = \sqrt{x + \sqrt{x + \sqrt{x}}}`, 求它的导函数 :math:`f'(x)`.

.. proof:solution::

    由复合函数求导的链式法则可得

    .. math::
        f'(x) = \dfrac{1}{2\sqrt{x + \sqrt{x + \sqrt{x}}}} \cdot \left(1 + \dfrac{1}{2\sqrt{x + \sqrt{x}}} \cdot \left(1 + \dfrac{1}{2\sqrt{x}}\right)\right)

2. 设函数 :math:`f(x)` 可微且函数值大于 :math:`0`. 令 :math:`g(x) = \ln f(\sin^2 x)`, 求函数 :math:`g(x)` 的微分.

.. proof:solution::

    .. math::
        \mathrm{d} y & = \mathrm{d} (\ln f(\sin^2 x)) \\
        & = \dfrac{1}{f(\sin^2 x)} \cdot f'(\sin^2 x) \cdot 2\sin x \cos x \cdot \mathrm{d} x \\
        & = \dfrac{\sin 2x}{f(\sin^2 x)} \cdot f'(\sin^2 x) \cdot \mathrm{d} x

3. 令 :math:`y(t) = \left( \dfrac{1}{t + 1} \right)^{\frac{1}{t}}`.

   (1). 求 :math:`\lim\limits_{t \to 0} y(t)` 以及 :math:`\lim\limits_{t \to 0} y'(t)`.

   (2). 求极限 :math:`\lim\limits_{x \to \infty} x \left( \dfrac{1}{e} - \left( \dfrac{x}{x + 1} \right)^x \right)`.

   提示: 利用带佩亚诺型余项的麦克劳林公式

   .. math::
        & \dfrac{1}{1 - t} = 1 + t + t^2 + o(t^2), \\
        & \ln (1 + t) = t - \dfrac{t^2}{2} + o(t^2).

.. proof:solution::

    (1). :math:`\lim\limits_{t \to 0} y(t) = \lim\limits_{t \to 0} \dfrac{1}{(t + 1)^{\frac{1}{t}}} = \dfrac{1}{e}`.

    对 :math:`\ln y(t) = - \dfrac{\ln (t + 1)}{t}` 两边求导可得

    .. math::
        \dfrac{y'(t)}{y(t)} = - \dfrac{\dfrac{t}{t + 1} - \ln (t + 1)}{t^2}.

    利用带佩亚诺型余项的麦克劳林公式

    .. math::
        & \dfrac{1}{1 - t} = 1 + t + t^2 + o(t^2), \\
        & \ln (1 + t) = t - \dfrac{t^2}{2} + o(t^2),

    可得

    .. math::
        \dfrac{y'(t)}{y(t)} & = - \dfrac{\dfrac{t}{t + 1} - \ln (t + 1)}{t^2} \\
        & = - \dfrac{t(1 - t + o(t)) - (t - \dfrac{t^2}{2} + o(t^2))}{t^2} \\
        & = \dfrac{1}{2} + o(1).

    于是, :math:`\lim\limits_{t \to 0} \dfrac{y'(t)}{y(t)} = \dfrac{1}{2}`, 从而有 :math:`\lim\limits_{t \to 0} y'(t) = \dfrac{1}{2e}`.

    (2). 由于

    .. math::
        \lim\limits_{x \to \infty} \left( \dfrac{x}{x + 1} \right)^x = \lim\limits_{x \to \infty} \dfrac{1}{\left( 1 + \dfrac{1}{x} \right)^x} = \dfrac{1}{e},

    所以 :math:`x \left( \dfrac{1}{e} - \left( \dfrac{x}{x + 1} \right)^x \right)` 是一个 :math:`\infty \cdot 0` 型的不定式 (:math:`x \to \infty`).
    令 :math:`t = \dfrac{1}{x}`, 则

    .. math::
        \lim\limits_{x \to \infty} x \left( \dfrac{1}{e} - \left( \dfrac{x}{x + 1} \right)^x \right)
        = \lim\limits_{t \to 0} \dfrac{\dfrac{1}{e} - \left( \dfrac{1}{1 + t} \right)^{\frac{1}{t}}}{t}
        = \lim\limits_{t \to 0} \dfrac{\dfrac{1}{e} - y(t)}{t}

    化为了 :math:`\dfrac{0}{0}` 型的不定式. 对上式利用洛必达法则可得

    .. math::
        \lim\limits_{t \to 0} \dfrac{\dfrac{1}{e} - y(t)}{t} = \lim\limits_{t \to 0} \dfrac{- y'(t)}{1} = - \dfrac{1}{2e}.

    因此, :math:`\lim\limits_{x \to \infty} x \left( \dfrac{1}{e} - \left( \dfrac{x}{x + 1} \right)^x \right) = - \dfrac{1}{2e}`.

4. 设 :math:`0 < a < b`, 证明存在 :math:`\xi \in (a, b)`, 使得

   .. math::
        a e^b - b e^a = (a - b) (1 - \xi)e^\xi.

   提示: 两边同时除以 :math:`ab`, 构造辅助函数, 并在区间 :math:`\left[ \dfrac{1}{b}, \dfrac{1}{a} \right]` 上利用拉格朗日中值定理.

.. proof:solution::

    对 :math:`a e^b - b e^a = (a - b) (1 - \xi)e^\xi` 两边同时除以 :math:`ab` 可得

    .. math::
        \dfrac{e^b}{b} - \dfrac{e^a}{a} = \left(\dfrac{1}{a} - \dfrac{1}{b}\right) \left( 1 - \xi \right) e^\xi.

    考虑函数 :math:`f(x) = x e^{\frac{1}{x}}`, 则 :math:`f'(x) = e^{\frac{1}{x}} \left(1 - \dfrac{1}{x}\right)`, 由拉格朗日中值定理可得,
    存在 :math:`\tau \in \left( \dfrac{1}{b}, \dfrac{1}{a} \right)`, 使得

    .. math::
        f(\dfrac{1}{b}) - f(\dfrac{1}{a}) = f'(\tau) \left(\dfrac{1}{b} - \dfrac{1}{a}\right).

    令 :math:`\xi = \dfrac{1}{\tau}`, 则 :math:`\xi \in (a, b)`, 且

    .. math::
        \dfrac{e^b}{b} - \dfrac{e^a}{a} = \left(\dfrac{1}{a} - \dfrac{1}{b}\right) \left(1 - \dfrac{1}{\tau}\right) e^{\frac{1}{\tau}}
        = \left(\dfrac{1}{a} - \dfrac{1}{b}\right) \left( 1 - \xi \right) e^\xi.

    上式两边同时乘以 :math:`ab` 即可得到题目要证明的结论:

    .. math::
        a e^b - b e^a = (a - b) (1 - \xi)e^\xi.

    .. note::
        另一种方法: 令 :math:`f(x) = \dfrac{e^x}{x}, g(x) = \dfrac{1}{x}`, 那么有

        .. math::
            f'(x) = \dfrac{e^x}{x^2} \left(x - 1\right), \quad g'(x) = - \dfrac{1}{x^2}.

        :math:`f(x), g(x)` 在闭区间 :math:`[a, b]` 上连续, 且在开区间 :math:`(a, b)` 上可导, 且 :math:`g(x) = \dfrac{1}{x}` 在闭区间 :math:`[a, b]` 上恒不为零,
        那么根据柯西中值定理可知, 存在 :math:`\xi \in (a, b)`, 使得

        .. math::
            \dfrac{f(b) - f(a)}{g(b) - g(a)} = \dfrac{f'(\xi)}{g'(\xi)},

        代入 :math:`f'(x), g'(x)` 的表达式可得

        .. math::
            \dfrac{e^b - e^a}{b - a} = \left.\dfrac{e^\xi}{\xi^2} \left(\xi - 1\right) \middle/ \left(- \dfrac{1}{\xi^2}\right) \right.
            = \left(1 - \xi\right) e^\xi.

5. 求函数 :math:`y = x^{1/x}, x > 0` 的极大值.

.. proof:solution::

    对 :math:`y = x^{1/x}, x > 0` 两边同时取对数可得

    .. math::
        \ln y = \dfrac{\ln x}{x}.

    令 :math:`f(x) = \dfrac{\ln x}{x}`, 则 :math:`f'(x) = \dfrac{1 - \ln x}{x^2}`. 令 :math:`f'(x) = 0` 解得 :math:`x = e`.
    由于 :math:`f''(x) = \dfrac{2 \ln x - 3}{x^3}`, 可得

    .. math::
        f''(e) = \dfrac{2 \ln e - 3}{e^3} = - \dfrac{1}{e^3} < 0.

    所以 :math:`x = e` 是极大值点, :math:`y = e^{1 / e}` 是相应的极大值.

    .. note::
        函数 :math:`y = x^{1/x}, x > 0` 的图像如下图所示

        .. tikz:: 函数 :math:`y = x^{1/x}, x > 0` 的图像
            :align: center
            :xscale: 60

            \begin{axis}[
                width=10cm, height=8cm, axis lines=middle,
                xlabel=$x$, ylabel=$y$,
                xmin=-0.3, xmax=5, ymin=-0.3, ymax=2,
                smooth, samples=200,
                domain=0.01:4.7, minor tick num=4
            ]
            \addplot[thick, cyan, smooth] {exp(ln(\x)/\x)};
            \end{axis}