第二章随堂测验答案解析

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

解

由复合函数求导的链式法则可得

\[f'(x) = \dfrac{1}{2\sqrt{x + \sqrt{x + \sqrt{x}}}} \cdot \left(1 + \dfrac{1}{2\sqrt{x + \sqrt{x}}} \cdot \left(1 + \dfrac{1}{2\sqrt{x}}\right)\right)\]

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

解

\[\begin{split}\mathrm{d} y & = \mathrm{d} (\ln f(\sin^2 x)) \\ & = \dfrac{1}{f(\sin^2 x)} \cdot f'(\sin^2 x) \cdot 2\sin x \cos x \cdot \mathrm{d} x \\ & = \dfrac{\sin 2x}{f(\sin^2 x)} \cdot f'(\sin^2 x) \cdot \mathrm{d} x\end{split}\]

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

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

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

   提示: 利用带佩亚诺型余项的麦克劳林公式

   \[\begin{split}& \dfrac{1}{1 - t} = 1 + t + t^2 + o(t^2), \\ & \ln (1 + t) = t - \dfrac{t^2}{2} + o(t^2).\end{split}\]

解

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

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

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

利用带佩亚诺型余项的麦克劳林公式

\[\begin{split}& \dfrac{1}{1 - t} = 1 + t + t^2 + o(t^2), \\ & \ln (1 + t) = t - \dfrac{t^2}{2} + o(t^2),\end{split}\]

可得

\[\begin{split}\dfrac{y'(t)}{y(t)} & = - \dfrac{\dfrac{t}{t + 1} - \ln (t + 1)}{t^2} \\ & = - \dfrac{t(1 - t + o(t)) - (t - \dfrac{t^2}{2} + o(t^2))}{t^2} \\ & = \dfrac{1}{2} + o(1).\end{split}\]

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

(2). 由于

\[\lim\limits_{x \to \infty} \left( \dfrac{x}{x + 1} \right)^x = \lim\limits_{x \to \infty} \dfrac{1}{\left( 1 + \dfrac{1}{x} \right)^x} = \dfrac{1}{e},\]

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

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

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

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

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

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

   \[a e^b - b e^a = (a - b) (1 - \xi)e^\xi.\]

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

解

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

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

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

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

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

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

上式两边同时乘以 \(ab\) 即可得到题目要证明的结论:

\[a e^b - b e^a = (a - b) (1 - \xi)e^\xi.\]

备注

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

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

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

\[\dfrac{f(b) - f(a)}{g(b) - g(a)} = \dfrac{f'(\xi)}{g'(\xi)},\]

代入 \(f'(x), g'(x)\) 的表达式可得

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

5. 求函数 \(y = x^{1/x}, x > 0\) 的极大值.

解

对 \(y = x^{1/x}, x > 0\) 两边同时取对数可得

\[\ln y = \dfrac{\ln x}{x}.\]

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

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

所以 \(x = e\) 是极大值点, \(y = e^{1 / e}\) 是相应的极大值.

备注

函数 \(y = x^{1/x}, x > 0\) 的图像如下图所示

函数 \(y = x^{1/x}, x > 0\) 的图像