第三章 导数与微分

第三章 导数与微分#

课后习题解答#

§3.1 导数的概念#

  1. 设 \(f(x)\) 在 \(x_0\) 处可导, 求下列极限值.

    1. \(\lim\limits_{\Delta x \to 0} \dfrac{f(x_0 + 2\Delta x) - f(x_0 - \Delta x)}{\Delta x}\).

解

由可导的定义, 有

\[\begin{split}f(x_0 + 2\Delta x) = f(x_0) + f'(x_0) \cdot 2\Delta x + o(\Delta x), \\ f(x_0 - \Delta x) = f(x_0) - f'(x_0) \cdot \Delta x + o(\Delta x).\end{split}\]

因此

\[\begin{split}& \lim\limits_{\Delta x \to 0} \dfrac{f(x_0 + 2\Delta x) - f(x_0 - \Delta x)}{\Delta x} \\ = & \lim\limits_{\Delta x \to 0} \dfrac{f(x_0) + f'(x_0) \cdot 2\Delta x + o(\Delta x) - \left( f(x_0) - f'(x_0) \cdot \Delta x + o(\Delta x) \right)}{\Delta x} \\ = & \lim\limits_{\Delta x \to 0} \dfrac{3 f'(x_0) \cdot \Delta x + o(\Delta x)}{\Delta x} = 3 f'(x_0).\end{split}\]

  1. 已知 \(f(x) = \begin{cases} \sin x, & x > 0, \\ e^x, & x \leqslant 0, \end{cases}\) 求 \(f'(x)\).

解

当 \(x > 0\) 时, \(f(x) = \sin x\), 因此 \(f'(x) = \cos x\).

当 \(x < 0\) 时, \(f(x) = e^x\), 因此 \(f'(x) = e^x\).

当 \(x = 0\) 时, \(\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} \sin x = 0\), \(\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} e^x = 1\), 因此 \(f(x)\) 在 \(x = 0\) 处不连续, 故不可导. 综上所述, 有

\[\begin{split}f'(x) = \begin{cases} \cos x, & x > 0, \\ \text{不存在}, & x = 0, \\ e^x, & x < 0. \end{cases}\end{split}\]

备注

这题要注意的地方是, 不要把导函数的单边极限 \(\lim\limits_{x \to 0^+} f'(x)\), \(\lim\limits_{x \to 0^-} f'(x)\), 和单边导数 \(f'_+(0)\), \(f'_-(0)\) 搞混了.

§3.2 求导法则#

§3.3 高阶导数#

  1. 证明: 函数

    \[\begin{split}f(x) = \begin{cases} e^{-\frac{1}{x^2}}, & x \neq 0, \\ 0, & x = 0, \end{cases}\end{split}\]

    在 \(x = 0\) 处 \(n\) 阶可导且 \(f^{(n)}(0) = 0\), 其中 \(n\) 是任意的正整数.

证明

由题意可知, 当 \(x \neq 0\) 时, \(f(x) = e^{-\frac{1}{x^2}}\) 是一个初等函数, 因此在 \(x \neq 0\) 时, \(f(x)\) 存在任意阶导数.

下面用数学归纳法证明 \(f(x)\) 在 \(x = 0\) 处 \(n\) 阶可导且 \(f^{(n)}(0) = 0\).

当 \(n = 1\) 时, 有

\[f'(0) = \lim\limits_{x \to 0} \dfrac{f(x) - f(0)}{x - 0} = \lim\limits_{x \to 0} \dfrac{e^{-\frac{1}{x^2}} - 0}{x} \xlongequal{t = \frac{1}{x}} \lim\limits_{t \to \infty} \dfrac{t}{e^{t^2}} = 0.\]

假设当 \(n = k\) 时, \(f^{(k)}(0) = 0\) 成立, 那么当 \(n = k + 1\) 时, 有

\[f^{(k+1)}(0) = \lim\limits_{x \to 0} \dfrac{f^{(k)}(x) - f^{(k)}(0)}{x - 0} = \lim\limits_{x \to 0} \dfrac{f^{(k)}(x) - 0}{x}.\]

注意到 \(f^{(k)}(x)\) 是由初等函数通过有限次求导得到的函数, 其中前几项为

\[\begin{split}f'(x) & = \dfrac{2}{x^3} e^{-\frac{1}{x^2}}, \\ f''(x) & = \left( \dfrac{4}{x^6} - \dfrac{6}{x^4} \right) e^{-\frac{1}{x^2}}, \\ f^{(3)}(x) & = \left( \dfrac{8}{x^9} - \dfrac{36}{x^7} + \dfrac{24}{x^5} \right) e^{-\frac{1}{x^2}}, \\ \cdots\end{split}\]

因此可以归纳地得到 (用数学归纳法验证), 对任意的正整数 \(k\), 有

\[f^{(k)}(x) = P_k\left( \dfrac{1}{x} \right) e^{-\frac{1}{x^2}},\]

其中 \(P_k(t)\) 是关于 \(t\) 的多项式. 验证 \(f^{(k+1)}(x) = P_{k+1}\left( \dfrac{1}{x} \right) e^{-\frac{1}{x^2}}\):

\[\begin{split}f^{(k+1)}(x) & = \dfrac{\mathrm{d}}{\mathrm{d} x} \left( P_k\left( \dfrac{1}{x} \right) e^{-\frac{1}{x^2}} \right) \\ & = P_k'\left( \dfrac{1}{x} \right) \cdot \left( -\dfrac{1}{x^2} \right) e^{-\frac{1}{x^2}} + P_k\left( \dfrac{1}{x} \right) \cdot \dfrac{2}{x^3} e^{-\frac{1}{x^2}} \\ & = \left( -\dfrac{1}{x^2} P_k'\left( \dfrac{1}{x} \right) + \dfrac{2}{x^3} P_k\left( \dfrac{1}{x} \right) \right) e^{-\frac{1}{x^2}} \\ & = P_{k+1}\left( \dfrac{1}{x} \right) e^{-\frac{1}{x^2}}.\end{split}\]

这里, \(P_{k+1}(t) = -t^2 P_k'(t) + 2t^3 P_k(t)\). 因此

\[\begin{split}f^{(k+1)}(0) & = \lim\limits_{x \to 0} \dfrac{P_k\left( \dfrac{1}{x} \right) e^{-\frac{1}{x^2}}}{x} \xlongequal{t = \frac{1}{x}} \lim\limits_{t \to \infty} t P_k(t) e^{-t^2} \\ & = \lim\limits_{t \to \infty} \dfrac{t P_k(t)}{e^{t^2}} = 0.\end{split}\]

§3.4 函数的微分#

补充内容#

§3.1 导数的概念#

  1. 处处连续, 但处处不可导的函数: Generalized Van der Waerden-Takagi 函数.

    该函数定义如下

    \[\begin{split}& \varphi(x) = d(x, \mathbb{Z}) = \min_{n \in \mathbb{Z}} |x - n|, \quad x \in \mathbb{R}, \\ & f(x) = \sum_{n=0}^{\infty} a^n \varphi(b^n x).\end{split}\]

    当 \(0 < a < 1\), \(b \in \mathbb{N}_{\geqslant 2}\), 且 \(ab \geqslant 1\) 时, \(f(x)\) 是一个在 \(\mathbb{R}\) 上处处不可导的连续函数.

    那么, 处处可导, 但导函数处处不连续的函数是否存在呢? 答案是不存在.

§3.3 高阶导数#

  1. 莱布尼茨公式 \((uv)^{(n)} = \sum\limits_{k=0}^n C_n^k u^{(k)} v^{(n-k)}\) 的证明:

证明

用数学归纳法证明. 当 \(n = 1\) 时, \((uv)' = u'v + uv'\), 成立.

假设当 \(n = k\) 时, \((uv)^{(k)} = \sum\limits_{i=0}^k C_k^i u^{(i)} v^{(k-i)}\) 成立, 那么 \(n = k + 1\) 时有

\[\begin{split}(uv)^{(k + 1)} & = \dfrac{\mathrm{d}}{\mathrm{d} x} \left( \sum\limits_{i=0}^k C_k^i u^{(i)} v^{(k-i)} \right) \\ & = \sum\limits_{i=0}^k C_k^i \dfrac{\mathrm{d}}{\mathrm{d} x} \left( u^{(i)} v^{(k-i)} \right) \\ & = \sum\limits_{i=0}^k C_k^i \left( u^{(i+1)} v^{(k-i)} + u^{(i)} v^{(k-i+1)} \right) \\ & = \sum\limits_{i=0}^k C_k^i u^{(i+1)} v^{(k-i)} + \sum\limits_{i=0}^k C_k^i u^{(i)} v^{(k-i+1)} \\ & = \sum\limits_{i=1}^{k+1} C_k^{i-1} u^{(i)} v^{(k-i+1)} + \sum\limits_{i=0}^k C_k^i u^{(i)} v^{(k-i+1)} \\ & = u^{(k+1)} v + \sum\limits_{i=1}^k \left( C_k^{i-1} + C_k^i \right) u^{(i)} v^{(k-i+1)} + u v^{(k+1)} \\ & = u^{(k+1)} v + \sum\limits_{i=1}^k C_{k+1}^i u^{(i)} v^{(k-i+1)} + u v^{(k+1)} \\ & = C_{k+1}^{k+1} u^{(k+1)} v + \sum\limits_{i=0}^k C_{k+1}^i u^{(i)} v^{(k-i+1)} + C_{k+1}^0 u v^{(k+1)} \\ & = \sum\limits_{i=0}^{k+1} C_{k+1}^i u^{(i)} v^{((k+1)-i)}\end{split}\]

于是当 \(n = k + 1\) 时, \((uv)^{(n)} = \sum\limits_{i=0}^n C_n^i u^{(i)} v^{(n-i)}\) 成立. 根据数学归纳法原理, 对于任意的 \(n \in \mathbb{N}\), \((uv)^{(n)} = \sum\limits_{i=0}^n C_n^i u^{(i)} v^{(n-i)}\) 成立.

目录