高数 02.03高阶导数

$\color{blue}{第二章 第三节 高阶导数}$

$\color{blue}{一、高阶导数的概念}$

$引例：变速直线运动 s = s(t) \\ 速度 v = \dfrac{ds}{dt}, 即 v = s^{\prime} \\ 加速度 a = \dfrac{dv}{dt} = \dfrac{d}{dt}(\dfrac{ds}{dt}) \\ 即 a = (s^{\prime})^{\prime} \\ $
$定义.若函数y = f(x)的导数y^{\prime} = f^{\prime}(x)可导，则称 \\ f^{\prime}(x)的导数为f(x)的二阶导数，记作y{\prime\prime}或\dfrac{d^2y}{dx^2},即 \\ y^{\prime \prime} = (y^{\prime})^{\prime} 或 \dfrac{d^2 y}{d x^2} = \dfrac{d}{dx}{(\dfrac{dy}{dx})} \\ 类似地，二阶导数的导数称为三阶导数，依次类推，\\ n - 1阶导数的导数称为n阶导数，分别记作 \\ y^{\prime \prime \prime}, y^{(4)}, \cdots, y^{(n)} \\ 或 \dfrac{d^3 y}{dx^3}, \dfrac{d^4 y}{d x^4}, \cdots, \dfrac{d^n y}{d x^n}$

$例1.设 y = ax + b,求 y^{\prime \prime}.$
$\color{blue}{解： y^{\prime} = a, \quad y^{\prime \prime} = 0 }$

$例2.设 s = \sin{\omega t},求 s^{\prime \prime}.$
$\color{blue}{解： s^{\prime} = \omega \cos{\omega t}, \quad s^{\prime \prime} = -\omega ^2 \sin{\omega t} }$

$例3.设 y = a_0 + a_1x + a_2 x^2 + \cdots + a_n x^n,求 y^{(n)}.$
$\color{blue}{解： y^{\prime} = a_1 + 2a_2 x^ 1 + \cdots + na_n x^{n - 1} \\ y^{(2)} = 2\cdot 1 a_2 + \cdots + n \cdot (n - 1)a_n x^{(n - 2)} \\ \cdots \\ y^{(n)} = n! \cdot a_n x^{(n - n)} = a_n \cdot n! }$

$思考：设 y = x^{\mu} (\mu 为任意常数)，问 y^{(n)} = ?$
$\color{blue}{(x^{\mu})^{(n)} = \mu (\mu - 1)(\mu - 2) \cdots (m - n + 1)x^{\mu - n} }$

$例4.设 y = e^{ax},求 y^{(n)}.$
$\color{blue}{解： y^{\prime} = a e^{ax}, \quad y^{\prime \prime} = a^2 e^{ax} \\ y^{(n)} = a^n e^ax \\ 特别有：(e^x)^{(n)} = e^x }$

$例5.设 y = \ln(1 + x),求 y^{(3)}.$
$\color{blue}{解： y^{\prime} = \dfrac{1}{1 + x} \\ y^{\prime \prime} = -(1 + x)^{-2} \\ y^{(3)} = \dfrac{2}{(1 + x)^3} \\ }$

$\color{blue}{二、高阶导数的运算法则}$

$设函数u = u(x) 及 v = v(x) 都有n阶导数,则\\ 1.(u \pm v)^{(n)} = u^{(n)} \pm v^{(n)} \\ 2.(Cu)^{(n)} = Cu^{(n)} (C为常数) \\ 3.(u v)^{(n)} = u^{(n)}v + n u^{(n - 1)}v^{\prime} + \dfrac{n(n - 1)}{2!} u^{(n - 2)}v^{\prime \prime} + \cdots + \dfrac{n(n - 1) \cdots (n -k + 1)}{k!} u^{(n -k)}v^{(k)} + \cdots + u v^{(n)} \\ 3也称为莱布尼兹(Leibniz)公式$

$例6.y = x^2 e^{2x}, 求y^{(20)}.$
$解：设u = e^{2x}, v = x^2,则 \\ u^{(k)} = 2^k e^{2x} (k = 1, 2, \cdots, 20) \\ v^{(1)} = 2x \\ v^{(2)} = 2 \\ v^{(k)} = 0 (k = 3, 4, \cdots, 20) \\ (u v)^{(20)} = u^{(20)}v + 20 u^{(19)}v^{\prime} + \dfrac{20(19)}{2!} u^{(18)}v^{\prime \prime}\\ = 2^{20}e^{2x} x^2 + 20 \cdot 2^{19} e^{2x} \cdot 2x + 190 \cdot 2^{18}e^{2x} \cdot 2 \\ = 2^{20}e^{2x}(x^2 + 20x + 95)$

内容小结：
高阶导数的求法
(1)逐阶求导法
(2)利用归纳法
(3)间接法–利用已知的高阶导数公式
$如, (\dfrac{1}{a + x})^{(n)} = (-1)^n \dfrac{n!}{(a + x)^{n + 1}}$
$(\dfrac{1}{a - x})^{(n)} = \dfrac{n!}{(a - x)^{n + 1}}$
(4)利用莱布尼兹公式

思考与练习
1.如何求下列函的n阶导数?
$(1) y = \dfrac{1 - x}{1 + x}$
$(2) y = \dfrac{x^3}{1 - x}$
$(3) y = \dfrac{1}{x^2 - 3x + 2}$
$\color{blue}{解： (1) y = -1 + 2\dfrac{1}{1 + x} \\ y^{(n)} = (-1)^{(n)} + 2 (\dfrac{1}{1 + x})^{(n)} \\ = 0 + 2 \cdot (-1)^n \cdot \dfrac{n!}{(1 + x)^{n + 1}} \\ = (-1)^n \cdot \dfrac{2n!}{(1 + x)^{n + 1}} \ \\ (2)y = \dfrac{x^3}{1 - x} \\ = -x^2 - x - 1 + \dfrac{1}{1 - x} \\ \ \\ y^{(1)} = -2x - 1 + \dfrac{1}{(1 - x)^2} \\ y^{(2)} = -2 + \dfrac{2}{(1 - x)^3} \\ y^{(n)} = \dfrac{n!}{(1 - x)^{n + 1}}, n \geq 3 }$

$\color{blue}{(3) y = \dfrac{1}{x^2 - 3x + 2} \\ = \dfrac{1}{x -2} - \dfrac{1}{x - 1} \\ y^{(n)} = (-1)^n n! [\dfrac{1}{(x -2)^{n + 1}} - \dfrac{1}{(x - 1)^{n + 1}}] }$

$(4) y = \sin^6{x} + \cos^6{x}$
$\color{blue}{解： y = (\sin^2{x})^3 + (\cos^2{x})^3 \\ = (\sin^2{x} + \cos^2{x})(\sin^4{x} -\sin^2{x}\cos^2{x} + \cos^4{x}) \\ =(\sin^2{x} + \cos^2{x})^2 - 3\sin^2{x}\cos^2{x} \\ = 1 - \dfrac{3}{4}\sin^2{2x} \\ = \dfrac{5}{8} + \dfrac{3}{8} \cos{4x} \\ y^{(n)} = \dfrac{3}{8} \cdot 4^n cos(4x + n\frac{\pi}{2}) }$