高数强化NO8|导数的计算|导数定义|公式|求导法则|低阶导数|高阶导数

一元函数导数的计算

利用导数定义

基本求导公式

$$\begin{array}{rl}& (x^a{)}^{\prime }=a{x}^{a-1},(\ln x{)}^{\prime }=\frac{1}{x},(e^x{)}^{\prime }=e^x,(\sin x{)}^{\prime }=\cos x,(\cos x{)}^{\prime }=-\sin x,(a^x{)}^{\prime }=a^x\ln a\\ & \\ & (\tan x{)}^{\prime }={\sec }^{2}x,(\cot x{)}^{\prime }=-{\csc }^{2}x,(\sec x{)}^{\prime }=\sec x\tan x,(\csc x{)}^{\prime }=-\csc x\cot x\\ & \\ & (\arctan x{)}^{\prime }=\frac{1}{1+x^2},(\text{arccot}x{)}^{\prime }=-\frac{1}{1+x^2},(\arcsin x{)}^{\prime }=\frac{1}{\sqrt{1-x^2}},(\arccos x{)}^{\prime }=-\frac{1}{\sqrt{1-x^2}}\end{array}$$

求导法则

1. 四则运算求导法则
   $$\begin{array}{rl}& \text{设函数}f(x)\text{与}g(x)\text{均可导，则}\\ & \\ & [f(x)\pm g(x){]}^{\prime }=f^{\prime }(x)\pm g^{\prime }(x),[f(x)g(x){]}^{\prime }=f^{\prime }(x)g(x)+f(x)g^{\prime }(x),\\ & \\ & {\left[\frac{f(x)}{g(x)}\right]}^{\prime }=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{g^2(x)}.\end{array}$$
2. 复合函数求导法则
   $$\begin{array}{rl}& \text{定理：设}y=f(u),u=g(x)，\text{如果}g(x)\text{在}x\text{处可导，且}f(u)\text{在对应的}u=g(x)\text{处可导，}\\ & \text{则复合函数}y=f(g(x))\text{在}x\text{处可导，且有：}\\ & \\ & [f(g(x)){]}^{\prime }=f^{\prime }(g(x))g^{\prime }(x)\text{或}\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}.\end{array}$$
3. 反函数求导法则
   $$\begin{array}{rl}& \text{定理：设函数可导且}{f}^{\prime }(x)\ne 0,\text{并令其反函数为}x=f^{-1}(y),\\ & (f^{-1}(y){)}^{\prime }=\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}=\frac{1}{f^{\prime }(x)}=\frac{1}{f^{\prime }(f^{-1}(y))}.\end{array}$$
4. 变上限积分求导
   $$\begin{array}{rl}& (1){\left({\int }_a^{x}f(t)dt\right)}^{\prime }=f(x);\\ & \\ & (2){\left({\int }_x^{b}f(t)dt\right)}^{\prime }=-f(x);\\ & \\ & (3){\left[{\int }_a^{u(x)}f(t)dt\right]}^{\prime }=f(u(x))u^{\prime }(x);\\ & \\ & (4){\left[{\int }_{v(x)}^{u(x)}f(t)dt\right]}^{\prime }=f(u(x))u^{\prime }(x)-f(v(x))v^{\prime }(x).\end{array}$$

多元函数导数的计算

利用偏导数定义

基本求导公式

复合函数求导法则

$$\begin{array}{rl}& \text{根据复合函数中间变量的不同形式，我们有如下求导公式：}\\ & \\ & (1)\text{如果}z=f(u,v)=f(\phi (t),\psi (t))，\text{则}\frac{dz}{dt}=\frac{\partial f}{\partial u}\frac{du}{dt}+\frac{\partial f}{\partial v}\frac{dv}{dt}；\\ & \\ & (2)\text{如果}z=f(u,v)=f(\phi (x,y),\psi (x,y))，\text{则}\frac{\partial z}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}；\\ & \\ & \frac{\partial z}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}；\end{array}$$

隐函数求导法则

$$\begin{array}{rl}& \text{定理:设}F(x,y,z)\text{在}(x_0,y_0,z_0)\text{的某邻域内有连续的一阶偏导数,且}F(x_0,y_0,z_0)=0,\\ & \text{若}{F}_z^{\prime }(x_0,y_0,z_0)\ne 0,\text{则方程}F(x,y,z)=0\text{在点}(x_0,y_0,z_0)\text{的某邻域内恒能唯一确定一}\\ & \text{个连续且具有连续偏导数的函数}z=z(x,y),\text{且}\\ & \\ & \frac{\partial z}{\partial x}=-\frac{F_x^{\prime }}{F_z^{\prime }},\frac{\partial z}{\partial y}=-\frac{F_y^{\prime }}{F_z^{\prime }},\\ & \\ & \text{其中}{F}_x^{\prime },F_y^{\prime },F_z^{\prime }\text{是三元函数}F(x,y,z)\text{对}x,y,z\text{的偏导数.}\end{array}$$

一元函数导数的计算

低阶导数的计算

$$\begin{array}{rl}& \text{解：}\ln \sqrt{\frac{e^{2x}}{e^{2x}+1}}=\frac{1}{2}\left(\ln e^{2x}-\ln (e^{2x}+1)\right)=\frac{1}{2}\left(2x-\ln (e^{2x}+1)\right)\\ & =x-\frac{1}{2}\ln (e^{2x}+1)\\ & \\ & y^{\prime }=\frac{e^x}{1+e^{2x}}-1+\frac{1}{2}\cdot \frac{2e^{2x}}{e^{2x}+1}\\ & \\ & y^{\prime }(1)=\frac{e}{1+e^2}-1+\frac{e^2}{1+e^2}=\frac{e-1-e^2+e^2}{1+e^2}=\frac{e-1}{1+e^2}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{如果函数为多项相乘，直接求导比较麻烦，可以考虑先取对数再求导。求对}\\ & \text{数函数}\ln |f(x)|\text{的导数，可以忽略绝对值的影响：}{\left(\ln |f(x)|\right)}^{\prime }=\frac{f^{\prime }(x)}{f(x)}。\\ & \\ & 2.\text{要计算幂指函数}u(x{)}^{v(x)}\text{的导数，基本思路是先用对数恒等变形，再运用求导法则计}\\ & \text{算其导数：}\\ & {\left[u(x{)}^{v(x)}\right]}^{\prime }={\left[e^{v(x)\ln u(x)}\right]}^{\prime }=e^{v(x)\ln u(x)}{\left[v(x)\ln u(x)\right]}^{\prime }=u(x{)}^{v(x)}\left[v^{\prime }(x)\ln u(x)+v(x)\frac{u^{\prime }(x)}{u(x)}\right]\\ & \\ & 3.\text{当遇到对数函数求导时，可以首先利用公式}\ln x^k=k\ln |x|(x>0)，\\ & \ln ab=\ln a+\ln b(a,b>0)\text{将对数函数简化。}\end{array}$$

$$\begin{array}{rl}& \text{解：在}y-x{e}^{y-1}=1①\text{两边对}x\text{求导：}{y}^{\prime }-e^{y-1}-x{e}^{y-1}\cdot y^{\prime }=0②\\ & \text{在}②\text{两边对}x\text{求导：}{y}^{\prime \prime }-e^{y-1}\cdot y^{\prime }-\left(e^{y-1}\cdot y^{\prime }\right)-x{\left(e^{y-1}\cdot y^{\prime }\right)}^{\prime }=0③\\ & \text{将}x=0\text{代入①，得}y(0)=1；x=0,y=1\text{代入②，得}{y}^{\prime }(0)=1；x=0,y=1,y^{\prime }=1\text{代入③，得}{y}^{\prime \prime }=2.\\ & \\ & {\frac{dz}{dx}|}_{x=0}=f^{\prime }\left(\ln y-\sin x\right)\cdot \left(\frac{y^{\prime }}{y}-\cos x\right){|}_{x=0}=0\\ & \\ & {\frac{d^{2}z}{d{x}^2}|}_{x=0}=\left[f^{\prime \prime }\left(\ln y-\sin x\right){\left(\frac{y^{\prime }}{y}-\cos x\right)}^2+f^{\prime }\left(\ln y-\sin x\right)\left(\frac{y^{\prime \prime }y-(y^{\prime }{)}^2}{y^2}+\sin x\right)\right]{|}_{x=0}\\ & =1\end{array}$$

$$\begin{array}{rl}& \text{变限积分求导}\\ & ①\phi (x)={\int }_0^{x}f(t)dt\text{直接用公式}\\ & ②\phi (x)={\int }_0^{x}xf(t)dt\text{提出去}\\ & ③\phi (x)={\int }_0^{x}f(t,x)dt\text{换元}\end{array}$$
$$\begin{array}{rl}& \text{解：}\phi (1)={\int }_0^{1^2}1\cdot f(t)dt={\int }_0^{1}f(t)dt.\\ & \\ & {\phi }^{\prime }(x)={\left(x{\int }_0^{x^2}f(t)dt\right)}^{\prime }\\ & ={\int }_0^{x^2}f(t)dt+x\cdot f(x^2)\cdot 2x\\ & \\ & {\phi }^{\prime }(1)={\int }_0^{1}f(t)dt+2f(1)\\ & =\phi (1)+2f(1)\\ & \\ & 5=1+2f(1)\\ & f(1)=2.\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{如果积分变限函数积分号下出现关于}x\text{的函数}u(x),\text{类似于本题，则可以将}\\ & u(x)\text{移至积分号外，再进行求导，即}{\int }_a^{x}u(x)f(t)dt=u(x){\int }_a^{x}f(t)dt,\text{故}\\ & \\ & {\left({\int }_a^{x}u(x)f(t)dt\right)}^{\prime }=u^{\prime }(x){\int }_a^{x}f(t)dt+u(x)f(x).\end{array}$$

$$\begin{array}{rl}& \text{解：}{\int }_0^{x}tf(x-t)dt\stackrel{u=x-t}{=}{\int }_x^0(x-u)f(u)d(x-u)={\int }_0^x(x-u)f(u)du\\ & =x{\int }_0^{x}f(u)du-{\int }_0^{x}uf(u)du\\ & \\ & \text{法1：两边对}x\text{求导：}{\int }_0^{x}f(u)du+xf(x)-xf(x)=\sin x\\ & {\int }_0^{x}f(u)du=\sin x\\ & \text{令}x=\frac{\pi }{2},{\int }_0^{\frac{\pi }{2}}f(u)du=\sin \frac{\pi }{2}=1\end{array}$$
$$\begin{array}{rl}& \text{不变限求导：}{\left({\int }_0^x(x-u)f(u)du\right)}^{\prime }=(x-x)f(x)\cdot (x{)}^{\prime }-(x-0)f(0)\cdot (0{)}^{\prime }\\ & +{\int }_0^x\frac{\partial \left[(x-u)f(u)\right]}{\partial x}du={\int }_0^{x}f(u)du.\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1、\text{如果}x\text{出现在积分号下被积函数的自变量中,类似于本题,则可以先进行换}\\ & \text{元,再进行求导,即形如}{\int }_a^{x}f(u(x,t))dt\text{的积分变限函数,令}u=u(x,t)\text{进行换元.}\\ & \\ & 2、\text{积分变限函数的三大题型：}\\ & ①\text{直接使用公式进行计算;}\\ & ②\text{将关于}x\text{的函数移至积分号外,再进行求导;}\\ & ③\text{先进行换元,再进行求导.}\end{array}$$

$$\begin{array}{rl}& \text{解：}{F}^{\prime }(t)={\left({\int }_1^t\left({\int }_y^{t}f(x)dx\right)dy\right)}^{\prime }={\int }_t^{t}f(x)dx\cdot (t{)}^{\prime }-{\int }_1^{t}f(x)dx\cdot (1{)}^{\prime }\\ & +{\int }_1^t\frac{\partial \left({\int }_y^{t}f(x)dx\right)}{\partial t}dy\\ & ={\int }_1^{t}f(t)dy=f(t)(t-1),F^{\prime }(2)=f(2)\\ & 选B\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{变限积分求导公式推广：设}\psi (x),\phi (x)\text{均可导，}f(x,t)\text{关于}x\text{的偏导数存}\\ & \text{在且连续，则}\\ & {\left({\int }_{\phi (x)}^{\psi (x)}f(x,t)dt\right)}^{\prime }=f(x,\psi (x)){\psi }^{\prime }(x)-f(x,\phi (x)){\phi }^{\prime }(x)+{\int }_{\phi (x)}^{\psi (x)}\frac{\partial f}{\partial x}dt。\end{array}$$

高阶导数的计算

$$\begin{array}{rl}& \text{解：}{f}^{\prime }(x)=2(x+1)+2f(x)\\ & f^{\prime \prime }(x)=2+2f^{\prime }(x)\\ & f^{\prime \prime \prime }(x)=2f^{\prime \prime }(x)\\ & f^{(4)}(x)=2f^{\prime \prime \prime }(x)\\ & \text{故}{f}^{(n)}(x)=2f^{(n-1)}(x)=2^2{f}^{(n-2)}(x)=\cdots =2^{n-2}{f}^{\prime \prime }(x)(n\ge 3)\\ & \\ & f^{\prime \prime }(0)=2+2f^{\prime }(0)=2+2\left(2+2f(0)\right)=10\\ & f^{(n)}(0)=2^{n-2}{f}^{\prime \prime }(0)=2^{n-2}\cdot 10=5\cdot 2^{n-1}(n\ge 3)\\ & \\ & \text{当}n=2\text{时，}{f}^{\prime \prime }(0)=10(n=2\text{也适合此等式})\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{如果考试中给出的是抽象函数,可以计算函数的一阶、二阶、三阶导数,进一步}\\ & \text{寻找规律.}\end{array}$$

$$\begin{array}{rl}& \text{解：}{f}^{(n)}(x)={\mathrm{C}}_n^0{x}^2\cdot {\left(\ln (1+x)\right)}^{(n)}+{\mathrm{C}}_n^1(x^2{)}^{\prime }{\left(\ln (1+x)\right)}^{(n-1)}\\ & +{\mathrm{C}}_n^2(x^2{)}^{\prime \prime }{\left(\ln (1+x)\right)}^{(n-2)}\\ & \\ & f^{(n)}(0)=\frac{n(n-1)}{2}\cdot 2\cdot (-1{)}^{n-3}{\frac{(n-3)!}{(1+x{)}^{n-2}}|}_{x=0}\end{array}$$
$$\begin{array}{rl}& =n(n-1)(n-3)!(-1{)}^{n-2}\\ & =\frac{n!}{n-2}(-1{)}^{n-1}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{常见的六个高阶导数公式}\\ & \begin{array}{rl}& (e^x{)}^{(n)}=e^x;\\ & (a^x{)}^{(n)}=a^x(\ln a{)}^n;\\ & (\sin x{)}^{(n)}=\sin \left(x+\frac{n}{2}\pi \right);\\ & (\cos x{)}^{(n)}=\cos \left(x+\frac{n}{2}\pi \right);\\ & (x^a{)}^{(n)}=a(a-1)\cdots (a-n+1)x^{a-n}\\ & (\ln x{)}^{(n)}=(-1{)}^{n-1}\frac{(n-1)!}{x^n};\end{array}\\ & \text{特例：}{\left(\frac{1}{x}\right)}^{(n)}=(-1{)}^n\frac{n!}{x^{n+1}}。\\ & \text{其次，考研往往喜欢将上述六个公式与一次函数}ax+b\text{复合进行考查，此时仅需在公式}\\ & \text{之后乘以}{a}^n，\text{即}(f(ax+b){)}^{(n)}=a^n{f}^{(n)}(ax+b).\\ & \\ & 2.\text{Leibniz公式, 即}\\ & \text{设函数}u(x)、v(x)\text{具有}n\text{阶导数，则}\\ & (u\cdot v{)}^{(n)}={\mathrm{C}}_n^0(u{)}^{(n)}\cdot v+{\mathrm{C}}_n^1(u{)}^{(n-1)}\cdot v^{\prime }+{\mathrm{C}}_n^2(u{)}^{(n-2)}\cdot v^{\prime \prime }+\cdots +{\mathrm{C}}_n^{n}u\cdot (v{)}^{(n)}.\\ & \\ & 1、\text{Leibniz公式可借助于二项式定理进行记忆；}\\ & 2、\text{在考研数学中，}u(x)、v(x)\text{的选择往往有一定的规律：}u(x)\text{往往是常见的六个高阶}\\ & \text{导数公式之一，}v(x)\text{往往最多是二次函数}a{x}^2+bx+c，\text{这样利用Leibniz公式展开最多}\\ & \text{只有3项.}\end{array}$$

$$\begin{array}{rl}& \text{解：一方面，当}x\to 0\text{时，}\\ & f(x)=x-\frac{x^3}{3}+o(x^3)-x\left(1-a{x}^2+o(x^2)\right)\\ & =\left(a-\frac{1}{3}\right)x^3+o(x^3)\\ & \\ & \text{另一方面，当}x\to 0\text{时，}f(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2}{x}^2+\frac{f^{\prime \prime \prime }(0)}{6}{x}^3+o(x^3)\\ & \text{由泰勒公式展开的唯一性可知，对应系数相等：}\\ & \frac{f^{\prime \prime \prime }(0)}{6}=a-\frac{1}{3}\\ & \text{又}\frac{1}{6}=a-\frac{1}{3}，\text{故}a=\frac{1}{2}\end{array}$$
$$\begin{array}{rl}& \text{解：法2. 忘记泰勒公式的补救办法.}\\ & \underset{x\to 0}{\lim }\frac{\arctan x-\frac{x}{(1+a{x}^2)}}{x^3}=\underset{x\to 0}{\lim }\frac{\arctan x(1+a{x}^2)-x}{x^3(1+a{x}^2)}\\ & =\underset{x\to 0}{\lim }\frac{\arctan x-x}{x^3}+\underset{x\to 0}{\lim }\frac{\arctan x\cdot a{x}^2}{x^3}\\ & =-\frac{1}{3}+a\\ & \\ & \text{故}\arctan x-\frac{x}{(1+a{x}^2)}\sim \left(a-\frac{1}{3}\right)x^3.\\ & \text{由}\frac{f^{\prime \prime \prime }(0)}{3!}=a-\frac{1}{3}\text{，得}a=\frac{1}{2}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{Taylor公式计算高阶导数的理论基础:}\\ & f(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)+\frac{f^{\prime \prime }(x_0)}{2!}(x-x_0{)}^2+\cdots +\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n+\dots .\\ & \\ & \text{由Taylor公式展开的唯一性可知，}{f}^{(n)}(x_0)\text{就等于}(x-x_0{)}^n\text{的系数乘以}n!\end{array}$$
$$f(x)\sim k{x}^3⟺f(0)=0,f^{\prime }(0)=0,f^{\prime \prime }(0)=0,f^{\prime \prime \prime }(0)=3!\cdot k$$
$$①\text{已知}f(0)=0,f^{\prime }(0)=0,f^{\prime \prime }(0)=2,\Rightarrow \text{当}x\to 0\text{时，}f(x)\sim \frac{f^{\prime \prime }(0)}{2!}{x}^2=x^2$$

$$\begin{array}{rl}& \text{解：}f(x)\text{为偶函数，}{f}^{\prime }(x)\text{为奇，}{f}^{\prime \prime }(x)\text{偶，}{f}^{\prime \prime \prime }(x)\text{奇.}\\ & f^{\prime \prime \prime }(0)=0,f(x)\text{周期为}2\pi ,f^{\prime \prime \prime }(x)\text{周期}2\pi .\\ & f^{\prime \prime \prime }(2\pi )=f^{\prime \prime \prime }(0)=0\end{array}$$
$$\begin{array}{rl}& \text{小结：偶函数的奇数阶导数为奇函数，}\\ & \text{奇函数的偶数阶导数为奇函数.}\end{array}$$
$$\begin{array}{rl}& \text{改：}g(x)=e^{\sin x}-e^{-\sin x}\\ & g^{(4)}(2\pi )=0\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{高阶导数方法总结}\\ & \text{高阶导数的计算一般有五种方法:}\\ & 第一、求出一阶、二阶、三阶导数,进行递推;\\ & 第二、使用常见的六个高阶导数公式;\\ & 第三、使用Leibniz公式;\\ & 第四、使用Taylor公式;\\ & 第五、借助函数奇偶性。\end{array}$$