【数学】一元函数微分（宇哥笔记）

一元函数微分学

导数\微分定义及其考法

定义

导数定义

$$\begin{array}{rl}& 假设一个普通教室在t=9:00时为u=20^{\circ }C,\check{t}=9:05时教室的温度\check{u}=25^{\circ }C\\ & 问教室里的温度在这5min中的平均变化率是多少？\\ & 很明显\frac{\Delta u}{\Delta t}=1^{\circ }C/min，但是把时间变成今天与一年前的今天，温度相同都是20^{\circ }C\\ & 用刚才的方法来算其平均变化率就成了0，很显然这个结果不能描述实际情况\\ & 如果我们令\Delta t\to 0,我们就能求某时刻的瞬时变化率，如下：\\ & f^{\prime }(x)=\underset{\Delta x\to 0}{\lim }\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}瞬时变化率\\ & [注1]换元，令x_0+\Delta x=x,则f^{\prime }(x_0)=\underset{\Delta x\to x_0}{\lim }\frac{f(x)-f(x_0)}{x-x_0}\\ & [注2]左右导数，f_{+}^{\prime }(x)=\underset{\Delta x\to 0^{+}}{\lim }\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}右导数，f_{-}^{\prime }(x)=\underset{\Delta x\to 0_{-}}{\lim }\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}左导数\\ & [注3]导数存在的条件，f^{\prime }(x_0)\exists ⟺f_{+}^{\prime }(x_0)=f_{-}^{\prime }(x_0)\end{array}$$

微分定义

$$\begin{array}{rl}& 幂的微分\{\begin{array}{l}d(x^2)=2xdx\\ d(x^n)=n{x}^{n-1}dx\end{array}\\ & 微分的幂\{\begin{array}{l}d{x}^2=(dx{)}^2\\ d{x}^n=(dx{)}^n\end{array}\\ & dy=y^{\prime }dx\end{array}$$

[外链图片转存失败(img-bCNaeZ06-1562482602646)(D:\文字\公众号\考研\数学\张宇高数十八讲\4.一元函数微分\图.png)]
$$\begin{array}{rl}& 如上图所示，f(x)=x^{2}f(x+\Delta x)=(x+\Delta x{)}^2=x^2+2x\Delta x+(\Delta x{)}^2\\ & 令\Delta y=f(x+\Delta x)-f(x)=2x\Delta x+(\Delta x{)}^2\\ & \Delta y=y^{\prime }(x)\cdot \Delta x+\circ (\Delta x),则y^{\prime }(x)\cdot \Delta x=dy称为线性全部\\ & \therefore \Delta x=1\cdot \Delta x+0⟹dx=\Delta x\\ & ⟹dy=y^{\prime }(x)dx⟹\frac{dy}{dx}=y^{\prime }(x)\end{array}$$

$$\begin{array}{rl}[例1]& 设y=e^{x^2},求\frac{dy}{dx},\frac{dy}{d(x^2)},\frac{d^{2}y}{d{x}^2}\\ & \frac{dy}{dx}=y^{\prime }=e^{x^2}\cdot 2x\\ & \frac{d^{2}y}{d{x}^2}=y^{{}^{\prime \prime }}=e^{x^2}\cdot 4x^2+e^{x^2}\cdot 2\\ & \frac{dy}{d(x^2)}=\frac{e^{x^2}\cdot 2xdx}{2xdx}=e^{x^2}\frac{dy}{2xdx}=\frac{1}{2x}\cdot e^{x^2}\cdot 2x=e^{x^2}\\ [例2]& y=f(x),f^{\prime }(x_0)=\frac{1}{2},\Delta x\to 0时,y=f(x)在x=x_0处得微分dy与\Delta x是(同阶非等价)\\ & dy=y^{\prime }(x_0)dx=\frac{1}{2}dx=\frac{1}{2}\Delta x\\ & \underset{\Delta x\to 0}{\lim }\frac{dx}{\Delta x}=\underset{\Delta x\to 0}{\lim }\frac{\frac{1}{2}\Delta x}{\Delta x}=\frac{1}{2}\\ [例3]& 设f(x)=(\cos x-4)\sin x+3x,求\frac{df(x)}{d(x^2)}\\ & df(a)=f^{\prime }(x)dx=(-{\sin }^{2}x+(\cos x-4)\cdot \cos x+3)dx\\ & d(x^2)=2xdx\\ \therefore & \frac{df(x)}{d(x^2)}=\frac{-{\sin }^{2}x+(\cos x-4)\cos x+3}{2x}=\frac{(\cos x-1{)}^2}{x}\\ [例4]& 设f^{\prime }(0)=1,f^{\prime \prime }(0)=0,求证：在x=0处，有\frac{d^2}{d{x}^2}f(x^2)=\frac{d^2}{d{x}^2}{f}^2(x)\\ & y_1^{\prime }=f^{\prime }(x^2)\cdot 2x,y_1^{{}^{\prime \prime }}{|}_0=f^{\prime \prime }(x^2)\cdot 2x\cdot 2x+f^{\prime }(x^2)\cdot 2{|}_{x=0}=2\\ & y_2^{\prime }=2f(x)\cdot f^{\prime }(x),y_2^{{}^{\prime \prime }}{|}_0=2f^{\prime }(x)f^{\prime }(x)+2f(x)f^{\prime \prime }(x){|}_{x=0}=2\end{array}$$

考法

抽象函数在一点（泛指x与特指x）

$$\begin{array}{rl}[例1]& 证明：若f(x)可导且为偶函数，请推f^{\prime }(x)为奇函数\\ & [分析]已知f(x)=f(-x)\\ & \therefore f^{\prime }(-x)=\underset{\Delta x\to 0}{\lim }\frac{f(-x+\Delta x)-f(-x)}{\Delta x}=-\underset{-\Delta x\to 0}{\lim }\frac{f(x+(-\Delta x))-f(x)}{-\Delta x}=-f^{\prime }(x)\\ [例2]& 证明f(x)可导，周期为T，请推f^{\prime }(x)的周期也是T\\ & [分析]已知f(x+T)=f(x)\\ & \therefore f^{\prime }(x+T)=\underset{\Delta x\to 0}{\lim }\frac{f(x+T+\Delta x)-f(x+T)}{\Delta x}=\underset{\Delta x\to 0}{\lim }\frac{f(x+\Delta x)-f(x)}{\Delta x}=f^{\prime }(x)\\ [例3]& f(x)二阶可导，T=2，奇函数，且f(\frac{1}{2})>0,f^{\prime }(x)>0,比较f(-\frac{1}{2}),f^{\prime }(\frac{3}{2}),f^{\prime \prime }(0)的大小\\ & \because 该函数为奇函数\therefore f(-x)=-f(x)\to f(-\frac{1}{2})=-f(\frac{1}{2})<0\\ & \because f(x):T=2\therefore f^{\prime }(x):T=2且f^{\prime }(x)为偶函数\\ & \therefore f^{\prime }(\frac{3}{2})=f^{\prime }(\frac{3}{2}-2)=f^{\prime }(-\frac{1}{2})=f^{\prime }(\frac{1}{2})>0\\ & \therefore f^{\prime \prime }(x)为奇函数即:f^{\prime \prime }(0)=0\\ & 得f(-\frac{1}{2})<f^{\prime \prime }(0)<f^{\prime }(\frac{3}{2})\\ [例4]& y=f(x)与y={\int }_0^{\arctan x}{e}^{-t^2}dt在(0,0)处切线相同，写出切线方程，求\underset{n\to \infty }{\lim }nf(\frac{2}{n})\\ & [分析]f^{\prime }(x_0)=k,切线方程为y-y_0=f^{\prime }(x_0)(x-x_0)\\ & ({\int }_0^{\arctan x}{e}^{-t^2}dt{)}_x^{\prime }=e^{-(\arctan x{)}^2}\cdot \frac{1}{1+x^2},令x=0,则f^{\prime }(0)=1,故切线方程为y=x\\ & \therefore \underset{n\to \infty }{\lim }nf(\frac{2}{n})=2\underset{\frac{2}{n}\to 0^{+}}{\lim }\frac{f(0+\frac{2}{n})-f(0)}{\frac{2}{n}}=2\cdot f^{\prime }(0)=2\\ [例5]& 设f^{\prime }(1)=1,则\underset{x\to 1}{\lim }\frac{f(x)-f(1)}{x^{10}-1}=\underline{}\\ & a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots +a{b}^{n-2}+b^{n-1})\\ & I=\underset{x\to 1}{\lim }\frac{f(x)-f(1)}{(x-1)(x^9+x^8+\cdots +x+1)}=f^{\prime }(1)\cdot \frac{1}{10}=\frac{1}{10}\\ [注]& f^{\prime }(x)=\underset{x\to x_0}{\lim }\frac{f(x)-f(x_0)}{x-x_0}=\underset{\Delta x\to 0}{\lim }\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\ [例6]& 设f(x)在(-\infty ,+\infty )内有定义(存在)且\forall x,x_1,x_2\in (-\infty ,+\infty ),\\ & 有f(x_1+x_2)=f(x_1)\cdot f(x_2),f(x)=1+xg(x),\underset{x\to 0}{\lim }g(x)=1,证明f(x)在(-\infty ,+\infty )内处处可导\\ & f^{\prime }(x)=\underset{\Delta x\to 0}{\lim }\frac{f(x+\Delta x)-f(x)}{\Delta x}=\underset{\Delta x\to 0}{\lim }\frac{f(x)\cdot f(\Delta x)-f(x)}{\Delta x}\\ & =f(x)\underset{\Delta x\to 0}{\lim }\frac{f(\Delta x)-1}{\Delta x}=f(x)\underset{\Delta x\to 0}{\lim }\frac{1+\Delta xg(\Delta x)-1}{\Delta x}=f(x)\\ & ⟹f^{\prime }(x)处处存在⟺f(x)处处可导\end{array}$$

分段函数在分段点（常见绝对值函数）

$$\begin{array}{rl}[例1]& 设f(x)在x=a处连续，F(x)=f(x)\cdot |x-a|,证明F(x)在x=a处可导的充要条件为f(a)=0(背过)\\ & [分析]F(x)=\{\begin{array}{l}-(x-a)f(x),x<a\\ 0,x=a\\ (x-a)f(x),x>a\end{array}\\ & F_{-}^{\prime }(x)=\underset{x\to a^{-}}{\lim }\frac{F(x)-F(a)}{x-a}=\underset{x\to a^{-}}{\lim }\frac{-(x-a)f(x)-0}{x-a}=-\underset{x\to a^{-}}{\lim }f(x)=-f(a)\\ & F_{+}^{\prime }(x)=\underset{x\to a^{+}}{\lim }\frac{F(x)-F(a)}{x-a}=\underset{x\to a^{+}}{\lim }\frac{(x-a)f(x)-0}{x-a}=\underset{x\to a^{+}}{\lim }f(x)=f(a)\\ & F^{\prime }(a)\exists ⟺F_{-}^{\prime }(a)=F_{+}^{\prime }(a)⟺f(a)=0\\ [例2]& 设f(x)=\mid x\mid e^{-\mid x-1\mid },求f^{\prime }(0),f^{\prime }(1)\\ & f(x)=\{\begin{array}{l}-x{e}^{x-1},x<0\\ x{e}^{x-1},0\le x<1\\ x{e}^{1-x},x\ge 1\end{array}\\ & f_{+}^{\prime }(0)=(e^{x-1}+x{e}^{x-1}){|}_{x=0}=e^{-1},再求f_{-}^{\prime }(0)=-e^{-1},即f^{\prime }(0)不存在\\ & f^{\prime }(1)=(e^{1-x}-x{e}^{1-x}){|}_{x=1}=1-1=0\\ & 同理，得f_{-}^{\prime }(1)=2,f_{+}^{\prime }(1)=0⟹f^{\prime }(1)不存在\end{array}$$

四则运算（不太复杂的点与不成立的点）

$$\begin{array}{rl}& [例1]f(x)=2\sqrt{1+x}+\arcsin \frac{1-x}{1+x^2},f^{\prime }(1)=?\\ & 设f_1=2\sqrt{1+x},f_2=\arcsin \frac{1-x}{1+x^2}\\ & f_1^{\prime }(1)=\frac{\sqrt{2}}{2},f_2(1)=\underset{x\to 1}{\lim }\frac{f_2(x)-f_2(1)}{x-1}=\underset{x\to 1}{\lim }\frac{\arcsin \frac{1-x}{1+x^2}-0}{x-1}=-\frac{1}{2}\\ & 故f^{\prime }(1)=\frac{\sqrt{2}}{2}-\frac{1}{2}\\ & {[例2]f(x)=\prod }_{n=1}^{100}(\tan \frac{\pi x^n}{4}-n)，则f^{\prime }(1)=?\\ & f(x)=(\tan \frac{\pi x}{4}-1)(\tan \frac{\pi x^2}{4}-2)\dots (\tan \frac{\pi x^{100}}{4}-100)\\ & 令g(x)=(\tan \frac{\pi x^2}{4}-2)\dots (\tan \frac{\pi x^{100}}{4}-100)⟹f(x)=(\tan \frac{\pi x}{4}-1)\cdot g(x)\\ & f^{\prime }(1)=\frac{\pi }{4}{\sec }^2\frac{\pi }{4}\cdot g(1)+(\tan \frac{\pi }{4}-1)g^{\prime }(1)=-\frac{\pi }{2}\cdot 99!\\ & [例3]f(x)=e^{10x}\cdot x(x+1)(x+2)\dots (x+10),求f^{\prime }(0)\\ & 令g(x)=e^{10x}\cdot (x+1)(x+2)\dots (x+10),则f(x)=x\cdot g(x)\\ & f^{\prime }(x)=g(x)+x\cdot g^{\prime }(x)=10!\\ & [例4]f(x)=\sqrt[3]{x^2}\sin x,求f^{\prime }(x)\\ & 1.x̸=0时，f^{\prime }(x)=(x^{\frac{2}{3}}\cdot \sin x{)}^{\prime }=\frac{2}{3}{x}^{-\frac{1}{3}}\sin x+x^{\frac{2}{3}}\cdot \cos x\\ & 2.x=0时,f^{\prime }(0)=\underset{x\to 0}{\lim }\frac{f(x)-f(0)}{x-0}=\underset{x\to 0}{\lim }\frac{x^{\frac{2}{3}}\sin x}{x}=\underset{x\to 0}{\lim }{x}^{\frac{2}{3}}=0\\ & 故f^{\prime }(x)=\{\begin{array}{l}\frac{2}{3}\frac{1}{\sqrt[3]{x}}\sin x+x^{\frac{2}{3}}\cdot \cos x,x̸=0\\ 0,x=0\end{array}\\ & [注](u,v{)}^{\prime }=u^{\prime }v+u{v}^{\prime }需两者处处可导，而x^{\frac{2}{3}}并非处处可导\end{array}$$

导数计算与应用

导数计算

基本求导公式表

$$\begin{array}{rl}& (x^k{)}^{\prime }=k{x}^{k-1}(\ln x{)}^{\prime }=\frac{1}{x}(\ln \mid x\mid {)}^{\prime }=\frac{1}{x}\\ & (e^x{)}^{\prime }=e^x(a^x{)}^{\prime }=a^x\ln a,a>0且̸=1\\ & (\sin x{)}^{\prime }=\cos x(\cos x{)}^{\prime }=-\sin x(\tan x{)}^{\prime }={\sec }^{2}x=\frac{1}{{\cos }^{2}x}\\ & (\cot x{)}^{\prime }=-{\csc }^{2}x(\sec x{)}^{\prime }=\sec x\cdot \tan x(\csc x{)}^{\prime }=-\csc x\cdot \cot x\\ & (\ln \mid \cos x\mid {)}^{\prime }=-\tan x(\ln \mid \sin x\mid {)}^{\prime }=\cot x\\ & (\ln \mid \sec x+\tan x\mid {)}^{\prime }=\sec x(\ln \mid \csc x-\cot x\mid {)}^{\prime }=\csc x\\ & (\arctan x{)}^{\prime }=\frac{1}{1+x^2}(arc\cot x{)}^{\prime }=-\frac{1}{1+x^2}\\ & (\arcsin x{)}^{\prime }=\frac{1}{\sqrt{1-x^2}}(\arccos x{)}^{\prime }=-\frac{1}{\sqrt{1-x^2}}\\ & (\ln (x+\sqrt{x^2+a^2}){)}^{\prime }=\frac{1}{\sqrt{x^2+a^2}}常见a=1(\ln (x-\sqrt{x^2-a^2}){)}^{\prime }=\frac{1}{\sqrt{x^2-a^2}},x>a\end{array}$$

复合、隐、参数、分段（含绝对值）、反函数等

$$\begin{array}{rl}[例1]& 设f(x)=x^3+2x-4,g(x)=f[f(x)],则g^{\prime }(0)=\underline{}\\ & 一层一层剥开她的心\\ & g^{\prime }(x)=f^{\prime }[f(x)]f^{\prime }(x)\\ & f^{\prime }(x)=3x^2+2,则f^{\prime }(0)=2,f^{\prime }(-4)=50\\ & g^{\prime }(0)=f^{\prime }[f(0)]f^{\prime }(0)=f^{\prime }(-4)f^{\prime }(0)=100\\ [例2]& 设y=x^3+3x+1,则\frac{dx}{dy}{|}_{y=1}=\underline{}\\ & \frac{dx}{dy}{|}_{y=1}=\frac{1}{y_x^{\prime }}{|}_{x=0}=\frac{1}{3x^2+3}{|}_{x=0}=\frac{1}{3}\\ [例3]& 已知可微函数y=y(x),由方程y=-y{e}^x+2e^y\sin x-7x所确定，求y^{\prime \prime }(0)\\ & y=-y{e}^x+2e^y\sin x-7x\\ & ⟹y^{\prime }=-y^{\prime }{e}^x-y{e}^x+2e^y\sin x\cdot y^{\prime }+2e^y\cdot \cos x-7\\ & ⟹y^{\prime \prime }=-y^{\prime \prime }{e}^x-y^{\prime }{e}^x-y^{\prime }{e}^x-y{e}^x+2e^y\cdot (y^{\prime }{)}^2\sin x+\\ & 2e^y\cos x\cdot y^{\prime }+2e^y\sin x\cdot y^{\prime \prime }+2e^y\cdot y^{\prime }\cos x-2e^y\sin x\\ & 由x=0代入，分别得：\{\begin{array}{l}y=0\\ y^{\prime }=-\frac{5}{2}\\ y^{\prime \prime }=-\frac{5}{2}\end{array}\\ [例4]& 设函数y=y(x)由参数方程\{\begin{array}{l}x=1+t^2\\ y=\cos t\end{array}所确定\\ & 求(1)\frac{dy}{dx}和\frac{d^{2}y}{d{x}^2};\\ & (2)\underset{x\to 1^{+}}{\lim }\frac{dy}{dx}和\underset{x\to 1^{+}}{\lim }\frac{d^{2}y}{d{x}^2}\\ (1)& \frac{dy}{dx}=\frac{y_t^{\prime }}{x_t^{\prime }}=\frac{-\sin t}{2t}\\ & \frac{d^{2}y}{d{x}^2}=\frac{d(\frac{dy}{dx})}{dx}=\frac{-\frac{1}{2}(\frac{t\cos t-\sin t}{t^2})}{2t}=-\frac{t\cos t-\sin t}{4t^3}\\ (2)& \underset{x\to 1^{+}}{\lim }\frac{-\sin t}{2t}=\underset{t\to 0}{\lim }-\frac{\sin t}{2t}=-\frac{1}{2}\\ & \underset{x\to 1^{+}}{\lim }=\underset{t\to 0}{\lim }\frac{\sin t-t\cos t}{4t^3}=\underset{t\to 0}{\lim }\frac{\cos t-\cos t+t\sin t}{12t^2}=\frac{1}{12}\\ [例5]& 设f(x)=\mid x\mid e^{-\mid x-1\mid },求f^{\prime }(x)\\ & (1)先写出f(x)=\{\begin{array}{l}-x{e}^{x-1},x<0\\ x{e}^{x-1},0\le x<1\\ x{e}^{1-x},x\ge 1\end{array}\\ & (2)f^{\prime }(0)不\exists ,f^{\prime }(1)不\exists (分段点用定义求，之前求过)\\ & (3)非分段点用公式求f^{\prime }(x)=\{\begin{array}{l}(-1-x)e^{x-1},x<0\\ (1+x)e^{x-1},0<x<1\\ (1-x)e^{1-x},x>1\end{array}\end{array}$$

$$\begin{array}{rl}[例6]& 设y=f(x)与x=g(y)互为反函数，y=f(x)可导且f^{\prime }(x)̸=0,f(3)=5,h(x)=f[\frac{1}{3}{g}^2(x^2+3x+1)],\\ & 求h^{\prime }(1)\\ & h^{\prime }(x)=f^{\prime }[\frac{1}{3}{g}^2(x^2+3x+1)]\frac{1}{3}2g(x^2+3x+1)\cdot g^{\prime }(x^2+3x+1)\cdot (2x+3)\\ & h^{\prime }(1)=f^{\prime }(\frac{1}{3}{g}^2(5))\cdot \frac{2}{3}g(5)\cdot g^{\prime }(5)\cdot 5\\ & 由y=f(x)与x=g(y)是反函数⟹\frac{dy}{dx}=f^{\prime }(x),\frac{dx}{dy}=g^{\prime }(y)\\ & ⟹\{\begin{array}{l}{f}^{\prime }(x)g^{\prime }(y)=1\\ f(3)=5,3=g(5)\\ f(x)=y,x=g(y)\end{array}⟹h^{\prime }(1)=f^{\prime }(3)\cdot \frac{2}{3}g(5)\cdot g^{\prime }(5)\cdot 5=10\\ [例7]& x=f(y)是函数y=x+\ln x的反函数，求\frac{d^{2}f}{d{y}^2}\\ & x_y^{\prime }={\frac{1}{y}}_x^{\prime },x_y^{\prime \prime }=-\frac{y_x^{\prime \prime }}{(y_x^{\prime }{)}^3}\\ & 有y_x^{\prime }=1+\frac{1}{x},y_x^{\prime \prime }=-\frac{1}{x^2},x_y^{\prime \prime }=-\frac{y_x^{\prime \prime }}{(y_x^{\prime }{)}^3}=-\frac{-1/x^2}{(1+\frac{1}{x}{)}^3}=\frac{x}{(1+x{)}^3}\end{array}$$

多项乘除开方乘方

$$\begin{array}{rl}[例1]& 设y=[(1+x)(3+x{)}^9{]}^{\frac{1}{2}}\cdot (2+x{)}^4,求y^{\prime }(0)\\ & 取对数，再求导\\ & \ln y=\frac{1}{2}\ln (1+x)+\frac{9}{2}\ln (3+x)+4\ln (2+x)\\ & ⟹\frac{1}{y}\cdot y^{\prime }=\frac{1}{2}\cdot \frac{1}{1+x}+\frac{9}{2}\cdot \frac{1}{3+x}+4\cdot \frac{1}{2+x}\\ & ⟹y^{\prime }(0)=(\frac{1}{2}+\frac{9}{6}+2)\cdot 3^{\frac{9}{2}}\cdot 2^4=2^6\cdot 3^{\frac{9}{2}}\\ [例2]& 设f(x)=\sqrt{\frac{(1+x)\sqrt{x}}{e^{x-1}}}+\arcsin \frac{1-x}{\sqrt{1+x^2}},求f^{\prime }(1)\\ & 令y_1=\sqrt{\frac{(1+x)\sqrt{x}}{e^{x-1}}}⟹\ln y_1=\frac{1}{2}(\ln (1+x)+\frac{1}{2}\ln x-(x-1))\\ & ⟹\frac{1}{y_1}\cdot y_1^{\prime \prime }=\frac{1}{2}(\frac{1}{1+x}+\frac{1}{2x}-1)\\ & ⟹y_1^{\prime }(1)代入0\\ & 令y_2=\arcsin \frac{1-x}{\sqrt{1+x^2}}⟹y_2^{\prime }(1)=\underset{x\to 1}{\lim }\frac{y_2(x)-y_2(1)}{x-1}\\ & =\underset{x\to 1}{\lim }\frac{\arcsin \frac{1-x}{\sqrt{1+x^2}}-0}{x-1}=\underset{x\to 1}{\lim }\frac{\frac{1-x}{\sqrt{1+x^2}}}{x-1}=-\frac{\sqrt{2}}{2}\\ & 故f^{\prime }(1)=-\frac{\sqrt{2}}{2}\end{array}$$

高阶导数

归纳法

莱布尼茨公式法

展开

$$\begin{array}{rl}[例1]& 求下列导数\\ (1)& y=\ln (1+x),求y^{(n)}\\ (2)& y=e^x\cos x,求y^{(4)}\\ (3)& 设f(x)=(x^2-3x+2{)}^n\cos \frac{\pi x^2}{16},则f^{(n)}(2)=\underline{}\\ (4)& 设f(x)=\frac{x}{1-2x^4},则f^{(101)}(0)=\underline{}\\ (1)& (\frac{1}{x}{)}^{\prime }=(x^{-1}{)}^{-1}=(-1)x^{-2},(\frac{1}{x}{)}^{\prime \prime }=(-1)(-2)x^{-3},(\frac{1}{x}{)}^n=(-1)n!x^{-(n+1)}\\ & (\ln x{)}^{\prime }=\frac{1}{x},(\ln x{)}^n=(\frac{1}{x}{)}^{n-1}=(-1{)}^{n-1}(n-1)!\\ & (\ln (1+x){)}^{(n)}=(-1{)}^{n-1}(n-1)!(1+x{)}^{-n}\\ [注]& 同理，(e^x{)}^{(n)}=e^x,(a^x{)}^{(n)}=a^x(\ln a{)}^n\\ & (\sin kx{)}^{(n)}=k^n\sin (kx+n\cdot \frac{\pi }{2}),(\cos kx{)}^{(n)}=k^n\cos (kx+n\cdot \frac{\pi }{2})\\ (2)& (u\cdot v{)}^n=\sum _{k=0}^n{C}_n^k{u}^{(k)}{v}^{}\\ & \\ (3)& \\ & \\ & \\ & \\ (4)& \\ & \\ & \\ & \\ & \\ [注]& \\ & \\ & \end{array}$$