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展开全部切线过原点,所以可设切线方程为y=kx对曲线y=lnx求导y'=1/x即曲线上任意一点(x0,y0)处满足y0=lnx0且通过该点的切线的斜率为k=1/x0因此有y0=lnx0k=1/x0y0=kx0因此y0=(1/x0)x0=1x0=e^y0=ek=1/x0=1/e因此所求切线方程为y=x/e
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导数\微分定义及其考法定义导数定义\begin{aligned} &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\to0,我们就能求某时刻的瞬时变化率，如下：\\ &\color{red}{f'(x)=\lim_{\Delta x\to0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\ \ 瞬时变化率}\\ &\color{grey}{[注1]换元，令x_0+\Delta x=x,则f'(x_0)=\lim_{\Delta x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}}\\ &\color{grey}{[注2]左右导数，f_+'(x)=\lim_{\Delta x\to0^+}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}右导数，f_-'(x)=\lim_{\Delta x\to0_-}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}左导数}\\ &\color{grey}{[注3]导数存在的条件，f'(x_0)\exists\iff f'_+(x_0)=f'_-(x_0)} \end{aligned} 微分定义\begin{aligned} &幂的微分\begin{cases}d(x^2)=2xdx\\d(x^n)=nx^{n-1}dx\end{cases}\\ &微分的幂\begin{cases}dx^2=(dx)^2\\dx^n=(dx)^n\end{cases}\\ &dy=y'dx \end{aligned} \begin{aligned} &如上图所示，f(x)=x^2\quad 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'(x)\cdot\Delta x+\circ(\Delta x),则y'(x)\cdot\Delta x=dy称为线性全部\\ &\therefore \Delta x=1\cdot\Delta x+0\implies dx=\Delta x\\ &\implies dy=y'(x)dx\implies \frac{dy}{dx}=y'(x)\\ \end{aligned} \begin{aligned} \ [例1]&\color{maroon}{设y=e^{x^2},求\frac{dy}{dx},\frac{dy}{d(x^2)},\frac{d^2y}{dx^2}}\\ &\frac{dy}{dx}=y'=e^{x^2}\cdot2x\\ &\frac{d^2y}{dx^2}=y^{''}=e^{x^2}\cdot4x^2+e^{x^2}\cdot2\\ &\frac{dy}{d(x^2)}=\frac{e^{x^2}\cdot2xdx}{2xdx}=e^{x^2}\qquad\frac{dy}{2xdx}=\frac1{2x}\cdot e^{x^2}\cdot2x=e{x^2}\\ [例2]&\color{maroon}{y=f(x),f'(x_0)=\frac12,\Delta x\to0时,y=f(x)在x=x_0处得微分dy与\Delta x是(同阶非等价)}\\ &dy=y'(x_0)dx=\frac12dx=\frac12\Delta x\\ &\lim_{\Delta x\to0}\frac{dx}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac12\Delta x}{\Delta x}=\frac12\\ [例3]&\color{maroon}{设f(x)=(\cos x-4)\sin x+3x,求\frac{df(x)}{d(x^2)}}\\ &df(a)=f'(x)dx=(-\sin^2x+(\cos x-4)\cdot\cos x+3)dx\\ &d(x^2)=2xdx\\ \therefore & \frac{df(x)}{d(x^2)}=\frac{-\sin^2x+(\cos x-4)\cos x+3}{2x}=\frac{(\cos x-1)^2}{x}\\ [例4]&\color{maroon}{设f'(0)=1,f''(0)=0,求证：在x=0处，有\frac{d^2}{dx^2}f(x^2)=\frac{d^2}{dx^2}f^2(x)}\\ &y_1'=f'(x^2)\cdot2x,y_1^{''}|_0=f''(x^2)\cdot2x\cdot2x+f'(x^2)\cdot2|_{x=0}=2\\ &y_2'=2f(x)\cdot f'(x),y_2^{''}|_0=2f'(x)f'(x)+2f(x)f''(x)|_{x=0}=2\\ \end{aligned} \begin{aligned} \ [例1]&\color{maroon}{设y=e^{x^2},求\frac{dy}{dx},\frac{dy}{d(x^2)},\frac{d^2y}{dx^2}}\\ &\frac{dy}{dx}=y'=e^{x^2}\cdot2x\\ &\frac{d^2y}{dx^2}=y^{''}=e^{x^2}\cdot4x^2+e^{x^2}\cdot2\\ &\frac{dy}{d(x^2)}=\frac{e^{x^2}\cdot2xdx}{2xdx}=e^{x^2}\qquad\frac{dy}{2xdx}=\frac1{2x}\cdot e^{x^2}\cdot2x=e{x^2}\\ [例2]&\color{maroon}{y=f(x),f'(x_0)=\frac12,\Delta x\to0时,y=f(x)在x=x_0处得微分dy与\Delta x是(同阶非等价)}\\ &dy=y'(x_0)dx=\frac12dx=\frac12\Delta x\\ &\lim_{\Delta x\to0}\frac{dx}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac12\Delta x}{\Delta x}=\frac12\\ [例3]&\color{maroon}{设f(x)=(\cos x-4)\sin x+3x,求\frac{df(x)}{d(x^2)}}\\ &df(a)=f'(x)dx=(-\sin^2x+(\cos x-4)\cdot\cos x+3)dx\\ &d(x^2)=2xdx\\ \therefore & \frac{df(x)}{d(x^2)}=\frac{-\sin^2x+(\cos x-4)\cos x+3}{2x}=\frac{(\cos x-1)^2}{x}\\ [例4]&\color{maroon}{设f'(0)=1,f''(0)=0,求证：在x=0处，有\frac{d^2}{dx^2}f(x^2)=\frac{d^2}{dx^2}f^2(x)}\\ &y_1'=f'(x^2)\cdot2x,y_1^{''}|_0=f''(x^2)\cdot2x\cdot2x+f'(x^2)\cdot2|_{x=0}=2\\ &y_2'=2f(x)\cdot f'(x),y_2^{''}|_0=2f'(x)f'(x)+2f(x)f''(x)|_{x=0}=2\\ \end{aligned} \begin{aligned} \ [例1]&\color{maroon}{设y=e^{x^2},求\frac{dy}{dx},\frac{dy}{d(x^2)},\frac{d^2y}{dx^2}}\\ &\frac{dy}{dx}=y'=e^{x^2}\cdot2x\\ &\frac{d^2y}{dx^2}=y^{''}=e^{x^2}\cdot4x^2+e^{x^2}\cdot2\\ &\frac{dy}{d(x^2)}=\frac{e^{x^2}\cdot2xdx}{2xdx}=e^{x^2}\qquad\frac{dy}{2xdx}=\frac1{2x}\cdot e^{x^2}\cdot2x=e^{x^2}\\ [例2]&\color{maroon}{y=f(x),f'(x_0)=\frac12,\Delta x\to0时,y=f(x)在x=x_0处得微分dy与\Delta x是(同阶非等价)}\\ &dy=y'(x_0)dx=\frac12dx=\frac12\Delta x\\ &\lim_{\Delta x\to0}\frac{dx}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac12\Delta x}{\Delta x}=\frac12\\ [例3]&\color{maroon}{设f(x)=(\cos x-4)\sin x+3x,求\frac{df(x)}{d(x^2)}}\\ &df(a)=f'(x)dx=(-\sin^2x+(\cos x-4)\cdot\cos x+3)dx\\ &d(x^2)=2xdx\\ \therefore & \frac{df(x)}{d(x^2)}=\frac{-\sin^2x+(\cos x-4)\cos x+3}{2x}=\frac{(\cos x-1)^2}{x}\\ [例4]&\color{maroon}{设f'(0)=1,f''(0)=0,求证：在x=0处，有\frac{d^2}{dx^2}f(x^2)=\frac{d^2}{dx^2}f^2(x)}\\ &y_1'=f'(x^2)\cdot2x,y_1^{''}|_0=f''(x^2)\cdot2x\cdot2x+f'(x^2)\cdot2|_{x=0}=2\\ &y_2'=2f(x)\cdot f'(x),y_2^{''}|_0=2f'(x)f'(x)+2f(x)f''(x)|_{x=0}=2\\ \end{aligned} 考法抽象函数在一点（泛指x与特指x）\begin{aligned} \ [例1]&\color{maroon}{证明：若f(x)可导且为偶函数，请推f'(x)为奇函数}\\ &\color{black}[分析]已知f(x)=f(-x)\\ &\therefore f'(-x)=\lim_{\Delta x\to0}\frac{f(-x+\Delta x)-f(-x)}{\Delta x}=-\lim_{-\Delta x\to0}\frac{f(x+(-\Delta x))-f(x)}{-\Delta x}=-f'(x)\\ [例2]&\color{maroon}{证明f(x)可导，周期为T，请推f'(x)的周期也是T}\\ &\color{black}[分析]已知f(x+T)=f(x)\\ &\therefore f'(x+T)=\lim_{\Delta x\to0}\frac{f(x+T+\Delta x)-f(x+T)}{\Delta x}=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=f'(x)\\ [例3]&\color{maroon}{f(x)二阶可导，T=2，奇函数，且f(\frac12)>0,f'(x)>0,比较f(-\frac12),f'(\frac32),f''(0)的大小}\\ &\color{black}\because该函数为奇函数\therefore f(-x)=-f(x)\rightarrow f(-\frac12)=-f(\frac12)<0\\ &\because f(x):T=2 \therefore f'(x):T=2且f'(x)为偶函数\\
&\therefore f'(\frac32)=f'(\frac32-2)=f'(-\frac12)=f'(\frac12)>0\\ &\therefore f''(x)为奇函数\quad 即:f''(0)=0\\ &得f(-\frac12)<f''(0)<f'(\frac32)\\ [例4]&\color{maroon}{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\lim_{n\to\infty}nf(\frac2n)}\\ &\color{black}[分析]f'(x_0)=k,切线方程为y-y_0=f'(x_0)(x-x_0)\\ &(\int_0^{\arctan x}e^{-t^2}dt)'_x=e^{-(\arctan x)^2}\cdot\frac{1}{1+x^2},令x=0,则f'(0)=1,故切线方程为y=x\\ &\therefore\lim_{n\to\infty}nf(\frac2n)=2\lim_{\frac2n\to0^+}\frac{f(0+\frac2n)-f(0)}{\frac2n}=2\cdot f'(0)=2\\ [例5]&\color{maroon}{设f'(1)=1,则\lim_{x\to1}\frac{f(x)-f(1)}{x^{10}-1}}=\underline{\quad}\\ &a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})\\ &I=\lim_{x\to1}\frac{f(x)-f(1)}{(x-1)(x^9+x^8+\cdots+x+1)}=f'(1)\cdot\frac1{10}=\frac1{10}\\ [注]&f'(x)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{\Delta x\to0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\ [例6]&\color{maroon}{设f(x)在(-\infty,+\infty)内有定义(存在)且\forall x,x_1,x_2\in(-\infty,+\infty),}\\ &\color{maroon}{有f(x_1+x_2)=f(x_1)\cdot f(x_2),f(x)=1+xg(x),\lim_{x\to0}g(x)=1,证明f(x)在(-\infty,+\infty)内处处可导}\\ &f'(x)=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim_{\Delta x\to0}\frac{f(x)\cdot f(\Delta x)-f(x)}{\Delta x}\\ &=f(x)\lim_{\Delta x\to0}\frac{f(\Delta x)-1}{\Delta x}=f(x)\lim_{\Delta x\to0}\frac{1+\Delta xg(\Delta x)-1}{\Delta x}=f(x)\\ &\implies f'(x)处处存在\iff f(x)处处可导\\ \end{aligned} \begin{aligned} \ [例1]&\color{maroon}{证明：若f(x)可导且为偶函数，请推f'(x)为奇函数}\\ &\color{black}[分析]已知f(x)=f(-x)\\ &\therefore f'(-x)=\lim_{\Delta x\to0}\frac{f(-x+\Delta x)-f(-x)}{\Delta x}=-\lim_{-\Delta x\to0}\frac{f(x+(-\Delta x))-f(x)}{-\Delta x}=-f'(x)\\ [例2]&\color{maroon}{证明f(x)可导，周期为T，请推f'(x)的周期也是T}\\ &\color{black}[分析]已知f(x+T)=f(x)\\ &\therefore f'(x+T)=\lim_{\Delta x\to0}\frac{f(x+T+\Delta x)-f(x+T)}{\Delta x}=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=f'(x)\\ [例3]&\color{maroon}{f(x)二阶可导，T=2，奇函数，且f(\frac12)>0,f'(x)>0,比较f(-\frac12),f'(\frac32),f''(0)的大小}\\ &\color{black}\because该函数为奇函数\therefore f(-x)=-f(x)\rightarrow f(-\frac12)=-f(\frac12)<0\\ &\because f(x):T=2 \therefore f'(x):T=2且f'(x)为偶函数\\
&\therefore f'(\frac32)=f'(\frac32-2)=f'(-\frac12)=f'(\frac12)>0\\ &\therefore f''(x)为奇函数\quad 即:f''(0)=0\\ &得f(-\frac12)<f''(0)<f'(\frac32)\\ [例4]&\color{maroon}{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\lim_{n\to\infty}nf(\frac2n)}\\ &\color{black}[分析]f'(x_0)=k,切线方程为y-y_0=f'(x_0)(x-x_0)\\ &(\int_0^{\arctan x}e^{-t^2}dt)'_x=e^{-(\arctan x)^2}\cdot\frac{1}{1+x^2},令x=0,则f'(0)=1,故切线方程为y=x\\ &\therefore\lim_{n\to\infty}nf(\frac2n)=2\lim_{\frac2n\to0^+}\frac{f(0+\frac2n)-f(0)}{\frac2n}=2\cdot f'(0)=2\\ [例5]&\color{maroon}{设f'(1)=1,则\lim_{x\to1}\frac{f(x)-f(1)}{x^{10}-1}}=\underline{\quad}\\ &a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})\\ &I=\lim_{x\to1}\frac{f(x)-f(1)}{(x-1)(x^9+x^8+\cdots+x+1)}=f'(1)\cdot\frac1{10}=\frac1{10}\\ [注]&f'(x)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{\Delta x\to0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\ [例6]&\color{maroon}{设f(x)在(-\infty,+\infty)内有定义(存在)且\forall x,x_1,x_2\in(-\infty,+\infty),}\\ &\color{maroon}有f(x_1+x_2)=f(x_1)\cdot f(x_2),f(x)=1+xg(x),\lim_{x\to0}g(x)=1,证明f(x)在(-\infty,+\infty)内处处可导\\ &f'(x)=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim_{\Delta x\to0}\frac{f(x)\cdot f(\Delta x)-f(x)}{\Delta x}\\ &=f(x)\lim_{\Delta x\to0}\frac{f(\Delta x)-1}{\Delta x}=f(x)\lim_{\Delta x\to0}\frac{1+\Delta xg(\Delta x)-1}{\Delta x}=f(x)\\ &\implies f'(x)处处存在\iff f(x)处处可导\\ \end{aligned} 分段函数在分段点（常见绝对值函数）\begin{aligned} \ [例1]&\color{maroon}{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)}\\ &\color{black}[分析]F(x)=\begin{cases}-(x-a)f(x),x< a\\0,x=a\\(x-a)f(x),x>a\end{cases}\\ &F'_-(x)=\lim_{x\to a^-}\frac{F(x)-F(a)}{x-a}=\lim_{x\to a^-}\frac{-(x-a)f(x)-0}{x-a}=-\lim_{x\to a^-}f(x)=-f(a)\\ &F'_+(x)=\lim_{x\to a^+}\frac{F(x)-F(a)}{x-a}=\lim_{x\to a^+}\frac{(x-a)f(x)-0}{x-a}=\lim_{x\to a^+}f(x)=f(a)\\ &F'(a)\exists\iff F'_-(a)=F'_+(a)\iff f(a)=0\\ [例2]&\color{maroon}{设f(x)=\mid x\mid e^{-\mid x-1\mid},求f'(0),f'(1)}\\ &f(x)=\begin{cases}-xe^{x-1},x<0\\xe^{x-1},0\leq x<1\\xe^{1-x},x\geq1\end{cases}\\ &f_+'(0)=(e^{x-1}+xe^{x-1})|_{x=0}=e^{-1},再求f_{-}'(0)=-e^{-1},即f'(0)不存在\\ &f'(1)=(e^{1-x}-xe^{1-x})|_{x=1}=1-1=0\\ &同理，得f_-'(1)=2,f_+'(1)=0\implies f'(1)不存在\\ \end{aligned} \begin{aligned} \ [例1]&\color{maroon}{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)}\\ &\color{black}[分析]F(x)=\begin{cases}-(x-a)f(x),x< a\\0,x=a\\(x-a)f(x),x>a\end{cases}\\ &F'_-(x)=\lim_{x\to a^-}\frac{F(x)-F(a)}{x-a}=\lim_{x\to a^-}\frac{-(x-a)f(x)-0}{x-a}=-\lim_{x\to a^-}f(x)=-f(a)\\ &F'_+(x)=\lim_{x\to a^+}\frac{F(x)-F(a)}{x-a}=\lim_{x\to a^+}\frac{(x-a)f(x)-0}{x-a}=\lim_{x\to a^+}f(x)=f(a)\\ &F'(a)\exists\iff F'_-(a)=F'_+(a)\iff f(a)=0\\ [例2]&\color{maroon}{设f(x)=\mid x\mid e^{-\mid x-1\mid},求f'(0),f'(1)}\\ &f(x)=\begin{cases}-xe^{x-1},x<0\\xe^{x-1},0\leq x<1\\xe^{1-x},x\geq1\end{cases}\\ &f_+'(0)=(e^{x-1}+xe^{x-1})|_{x=0}=e^{-1},再求f_{-}'(0)=-e^{-1},即f'(0)不存在\\ &f'(1)=(e^{1-x}-xe^{1-x})|_{x=1}=1-1=0\\ &同理，得f_-'(1)=2,f_+'(1)=0\implies f'(1)不存在\\ \end{aligned} 四则运算（不太复杂的点与不成立的点）\begin{aligned} &\color{maroon}{[例1]f(x)=2\sqrt{1+x}+\arcsin\frac{1-x}{1+x^2},f'(1)=?}\\ &\color{black}设f_1=2\sqrt{1+x},f_2=\arcsin\frac{1-x}{1+x^2}\\ &f'_1(1)=\frac{\sqrt2}2,f_2(1)=\lim_{x\to1}\frac{f_2(x)-f_2(1)}{x-1}=\lim_{x\to1}\frac{\arcsin\frac{1-x}{1+x^2}-0}{x-1}=-\frac12\\ &故f'(1)=\frac{\sqrt2}2-\frac12\\ &\color{maroon}{[例2]f(x)=\prod_{n=1}^{100}(\tan\frac{\pi x^n}4-n)，则f'(1)=?}\\ &\color{black}f(x)=(\tan\frac{\pi x}4-1)(\tan\frac{\pi x^2}4-2)\ldots(\tan\frac{\pi x^{100}}4-100)\\ &令g(x)=(\tan\frac{\pi x^2}4-2)\ldots(\tan\frac{\pi x^{100}}4-100)\implies f(x)=(\tan\frac{\pi x}4-1)\cdot g(x)\\ &f'(1)=\frac\pi4 \sec^2\frac{\pi}4\cdot g(1)+(\tan\frac{\pi}4-1)g'(1)=\frac\pi2\cdot99!\\ &\color{maroon}{[例3]f(x)=e^{10x}\cdot x(x+1)(x+2)\ldots(x+10),求f'(0)}\\ &\color{black}令g(x)=e^{10x}\cdot(x+1)(x+2)\ldots(x+10),则f(x)=x\cdot g(x)\\ &f'(x)=g(x)+x\cdot g'(x)=10!\\ &\color{maroon}{[例4]f(x)=\sqrt[3]{x^2}\sin x,求f'(x)}\\ &\color{black}1.x\neq0时，f'(x)=(x^{\frac23}\cdot \sin x)'=\frac23x^{-\frac13}\sin x+x^{\frac23}\cdot \cos x\\ &2.x=0时,f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{x^\frac23\sin x}{x}=\lim_{x\to0}x^\frac23=0\\ &故f'(x)=\begin{cases}\frac23\frac1{\sqrt[3]{x}}\sin x+x^\frac23\cdot \cos x,x\neq0\\0,x=0\end{cases}\\ &[注](u,v)'=u'v+uv'需两者处处可导，而x^{\frac23}并非处处可导\\ \end{aligned} \begin{aligned} &\color{maroon}{[例1]f(x)=2\sqrt{1+x}+\arcsin\frac{1-x}{1+x^2},f'(1)=?}\\ &\color{black}设f_1=2\sqrt{1+x},f_2=\arcsin\frac{1-x}{1+x^2}\\ &f'_1(1)=\frac{\sqrt2}2,f_2(1)=\lim_{x\to1}\frac{f_2(x)-f_2(1)}{x-1}=\lim_{x\to1}\frac{\arcsin\frac{1-x}{1+x^2}-0}{x-1}=-\frac12\\ &故f'(1)=\frac{\sqrt2}2-\frac12\\ &\color{maroon}{[例2]f(x)=\prod}_{n=1}^{100}(\tan\frac{\pi x^n}4-n)，则f'(1)=?\\ &\color{black}f(x)=(\tan\frac{\pi x}4-1)(\tan\frac{\pi x^2}4-2)\ldots(\tan\frac{\pi x^{100}}4-100)\\ &令g(x)=(\tan\frac{\pi x^2}4-2)\ldots(\tan\frac{\pi x^{100}}4-100)\implies f(x)=(\tan\frac{\pi x}4-1)\cdot g(x)\\ &f'(1)=\frac\pi4 \sec^2\frac{\pi}4\cdot g(1)+(\tan\frac{\pi}4-1)g'(1)=-\frac\pi2\cdot99!\\ &\color{maroon}{[例3]f(x)=e^{10x}\cdot x(x+1)(x+2)\ldots(x+10),求f'(0)}\\ &\color{black}令g(x)=e^{10x}\cdot(x+1)(x+2)\ldots(x+10),则f(x)=x\cdot g(x)\\ &f'(x)=g(x)+x\cdot g'(x)=10!\\ &\color{maroon}{[例4]f(x)=\sqrt[3]{x^2}\sin x,求f'(x)}\\ &\color{black}1.x\neq0时，f'(x)=(x^{\frac23}\cdot \sin x)'=\frac23x^{-\frac13}\sin x+x^{\frac23}\cdot \cos x\\ &2.x=0时,f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{x^\frac23\sin x}{x}=\lim_{x\to0}x^\frac23=0\\ &故f'(x)=\begin{cases}\frac23\frac1{\sqrt[3]{x}}\sin x+x^\frac23\cdot \cos x,x\neq0\\0,x=0\end{cases}\\ &[注](u,v)'=u'v+uv'需两者处处可导，而x^{\frac23}并非处处可导\\ \end{aligned} 导数计算与应用导数计算基本求导公式表\begin{aligned} &(x^k)'=kx^{k-1}\qquad (\ln x)'=\frac1x\qquad(\ln\mid x\mid)'=\frac1x\\ &(e^x)'=e^x\qquad (a^x)'=a^x\ln a,a>0且\not=1\\ &(\sin x)'=\cos x\qquad(\cos x)'=-\sin x\qquad(\tan x)'=\sec^2x=\frac1{\cos^2x}\\ &(\cot x)'=-\csc^2x\qquad(\sec x)'=\sec x\cdot \tan x\qquad(\csc x)'=-\csc x\cdot \cot x\\ &(\ln\mid\cos x\mid)'=-\tan x\qquad(\ln\mid\sin x\mid)'=\cot x\\ &(\ln\mid\sec x+\tan x\mid)'=\sec x\qquad(\ln\mid\csc x-\cot x\mid)'=\csc x\\ &(\arctan x)'=\frac1{1+x^2}\qquad(arc\cot x)'=-\frac1{1+x^2}\\ &(\arcsin x)'=\frac1{\sqrt{1-x^2}}\qquad(\arccos x)'=-\frac1{\sqrt{1-x^2}}\\ &(\ln(x+\sqrt{x^2+a^2}))'=\frac1{\sqrt{x^2+a^2}}常见a=1\qquad(\ln(x-\sqrt{x^2-a^2}))'=\frac1{\sqrt{x^2-a^2}},x>a\\ \end{aligned} \begin{aligned} &(x^k)'=kx^{k-1}\qquad (\ln x)'=\frac1x\qquad(\ln\mid x\mid)'=\frac1x\\ &(e^x)'=e^x\qquad (a^x)'=a^x\ln a,a>0且\not=1\\ &(\sin x)'=\cos x\qquad(\cos x)'=-\sin x\qquad(\tan x)'=\sec^2x=\frac1{\cos^2x}\\ &(\cot x)'=-\csc^2x\qquad(\sec x)'=\sec x\cdot \tan x\qquad(\csc x)'=-\csc x\cdot \cot x\\ &(\ln\mid\cos x\mid)'=-\tan x\qquad(\ln\mid\sin x\mid)'=\cot x\\ &(\ln\mid\sec x+\tan x\mid)'=\sec x\qquad(\ln\mid\csc x-\cot x\mid)'=\csc x\\ &(\arctan x)'=\frac1{1+x^2}\qquad(arc\cot x)'=-\frac1{1+x^2}\\ &(\arcsin x)'=\frac1{\sqrt{1-x^2}}\qquad(\arccos x)'=-\frac1{\sqrt{1-x^2}}\\ &(\ln(x+\sqrt{x^2+a^2}))'=\frac1{\sqrt{x^2+a^2}}常见a=1\qquad(\ln(x-\sqrt{x^2-a^2}))'=\frac1{\sqrt{x^2-a^2}},x>a\\ \end{aligned} 复合、隐、参数、分段（含绝对值）、反函数等\begin{aligned} \ [例1]&\color{maroon}{设f(x)=x^3+2x-4,g(x)=f[f(x)],则g'(0)=\underline{\quad}}\\ &一层一层剥开她的心\\ &g'(x)=f'[f(x)]f'(x)\\ &f'(x)=3x^2+2,则f'(0)=2,f'(-4)=50\\ &g'(0)=f'[f(0)]f'(0)=f'(-4)f'(0)=100\\ [例2]&\color{maroon}{设y=x^3+3x+1,则\frac{dx}{dy}|_{y=1}=\underline{\qquad}}\\ &\frac{dx}{dy}|_{y=1}=\frac1{y'_x}|_{x=0}=\frac1{3x^2+3}|_{x=0}=\frac13\\ [例3]&\color{maroon}{已知可微函数y=y(x),由方程y=-ye^x+2e^y\sin x-7x所确定，求y''(0)}\\ &y=-ye^x+2e^y\sin x-7x\\ &\implies y'=-y'e^x-ye^x+2e^y\sin x\cdot y'+2e^y\cdot \cos x-7\\ &\implies y''=-y''e^x-y'e^x-y'e^x-ye^x+2e^y\cdot(y')^2\sin x+\\ &2e^y\cos x\cdot y'+2e^y\sin x\cdot y''+2e^y\cdot y'\cos x-2e^y\sin x\\ &由x=0代入，分别得：\begin{cases}y=0\\y'=-\frac52\\y''=-\frac52\end{cases}\\ [例4]&\color{maroon}{设函数y=y(x)由参数方程\begin{cases}x=1+t^2\\y=\cos t\end{cases}所确定}\\ &\color{maroon}{求(1)\frac{dy}{dx}和\frac{d^2y}{dx^2};}\\ &\color{maroon}{(2)\lim_{x\to1^+}\frac{dy}{dx}和\lim_{x\to1^+}\frac{d^2y}{dx^2}}\\ (1)&\frac{dy}{dx}=\frac{y'_t}{x'_t}=\frac{-\sin t}{2t}\\ &\frac{d^2y}{dx^2}=\frac{d(\frac{dy}{dx})}{dx}=\frac{-\frac12(\frac{t\cos t-\sin t}{t^2})}{2t}=-\frac{t\cos t-\sin t}{4t^3}\\ (2)&\lim_{x\to1^+}\frac{-\sin t}{2t}=\lim_{t\to0}-\frac{\sin t}{2t}=-\frac12\\ &\lim_{x\to1^+}=\lim_{t\to0}\frac{\sin t-t\cos t}{4t^3}=\lim_{t\to0}\frac{\cos t-\cos t+t\sin t}{12t^2}=\frac1{12}\\ [例5]&\color{maroon}{设f(x)=\mid x\mid e^{-\mid x-1\mid},求f'(x)}\\ &(1)先写出f(x)=\begin{cases}-xe^{x-1},x<0\\xe^{x-1},0\leq x<1\\xe^{1-x},x\geq1\end{cases}\\ &(2)f'(0)不\exists,f'(1)不\exists(分段点用定义求，之前求过)\\ &(3)非分段点用公式求f'(x)=\begin{cases}(-1-x)e^{x-1},x<0\\(1+x)e^{x-1},0< x<1\\(1-x)e^{1-x},x>1\end{cases}\\ \end{aligned} \begin{aligned} \ [例1]&\color{maroon}{设f(x)=x^3+2x-4,g(x)=f[f(x)],则g'(0)=\underline{\quad}}\\ &一层一层剥开她的心\\ &g'(x)=f'[f(x)]f'(x)\\ &f'(x)=3x^2+2,则f'(0)=2,f'(-4)=50\\ &g'(0)=f'[f(0)]f'(0)=f'(-4)f'(0)=100\\ [例2]&\color{maroon}{设y=x^3+3x+1,则\frac{dx}{dy}|_{y=1}=\underline{\qquad}}\\ &\frac{dx}{dy}|_{y=1}=\frac1{y'_x}|_{x=0}=\frac1{3x^2+3}|_{x=0}=\frac13\\ [例3]&\color{maroon}{已知可微函数y=y(x),由方程y=-ye^x+2e^y\sin x-7x所确定，求y''(0)}\\ &y=-ye^x+2e^y\sin x-7x\\ &\implies y'=-y'e^x-ye^x+2e^y\sin x\cdot y'+2e^y\cdot \cos x-7\\ &\implies y''=-y''e^x-y'e^x-y'e^x-ye^x+2e^y\cdot(y')^2\sin x+\\ &2e^y\cos x\cdot y'+2e^y\sin x\cdot y''+2e^y\cdot y'\cos x-2e^y\sin x\\ &由x=0代入，分别得：\begin{cases}y=0\\y'=-\frac52\\y''=-\frac52\end{cases}\\ [例4]&\color{maroon}{设函数y=y(x)由参数方程\begin{cases}x=1+t^2\\y=\cos t\end{cases}所确定}\\ &\color{maroon}{求(1)\frac{dy}{dx}和\frac{d^2y}{dx^2};}\\ &\color{maroon}{(2)\lim_{x\to1^+}\frac{dy}{dx}和\lim_{x\to1^+}\frac{d^2y}{dx^2}}\\ (1)&\frac{dy}{dx}=\frac{y'_t}{x'_t}=\frac{-\sin t}{2t}\\ &\frac{d^2y}{dx^2}=\frac{d(\frac{dy}{dx})}{dx}=\frac{-\frac12(\frac{t\cos t-\sin t}{t^2})}{2t}=-\frac{t\cos t-\sin t}{4t^3}\\ (2)&\lim_{x\to1^+}\frac{-\sin t}{2t}=\lim_{t\to0}-\frac{\sin t}{2t}=-\frac12\\ &\lim_{x\to1^+}=\lim_{t\to0}\frac{\sin t-t\cos t}{4t^3}=\lim_{t\to0}\frac{\cos t-\cos t+t\sin t}{12t^2}=\frac1{12}\\ [例5]&\color{maroon}{设f(x)=\mid x\mid e^{-\mid x-1\mid},求f'(x)}\\ &(1)先写出f(x)=\begin{cases}-xe^{x-1},x<0\\xe^{x-1},0\leq x<1\\xe^{1-x},x\geq1\end{cases}\\ &(2)f'(0)不\exists,f'(1)不\exists(分段点用定义求，之前求过)\\ &(3)非分段点用公式求f'(x)=\begin{cases}(-1-x)e^{x-1},x<0\\(1+x)e^{x-1},0< x<1\\(1-x)e^{1-x},x>1\end{cases}\\ \end{aligned} \begin{aligned} \ [例6]&\color{maroon}{设y=f(x)与x=g(y)互为反函数，y=f(x)可导且f'(x)\not=0,f(3)=5,h(x)=f[\frac13g^2(x^2+3x+1)],}\\ &\color{maroon}{求h'(1)}\\ &h'(x)=f'[\frac13g^2(x^2+3x+1)]\frac132g(x^2+3x+1)\cdot g'(x^2+3x+1)\cdot(2x+3)\\ &h'(1)=f'(\frac13g^2(5))\cdot\frac23g(5)\cdot g'(5)\cdot5\\ &由y=f(x)与x=g(y)是反函数\implies\frac{dy}{dx}=f'(x)\cdot\frac{dx}{dy}=g'(y)\\ &\implies\begin{cases}f'(x)g'(y)=1\\f(3)=5,3=g(5)\\f(x)=y,x=g(y)\end{cases}\implies h'(1)=f'(3)\cdot\frac23g(5)\cdot g'(5)\cdot5=10\\ [例7]&\color{maroon}{x=f(y)是函数y=x+\ln x的反函数，求\frac{d^2f}{dy^2}}\\ &x'_y=\frac1y'_x,x''_y=-\frac{y''_x}{(y'_x)^3}\\ &有y'_x=1+\frac1x,y''_x=-\frac1{x^2},x''_y=-\frac{y''_x}{(y'_x)^3}=-\frac{-1/x^2}{(1+\frac1x)^3}=\frac{x}{(1+x)^3}\\ \end{aligned} \begin{aligned} \ [例6]&\color{maroon}{设y=f(x)与x=g(y)互为反函数，y=f(x)可导且f'(x)\not=0,f(3)=5,h(x)=f[\frac13g^2(x^2+3x+1)],}\\ &\color{maroon}{求h'(1)}\\ &h'(x)=f'[\frac13g^2(x^2+3x+1)]\frac132g(x^2+3x+1)\cdot g'(x^2+3x+1)\cdot(2x+3)\\ &h'(1)=f'(\frac13g^2(5))\cdot\frac23g(5)\cdot g'(5)\cdot5\\ &由y=f(x)与x=g(y)是反函数\implies\frac{dy}{dx}=f'(x),\frac{dx}{dy}=g'(y)\\ &\implies\begin{cases}f'(x)g'(y)=1\\f(3)=5,3=g(5)\\f(x)=y,x=g(y)\end{cases}\implies h'(1)=f'(3)\cdot\frac23g(5)\cdot g'(5)\cdot5=10\\ [例7]&\color{maroon}{x=f(y)是函数y=x+\ln x的反函数，求\frac{d^2f}{dy^2}}\\ &x'_y=\frac1y'_x,x''_y=-\frac{y''_x}{(y'_x)^3}\\ &有y'_x=1+\frac1x,y''_x=-\frac1{x^2},x''_y=-\frac{y''_x}{(y'_x)^3}=-\frac{-1/x^2}{(1+\frac1x)^3}=\frac{x}{(1+x)^3}\\ \end{aligned} 多项乘除开方乘方\begin{aligned} \ [例1]&\color{maroon}{设y=[(1+x)(3+x)^9]^{\frac12}\cdot(2+x)^4,求y'(0)}\\ &取对数，再求导\\ &\ln y=\frac12\ln(1+x)+\frac92\ln(3+x)+4\ln(2+x)\\ &\implies\frac1y\cdot y'=\frac12\cdot\frac1{1+x}+\frac92\cdot\frac1{3+x}+4\cdot\frac1{2+x}\\ &\implies y'(0)=(\frac12+\frac96+2)\cdot3^{\frac92}\cdot2^4=2^6\cdot3^{\frac92}\\ [例2]&\color{maroon}{设f(x)=\sqrt{\frac{(1+x)\sqrt x}{e^{x-1}}}+\arcsin\frac{1-x}{\sqrt{1+x^2}},求f'(1)}\\ &令y_1=\sqrt{\frac{(1+x)\sqrt x}{e^{x-1}}}\implies\ln{y_1}=\frac12(\ln(1+x)+\frac12\ln x-(x-1))\\ &\implies\frac1{y_1}\cdot y_1''=\frac12(\frac1{1+x}+\frac1{2x}-1)\\ &\implies y'_1(1)\underrightarrow{代入}0\\ &令y_2=\arcsin\frac{1-x}{\sqrt{1+x^2}}\implies y_2'(1)=\lim_{x\to1}\frac{y_2(x)-y_2(1)}{x-1}\\ &=\lim_{x\to1}\frac{\arcsin\frac{1-x}{\sqrt{1+x^2}}-0}{x-1}=\lim_{x\to1}\frac{\frac{1-x}{\sqrt{1+x^2}}}{x-1}=-\frac{\sqrt2}{2}\\ &故f'(1)=-\frac{\sqrt2}2 \end{aligned} \begin{aligned} \ [例1]&\color{maroon}{设y=[(1+x)(3+x)^9]^{\frac12}\cdot(2+x)^4,求y'(0)}\\ &取对数，再求导\\ &\ln y=\frac12\ln(1+x)+\frac92\ln(3+x)+4\ln(2+x)\\ &\implies\frac1y\cdot y'=\frac12\cdot\frac1{1+x}+\frac92\cdot\frac1{3+x}+4\cdot\frac1{2+x}\\ &\implies y'(0)=(\frac12+\frac96+2)\cdot3^{\frac92}\cdot2^4=2^6\cdot3^{\frac92}\\ [例2]&\color{maroon}{设f(x)=\sqrt{\frac{(1+x)\sqrt x}{e^{x-1}}}+\arcsin\frac{1-x}{\sqrt{1+x^2}},求f'(1)}\\ &令y_1=\sqrt{\frac{(1+x)\sqrt x}{e^{x-1}}}\implies\ln{y_1}=\frac12(\ln(1+x)+\frac12\ln x-(x-1))\\ &\implies\frac1{y_1}\cdot y_1''=\frac12(\frac1{1+x}+\frac1{2x}-1)\\ &\implies y'_1(1)\underrightarrow{代入}0\\ &令y_2=\arcsin\frac{1-x}{\sqrt{1+x^2}}\implies y_2'(1)=\lim_{x\to1}\frac{y_2(x)-y_2(1)}{x-1}\\ &=\lim_{x\to1}\frac{\arcsin\frac{1-x}{\sqrt{1+x^2}}-0}{x-1}=\lim_{x\to1}\frac{\frac{1-x}{\sqrt{1+x^2}}}{x-1}=-\frac{\sqrt2}{2}\\ &故f'(1)=-\frac{\sqrt2}2 \end{aligned} 高阶导数归纳法莱布尼茨公式法展开\begin{aligned} \ [例1]&\color{maroon}{求下列导数}\\ (1)&\color{maroon}{y=\ln(1+x),求y^{(n)}}\\ (2)&\color{maroon}{y=e^x\cos x,求y^{(4)}}\\ (3)&\color{maroon}{设f(x)=(x^2-3x+2)^n\cos\frac{\pi x^2}{16},则f^{(n)}(2)=\underline{\quad}}\\ (4)&\color{maroon}{设f(x)=\frac{x}{1-2x^4},则f^{(101)}(0)=\underline{\quad}}\\ (1)&(\frac1x)'=(x^{-1})^{-1}=(-1)x^{-2},(\frac1x)''=(-1)(-2)x^{-3},(\frac1x)^n=(-1)n!x^{-(n+1)}\\ &(\ln x)'=\frac1x,(\ln x)^n=(\frac1x)^{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}^nC_n^ku^{(k)}v^{(n-k)}=C_n^0uv^{(n)}+C_n^1u'v^{(n-1)}+\cdots+C_n^nu^{(n)}v^{(0)}\\ &\cos x(e^x)^4+4(-\sin x)(e^x)'''+6(-\cos x)(e^x)''+4\sin x(e^x)'+\cos x\cdot e^x\\ (3)&(x-2)^3\to3(x-2)^2\to6(x-2)\to6\to3!\\ &(x-x_0)^n求n阶导数得到n!\\ &故f(x)=(x-2)^n(x-1)^n\cdot\cos\frac{\pi x^2}{16}\\ &f^{(n)}(2)=(x-2)^n|_{x=2}+(x-2)^{n-1}|_{x=2}+\cdots+(x-2)|_{x=2}+1\cdot n!\cdot\frac{\sqrt2}2\\ (4)&\begin{cases}1.抽象展开f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n\\2.具体展开式f(x)=\\3.展开式唯一\implies1=2\implies f^{(n)}(0)\end{cases}\\ &1.f(x)=\sum_{m=0}^{\infty}\frac{f^{(m)}(0)}{m!}x^m\\ &2.f(x)=x\cdot\sum_{n=0}^{\infty}2^n\cdot x^{4n}=\sum_{n=0}^{\infty}2^n\cdot x^{4n+1}\\ &3.m=101\implies \frac{f^{(101)}(0)}{(101)!}x^{101}\implies 4n+1=101\implies n=25\\ &\implies 2^{25}x^{101}即f^{(101)}(0)=2^{25}(101)!\\ [注]&\frac1{1-x}=1+x+x^2+cdots+x^n+\cdots=\sum_{n=0}^{\infty}x^n,\mid x\mid<1\\ &\frac1{1-狗}=\sum_{n=0}^{\infty}狗^n,\mid 狗\mid<1\\ &\frac1{1-2x^4}=\sum_{n=0}^{\infty}(2x^4)^n,\mid 2x^4\mid<1\\ \end{aligned} \begin{aligned} \ [例1]&\color{maroon}{求下列导数}\\ (1)&\color{maroon}{y=\ln(1+x),求y^{(n)}}\\ (2)&\color{maroon}{y=e^x\cos x,求y^{(4)}}\\ (3)&\color{maroon}{设f(x)=(x^2-3x+2)^n\cos\frac{\pi x^2}{16},则f^{(n)}(2)=\underline{\quad}}\\ (4)&\color{maroon}{设f(x)=\frac{x}{1-2x^4},则f^{(101)}(0)=\underline{\quad}}\\ (1)&(\frac1x)'=(x^{-1})^{-1}=(-1)x^{-2},(\frac1x)''=(-1)(-2)x^{-3},(\frac1x)^n=(-1)n!x^{-(n+1)}\\ &(\ln x)'=\frac1x,(\ln x)^n=(\frac1x)^{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}^nC_n^ku^{(k)}v^{(n-k)}=C_n^0uv^{(n)}+C_n^1u'v^{(n-1)}+\cdots+C_n^nu^{(n)}v^{(0)}\\ &\cos x(e^x)^4+4(-\sin x)(e^x)'''+6(-\cos x)(e^x)''+4\sin x(e^x)'+\cos x\cdot e^x\\ (3)&(x-2)^3\to3(x-2)^2\to6(x-2)\to6\to3!\\ &(x-x_0)^n求n阶导数得到n!\\ &故f(x)=(x-2)^n(x-1)^n\cdot\cos\frac{\pi x^2}{16}\\ &f^{(n)}(2)=(x-2)^n|_{x=2}+(x-2)^{n-1}|_{x=2}+\cdots+(x-2)|_{x=2}+1\cdot n!\cdot\frac{\sqrt2}2\\ (4)&\begin{cases}1.抽象展开f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n\\2.具体展开式f(x)=\\3.展开式唯一\implies1=2\implies f^{(n)}(0)\end{cases}\\ &1.f(x)=\sum_{m=0}^{\infty}\frac{f^{(m)}(0)}{m!}x^m\\ &2.f(x)=x\cdot\sum_{n=0}^{\infty}2^n\cdot x^{4n}=\sum_{n=0}^{\infty}2^n\cdot x^{4n+1}\\ &3.m=101\implies \frac{f^{(101)}(0)}{(101)!}x^{101}\implies 4n+1=101\implies n=25\\ &\implies 2^{25}x^{101}即f^{(101)}(0)=2^{25}(101)!\\ [注]&\frac1{1-x}=1+x+x^2+cdots+x^n+\cdots=\sum_{n=0}^{\infty}x^n,\mid x\mid<1\\ &\frac1{1-狗}=\sum_{n=0}^{\infty}狗^n,\mid 狗\mid<1\\ &\frac1{1-2x^4}=\sum_{n=0}^{\infty}(2x^4)^n,\mid 2x^4\mid<1\\ \end{aligned} 导数的几何应用研究对象\begin{aligned} &1.祖孙三代\begin{cases}f(x),f_n(x),f_1f_2\ldots f_n\\f'(x),\frac{df(x)}{dx^2}=\frac1{2x}f'(x)\\\int_a^xf(t)dt\\\sum a_nx^n\end{cases}\\ &2.分段函数（含绝对值函数）\\ &3.用参数表示函数\begin{cases}x=x(t)\\y=y(t)\end{cases}\\ &4.隐函数F(x,y)=0 \end{aligned} \begin{aligned} &1.祖孙三代\begin{cases}f(x),f_n(x),f_1f_2\ldots f_n\\f'(x),\frac{df(x)}{dx^2}=\frac1{2x}f'(x)\\\int_a^xf(t)dt\\\sum a_nx^n\end{cases}\\ &2.分段函数（含绝对值函数）\\ &3.用参数表示函数\begin{cases}x=x(t)\\y=y(t)\end{cases}\\ &4.隐函数F(x,y)=0 \end{aligned} 研究内容斜率、切线、法线、截距\begin{aligned} &\color{blue}{y=y(x)\implies y'(x)\implies k\iff切线k,法线-\frac1k}\\ [例1]&\color{maroon}{曲线(2-x^{n^2})y=1在点(1,1)处的切线与x轴的交点为(x_n,0),n=1,2\ldots,则\lim_{n\to\infty}x_n^{\frac{n^2}2}=?}\color{green}{(f_n(x)\to k)}\\ &y=\frac1{2-x^{n^2}},y'=-\frac{-n^2x^{n^2-1}}{(2-x^{n^2})^2},y'(1)=n^2=k_n\\ &切线y-1=n^2(x-1),令y=0,x_n=1-\frac1{n^2}\\ &故\lim_{n\to\infty}x_n^{\frac{n^2}2}=\lim_{n\to\infty}(1-\frac1{n^2})^{\frac{n^2}2}=e^{\lim_n\to\infty n^2/2\cdot(-1/n^2)}=e^{-\frac12}\\ [例2]&\color{maroon}{使曲线f(x)=x^n在点(1,1)处的切线与x轴的交点为(x_n,0),n=1,2,\cdots,求\lim_{n\to\infty}f(x_n)}\\ &f'(x)=nx^{n-1}\implies k=n,故y-1=n(x-1)\implies x_n=1-\frac1n\\ &故I=\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}(1-\frac1n)^n=e^A=e^{-1}\\ &其中A=\lim_{n\to\infty}n(1-\frac1n-1)=-1\\ [例3]&\color{maroon}{若曲线C:f(x)由方程2x-y=2\arctan(y-x)确定，则曲线在点(1+\frac2\pi,2+\frac\pi2)的切线方程是\underline{\qquad}}\\ &2-y'=\frac2{1+(y-x)^2}(y'-1)\implies k=y'|_p=\frac32\\ &\implies y-(2+\frac{\pi}2)=\frac32(x-(1+\frac\pi2))\\ [例4]&\color{maroon}{已知两条曲线由y=f(x)与xy+e^{x+y}=1所确定，且在点(0,0)处的切线相同，}\\ &\color{maroon}{写出此切线方程，求极限\lim_{n\to0}nf(\frac2n）}\\ &由xy+e^{x+y}=1,知y+xy'+e^{x+y}(1+y')=0\\ &\implies y'(0)=-1=k,\therefore y-0=-x\\ &\implies I=\lim_{n\to\infty}\frac{f(\frac2n)}{\frac1n}=\lim_{n\to\infty}\frac{f(0+\frac2n)-f(0)}{\frac2n}\cdot2=2f'(0)=-2\\ \end{aligned} \begin{aligned} &\color{blue}{y=y(x)\implies y'(x)\implies k\iff切线k,法线-\frac1k}\\ [例1]&\color{maroon}{曲线(2-x^{n^2})y=1在点(1,1)处的切线与x轴的交点为(x_n,0),n=1,2\ldots,则\lim_{n\to\infty}x_n^{\frac{n^2}2}=?}\color{green}{(f_n(x)\to k)}\\ &y=\frac1{2-x^{n^2}},y'=-\frac{-n^2x^{n^2-1}}{(2-x^{n^2})^2},y'(1)=n^2=k_n\\ &切线y-1=n^2(x-1),令y=0,x_n=1-\frac1{n^2}\\ &故\lim_{n\to\infty}x_n^{\frac{n^2}2}=\lim_{n\to\infty}(1-\frac1{n^2})^{\frac{n^2}2}=e^{\lim_n\to\infty n^2/2\cdot(-1/n^2)}=e^{-\frac12}\\ [例2]&\color{maroon}{使曲线f(x)=x^n在点(1,1)处的切线与x轴的交点为(x_n,0),n=1,2,\cdots,求\lim_{n\to\infty}f(x_n)}\\ &f'(x)=nx^{n-1}\implies k=n,故y-1=n(x-1)\implies x_n=1-\frac1n\\ &故I=\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}(1-\frac1n)^n=e^A=e^{-1}\\ &其中A=\lim_{n\to\infty}n(1-\frac1n-1)=-1\\ [例3]&\color{maroon}{若曲线C:f(x)由方程2x-y=2\arctan(y-x)确定，则曲线在点(1+\frac\pi2,2+\frac\pi2)的切线方程是\underline{\qquad}}\\ &2-y'=\frac2{1+(y-x)^2}(y'-1)\implies k=y'|_p=\frac32\\ &\implies y-(2+\frac{\pi}2)=\frac32(x-(1+\frac\pi2))\\ [例4]&\color{maroon}{已知两条曲线由y=f(x)与xy+e^{x+y}=1所确定，且在点(0,0)处的切线相同，}\\ &\color{maroon}{写出此切线方程，求极限\lim_{n\to0}nf(\frac2n）}\\ &由xy+e^{x+y}=1,知y+xy'+e^{x+y}(1+y')=0\\ &\implies y'(0)=-1=k,\therefore y-0=-x\\ &\implies I=\lim_{n\to\infty}\frac{f(\frac2n)}{\frac1n}=\lim_{n\to\infty}\frac{f(0+\frac2n)-f(0)}{\frac2n}\cdot2=2f'(0)=-2\\ \end{aligned} 极值、单调性\begin{aligned} &\color{blue}{极值点：若存在x_0的某个邻域，使得在该邻域内任意一点x，则\begin{cases}f(x)\leq f(x_0)\to极大值\\f(x)\geq f(x_0)\to极小值\end{cases}}\\ &\color{blue}{单调性：若y=f(x)在区间I上有f'(x)>0,则y=f(x)在I上单调增加，若f'(x)<0,则单调减少}\\ &\color{blue}{判别: \begin{cases}1.令f'(x)=0\implies x_0驻点（不\exists\implies不可导点）\\2.判别(x_0-\delta,x_0)时,f'(x)?0,(x_0,x_0+\delta)时,f'(x)?0,\to x_0是否为极值点\end{cases}}\\ [例1]&\color{maroon}{求y=\sum_{k=0}^n\frac{x^k}{k!}\cdot e^{-x}的极值}\color{green}{(\sum a_nx^n\to极值/单调性)}\\ &y=\sum_{k=0}^n\frac{x^k}{k!}\cdot e^{-x}=(1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!})\cdot e^{-x}\\ &1.y'=(1+x+\frac{x^2}{2!}+\cdots+\frac{x^{n-1}}{n-1})\cdot e^{-x}+(1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!})\cdot e^{-x}\cdot(-1)=-\frac{x^n}{n!}\cdot e^{-x}\\ &令y'=0\implies x=0(驻点)\\ &2.\begin{cases}n为偶数\begin{cases}x<0\\x>0\end{cases}\to y'(x)<0\to x=0不是极值点\\n为奇数\begin{cases}x<0\to y'(x)>0\\x>0\to y'(x)<0\end{cases}\to x=0是极大值点\end{cases}\\ [例2]&\color{maroon}{\begin{cases}x^{2x},x>0\\xe^x+1,x\leq0\end{cases}求f'(x)并求f(x)的极值}\\ &1.x>0,f'(x)=(x^{2x})'=(e^{2x\ln x})'=e^{2x\ln x}\cdot(2\ln x+x)=2x^{2x}\cdot(\ln x+1)\\ &x<0,f'(x)=e^x+x\cdot e^x=(1+x)e^x\\ &x=0,f'_+(0)=\lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0^+}\frac{e^{2x\ln x}-1}{x}=-\infty,不存在\\ &则f'(x)=\begin{cases}2x^{2x}\cdot(\ln x+1),x>0\\(1+x)e^x,x<0\end{cases}\\ &2.令2x^{2x}\cdot(\ln x+1)=0\implies x=\frac1e\iff0< x<\frac1e\to f'(x)<0\qquad x>\frac1e\to f'(x)>0\\ &\lim_{x\to0^+}x^{2x}=e^0=1=f(0),\therefore 极小值f(\frac1e)=e^{-\frac2e}\\ &令(1+x)e^x=0\implies x=-1\iff x<-1\to f'(x)<0\qquad x>-1\implies f'(x)>0\\ &\therefore极小值f(-1)=1-\frac1e且极大值f(0)=1\\ [例3]&\color{maroon}{求双曲线y_1=\frac1x与抛物线y_2=\sqrt x的交角}\\ &交点(1,1),y_1'(1)=(-\frac1{x^2})|_{x=1}=-1=\tan\alpha\\ &y_2'(1)=(\frac1{2\sqrt x})|_{x=1}=\frac12=\tan\beta\\ &\implies r=\alpha-\beta=\frac34\pi-\arctan\frac12\\ [例4]&\color{maroon}{求函数f(x)=|x|e^{-|x-1|}的极值}\\ &f(x)=\begin{cases}-xe^{x-1},x<0\\0,x=0\\xe^{x-1},0< x<1\\ 1,x=1\\xe^{1-x},x>1\end{cases}\qquad f'(x)=\begin{cases}-e^{x-1}-xe^{x-1},x<0\\e^{x-1}+xe^{x-1},0< x<1\\ e^{1-x}-xe^{1_x},x>1\end{cases}\\ &f'_(0)=-e^{-1},f'_+(0)=e^{-1},\therefore f'(0)不存在\\ &f'_(1)=2,f'_+(1)=0,\therefore f'(1)不存在\\ &知x_1=-1,x_2=0,x_3=1,则x_1=-1为极大点，x_2=0为极小点，x_3=1为极大点\\ [例5]&\color{maroon}{设正值函数f(x)在(1,+\infty)内连续，求函数F(x)=\int_1^x[(\frac2x+\ln x)-(\frac2t+\ln t)]f(t)dt的最小值点}\\ &F(x)=\int_1^x(\frac2x+\ln x)f(t)dt-\int_1^x(\frac2t+\ln t)f(t)dt\\ &=(\frac2x+\ln x)\int_1^xf(t)dt-\int_1^x(\frac2t+\ln t)f(t)dt\\ & \implies F'(x)=(-\frac2{x^2}+\frac1x)\int_1^xf(t)dt+(\frac2x+\ln x)F(x)-(\frac2x+\ln x)f(x)\\ &由F'(x)=0知x=2是唯一极小值点,\therefore x=2是最小值点\\ [例6]&\color{maroon}{设f(x)=\begin{cases}\lim_{n\to\infty}\frac1n(1+\cos\frac xn+\cos\frac{2x}n+\cdots+\cos\frac{n-1}nx),x>0\\1,x=0\\f(-x),x<0\end{cases}}\\ &\color{maroon}{(1)求f'(0)\qquad(2)求f(x)在[-\pi,\pi]上的最大值}\\ (1)&x>0时,f(x)=\lim_{n\to\infty}\sum_{i=0}^{n-1}\cos\frac inx\cdot\frac1n=\lim_{n\to\infty}\sum_{i=0}^{n-1}\cos\frac xni\cdot\frac
xn\cdot\frac1x\\ &=\frac1x\int_0^x\cos tdt=\frac{\sin x}{x}\implies f(x)=\begin{cases}\frac{\sin x}{x},x>0\\1,x=0\\ \frac{\sin x}x,x<0\end{cases}为偶函数\\ &f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{\frac{\sin x}x-1}{x}=\lim_{x\to0}\frac{\sin x-x}{x^2}=0\\ (2)&只研究[0,\pi],f'(x)=(\frac{\sin x}{x})'=\frac{x\cos x-\sin x}{x^2}\\ &令g(x)=x\cos x-\sin x\\ &则g'(x)=\cos x+x(-\sin x)-\cos x=-x\sin x\leq0\\ &\implies g(x)单调递减，g(0)=0\implies g(x)<0\\ &\implies f'(x)<0\implies f(x)单调递减,f(0)=1,f(\pi)=0\implies [0,1]\\ [例7]&\color{maroon}{已知f'(-x)=x[f'(x)+1],求f(x)的极值点，并说明是极大值点还是极小值点}\\ &f'(-1x)=x[f'(x)+1]\implies f'(x)=-x[f'(x)+1]\\ &代入,得f'(x)=-x[x[f'(x)+1]+1]\implies f'(x)=\frac{-x^2-x}{1+x^2}\\ &由f'(x)=0,知x_1=0,x_2=-1\\ &\therefore x_1=0是极大值点，x_2=-1是极小值点\\ \end{aligned} 拐点、凹凸性\begin{aligned} &\color{blue}{凹凸性判别：设函数f(x)在I上二阶可导，则f''(x)>0为凹，f''(x)<0为凸}\\ &\color{blue}{拐点：设f''(x_0)存在，且点(x_0,f(x_0))为曲线上的拐点，则f''(x_0)=0}\\ &\color{blue}{参数方程求导：\begin{cases}x=x(t)\\y=y(t)\end{cases}\to\begin{cases}\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{y'(t}{x'(t)}=f(t)\\ \frac{d^2y}{dx^2}=\frac{d(dy/dx)}{dx}=\frac{df(t)/dt}{dx/dt}\end{cases}}\\ &\color{maroon}{[例]y(x)=\begin{cases}x=\frac13t^3+t+\frac13\\y=\frac13t^3-t+\frac13\end{cases}，求y=y(x)的极值、凹凸性和拐点}\color{green}{(参数方程\to凹凸性/拐点)}\\ &\color{black}[分析]\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{t^2-1}{t^2+1}\\ &\frac{d^2y}{dx^2}=\frac{f'(t)}{x'(t)}=\frac{[2t(t^2+1)-(t^2-1)\cdot2t]/(t^2+1)^2}{t^2+1}=\frac{2t^3+2t-2t^3+2t}{(t^2+1)^2}\cdot\frac1{(t^2+1)}=\frac{4t}{(t^2+1)^3}\\ &令\frac{dy}{dx}=0\to t=\pm1\qquad\frac{d^2y}{dx^2}\to t=0\\&
\begin{array}{c|c|c|c|c|c|c|c}
t & (-\infty,-1) & -1 & (-1,0) & 0 & (0,1) & 1 & (1,+\infty) \\
\hline
x & (-\infty,-1) & -1 & (-1,\frac13) & \frac13 & (\frac13,\frac35) & \frac35 & (\frac35,+\infty)
\\
\hline
y' & + & 0 & - & -1 & - & 0 & +
\\
\hline
y'' & - & - & - & 0 & + & + & +
\end{array}\qquad\qquad\qquad\qquad\qquad\quad\qquad \end{aligned} 渐近线\begin{aligned} &\color{blue}{判别\begin{cases}水平渐近线:\lim_{x\to\infty}f(x)=A\\ 铅锤渐近线:\lim_{x\to x_0}=\infty\\ 斜渐近线:\lim_{x\to+\infty}\frac{y(x)}x=k(同阶才可以)，若k\neq0,则求b=\lim_{x\to+\infty}[y(x)-kx],得y=kx+b\end{cases}}\\ [例1]&\color{maroon}{求y=\sqrt{4x^2+x}\cdot\ln(2+\frac1x)的全部渐近线}\color{green}{(f(x)复杂\to渐近线)}\\ &\color{black}[分析]由两个分部得定义域x\in(-\infty,-\frac12)\bigcup(0,+\infty)\\ 1.&\lim_{x\to-\frac12-0}\sqrt{4x^2+x}\cdot\ln(2+\frac1x)=-\infty\iff x=-\frac12是一条铅锤渐近线\\ &\lim_{x\to0^+}\sqrt{4x^2+x}\cdot\ln(2+\frac1x)=\lim_{x\to0^+}\sqrt{4x^2+x}\cdot\ln(2x+1)-\lim_{x\to0^+}\sqrt{4x^2+x}\cdot\ln x\\ &=-\lim_{x\to0^+}\sqrt{4x+1}\cdot\sqrt x\cdot\ln x=0\\ 2.&\lim_{x\to+\infty}\sqrt{4x^2+x}\cdot\ln(2+\frac1x)=+\infty,没有水平渐近线\\ 3.&\lim_{x\to+\infty}\frac yx=\lim_{x\to+\infty}\sqrt{4+\frac1x}\cdot\ln(2+\frac1x)=2\ln 2\neq0,故k\ \exists\\ &b=\lim_{x\to+\infty}[y-kx]=\lim_{x\to+\infty}(\sqrt{4x^2+x}\ln(2+\frac1x)-2\ln2\cdot x),令x=\frac1t\\ &b=\lim_{t\to0^+}(\sqrt{\frac{4+t}{t^2}}\ln(2+t)-\frac{2\ln2}{t})=\lim_{t\to0^+}\frac{\sqrt{4+t}\ln(2+t)-2\ln2}{t}\\ &=\lim_{t\to0^+}\frac{1}{2\sqrt{4+t}}\ln(2+t)+\frac{\sqrt{4+t}}{2+t}=\frac14\ln2+1\\ &y=2\ln2\cdot x+\frac14\ln2+1为斜渐近线\\ &当x\to-\infty时，y=-ax-b=-2\ln2\cdot x-(\frac14\ln2+1)为另一斜渐近线\\ &\color{red}{[注]\lim_{x\to0^+}x^{\alpha}\ln x=\lim_{x\to0^+}\frac{\ln x}{x^{-\alpha}}=\lim_{x\to0^+}\frac{x^{-1}}{-\alpha x^{-\alpha-1}}=-\frac1{\alpha}\lim_{x\to0^+}x^\alpha=0}\\ [例2]&\color{maroon}{求下列各题}\\ (1)&\color{maroon}{x>0,y=x\sin\frac1x,求其水平渐近线}\\ (2)&\color{maroon}{x>0,y=x+\sin\frac1x,求其斜渐近线}\\ (3)&\color{maroon}{y=\ln(e-\frac1x),求其铅垂渐近线}\\ (1)&\lim_{x\to+\infty}x\sin\frac1x=\lim_{x\to+\infty}\frac{\sin\frac1x}{\frac1x}=1\implies y=1为水平渐近线\\ (2)&\lim_{x\to+\infty}\frac{y}x=\lim_{x\to+\infty}\frac{x+\sin\frac1x}x=1=a\\ &b=\lim_{x\to+\infty}[y(ax)-ax]=\lim_{x\to+\infty}\sin\frac1x=0\implies y=ax+b为斜渐近线\\ (3)&有e-\frac1x>0,x>\frac1e\\ &故\lim_{x\to0^-}\ln(e-\frac1x)=+\infty\implies x=0是铅垂渐近线\\ &\lim_{x\to\frac1e^{+0}}\ln(e-\frac1x)=-\infty\implies x=\frac1e是铅垂渐近线\\ \end{aligned} 导数的证明性应用中值定理\begin{aligned} &设f(x)在[a,b]上连续\color{grey}{(涉及函数f(x)的中值定理)}\\ &\color{black}1.有界性定理：
f(x)|\leq k\\ &2.最值定理： m\leq f(x)\leq M\\ &3.介值定理： 若m\leq \mu\leq M,\exists\xi\in[a,b],使f(\xi)=\mu\\ &4.零点定理：若f(a),f(b)<0\implies\exists\xi\in(a,b),使f(\xi)=0\\ &\color{grey}{涉及到导数f'(x)的中值定理}\\ &\color{black}5.费马定理：设f(x)在x_0处\begin{cases}可导\\取极值\end{cases}\implies f'(x_0)=0\\ &6.罗尔定理：设f(x)\begin{cases}[a,b]连续\\(a,b)可导\\f(a)=f(b)\end{cases}\implies \exists f'(\xi)=0,\xi\in(a,b)\\ &7.拉格朗日中值定理：设f(x)\begin{cases}[a,b]连续\\(a,b)可导\end{cases}\implies f(b)-f(a)=f'(\xi)(b-a),\exists\xi\in(a,b)\\ &8.柯西中值定理：设f(x),g(x)\begin{cases}[a,b]连续\\(a,b)可导\\g'(x)\not=0\end{cases}\implies\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(\xi)}{g'(\xi)},\exists\xi\in(a,b)\\ &9.泰勒公式：\\ &f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\cdots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+\begin{cases}\circ((x=x_0)^n)\\\frac{f^{(n)}(\xi)}{n!}(x-x_0)^{n+1}\end{cases}\\ &麦克劳林公式：f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots+\frac{f^{(n)}(0)}{n!}x^n+\begin{cases}\circ(x^n)佩亚诺余项\\\frac{f^{n+1}(\xi)}{(n+1)!}x^{n+1}拉格朗日余项\end{cases}\\ &10.积分中值定理：设f(x)在[a,b]上连续\implies\int_a^bf(x)dx=f(\xi)(b-a),\exists\xi\in[a,b]\\ &\color{grey}{[注]称\begin{cases}\hat f=\frac1{b-a}\int_a^bf(x)dx\\\hat f=\frac1n\sum_{i=1}^nf(x_i)\end{cases}叫f(x)在[a,b]上的平均值} \end{aligned} 1.确定区间2.确定辅助函数3.确定点的值4.确定所用定理确定辅助函数\begin{aligned} (1)&简单情形：题设f(x)即为辅助函数\\ (2)&复杂情形\\ &\begin{cases}1.乘积求导公式逆用\begin{cases}(uv)'=u'v+uv'\\(uv)''=u''v+2u'v'+uv''\end{cases}\\2.商的求导公式逆用\begin{cases}1.(\frac{f(x)}{x})'=\frac{f'(x)x-f(x)}{x^2}\\2.[\frac{f'(x)}{f(x)}]'=\frac{f''(x)f(x)-[f'(x)]^2}{f^2(x)}\\3.[\ln f(x)]''=\frac{f''(x)f(x)-[f'(x)]^2}{f^2(x)}\end{cases}\\3.变现积分:若出现\int_a^bf(x)dx,可能是令F(x)=\int_a^xf(t)dt\\4.若复杂到作不出辅助函数，则题设给出F(x)或F(a)，提示考生令其为辅助函数\end{cases} \end{aligned} 确定使用定理\begin{aligned} 1.&介值定理:证f(\xi)=\mu\\ 2.&费马定理:证f'(\xi)=0(最/极值，区间内)\\ 3.&罗尔定理:证f'(\xi)=0(f(a)=f(b))\\ 4.&拉格朗日:\begin{cases}1.题设中有f,f'得关系，或f-f\\2.证f(x)\geq A或f'(\xi)\geq A\\3.证f^{(n)}(\xi)\geq A\\4.\theta\\5.单调性(f,f')\end{cases}\\ 5.&泰勒:\begin{cases}1.题设中有f与f^{(n)},n>1的关系\\2.证f^{(n)}(\xi)\geq A\end{cases}\\ 6.&柯西 \end{aligned} 确定点的信息\begin{aligned} 1.&用题设告知，如f(a)=0\\ 2.&用极限:\begin{cases}1.连续定义\lim_{x\to x_0}f(x)=f(x_0)\\2.导数定义f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\\3.保号(不等式脱帽法)\\4.取极限\end{cases}\\ 3.&用积分:\begin{cases}1.\overline{f}=\frac{\int_a^bf(x)dx}{b-a}=f(\xi)(均值定义)\\2.F(x)=\int_a^xf(t)dt(原函数定义)\\3.保号:若f(x)\geq0不恒=0\implies\int_a^bf(x)dx>0\\4.取积分\end{cases}\\ 4.&用介值：f(a)=A,f(b)=B,A<\mu<B\implies f(\xi)=\mu\\ 5.&用费马\implies f'(\xi)=0\\ 6.&用奇偶\implies \begin{cases}若f(x)奇\implies f(0)=0\\若f(x)偶\implies f'(x)奇\implies f'(0)=0\end{cases}\\ 7.&用几何:\begin{cases}1.存在相等的最大值：f(a)=f(b)\\2.f(x)与g(x)交于a点\implies F(a)=f(a)-g(a)=0\\3.f(x)与g(x)在a点处有公切线\implies F'(a)=f'(a)-g'(a)=0\end{cases}\\ 8.&用行列式:如f(x)=\begin{vmatrix}1&x&4\\2&2x&7\\3&3x&9\end{vmatrix}\implies f(0)=0,f(1)=0\implies 罗尔定理的f'(\xi)=0\\
\end{aligned} 例题\begin{aligned} \ [例1]&\color{maroon}{设lim_{x\to0}\frac{f(x)}{x}=1,f''(x)>0,证明f(x)\geq x}\\ &f(0)=\lim_{x\to0}f(x)=\lim_{x\to0}\frac{f(x)}x=0,且1=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=f'(0)\\ &f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(\xi)}2(x-x_0)^2\\ &即f(x)=f(0)+f'(0)(x-0)+\frac{f''(\xi)}2(x-0)^2\\ &f(x)=x+\circ\geq x\\ [Th]&若f(x)在x=x_0处连续且\lim_{x\to x_0}\frac{f(x)}{x-x_0}=A,则f(x_0)=0,f'(x_0)=A\\ [证]&f(x_0)=\lim_{x\to x_0}f(x)=\lim_{x\to x_0}\frac{f(x)}{x-x_0}(x-x_0)=0且f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=A\\ [例2]&\color{maroon}{设f(x)在[a,b]上连续,(a.b)内可导且f(a)\not=f(b),证明\exists\xi,\eta\in(a,b),使得\frac{f'(\xi)}{2\xi}=\frac{f'(\eta)}{b+a}.a>0}\\ &1.\frac{f(b)-f(a)}{b^2-a^2}=\frac{f'(\xi)}{2\xi}\\ &2.f(b)-f(a)=f'(\eta)(b-a)\implies f(b)-f(a)=\frac{f'(\xi)}{2\xi}(b^2-a^2)=f(b)-f(a)=f'(\eta)(b-a)\\ &\implies \frac{f'(\xi)}{2\xi}(b+a)(b-a)=f'(\eta)(b-a)\\ [例3]&\color{maroon}{设f(x)在[0,4]上一阶可导，且f'(x)\geq\frac14,f(2)\geq0,则在\underline{[3,4]}上必有f(x)\geq\frac14}\\ &f(b)-f(a)=f'(\xi)(b-a)\\ &f(3)-f(2)=f'(3)(3-2)\implies f(3)=f(2)+f'(3)\geq\frac14\\ [例4]&\color{maroon}{设x>0,证明(1)\sqrt{x+1}-\sqrt{X}=\frac{1}{2\sqrt{x+\theta(x)}},0<\theta(x)<1,(2)求\lim_{x\to+\infty}\theta(x)}\\ (1)&令f(t)=\sqrt t,则由拉氏定理得\\ &\sqrt{x+1}-\sqrt x=\frac1{2\sqrt{x+\theta(x)}}，其中0<\theta(x)=\xi-x<1\\ (2)&2\sqrt{x+\theta(x)}=\frac1{\sqrt{x+1}-\sqrt{x}}=\sqrt{x+1}+\sqrt{x}\\ &4(x+\theta(x))=x+1+2\sqrt{(x+1)x}+x\\ &\theta(x)=\frac{x}2+\frac14+\frac12\sqrt{(x+1)x}-x=\frac14+\frac12\sqrt{(x+1)x}-\frac{x}2\\ &\lim_{x\to+\infty}\theta(x)=\frac14+\frac12\lim_{x\to+\infty}(\sqrt{(x+1)x}-x)=\frac14+\frac12\lim_{x\to+\infty}\frac{1\cdot x}{\sqrt{(x+1)x}+x}=\frac12\\ [例5]&\color{maroon}{设f(x)=\arcsin x,\xi为f(x)在[0,b]上拉氏中值定理得中值点，0< b<1,求\lim_{b\to0^+}\frac{\xi}{b}}\\ &\arcsin b-\arcsin0=\frac{1}{\sqrt{1-b^2}}\cdot b\implies \xi=\sqrt{1-(\frac{b}{\arcsin b})^2}\\ &\lim_{b\to0^+}\frac{\xi}{b}=\lim_{b\to0^+}\frac{\sqrt{1-(\frac{b}{\arcsin b})^2}}{b}\\ &令b=\sin t则I=\lim_{t\to0^+}\frac{\sqrt{1-(\frac{\sin t}{t})^2}}{\sin t}=\lim_{t\to0^+}\frac{\sqrt{t^2-(\sin t)^2}}{t\sin t}=\lim_{t\to0^+}\frac{\frac1{\sqrt3}t^2}{t^2}=\frac{\sqrt3}{3}\\ \end{aligned}