前置例题
1.
S1=1+2+4+8+16+···
2S1= 2+4+8+16+···
2S1-S1=-1
S1=-12.
x=0.9999···
10x=9.9999···
9x=9
x=1函数的导数和微分
函数的不定积分和定积分
知识点1 平均变化率
一般地，已知函数\(y=f\left( x\right)\),\(x_{0},x_{1}\)是定义域内不同的两点，记\(\Delta x=x_{1}-x_{0}\),\(\Delta y=y_{1}-y_{0}=f\left( x_{1}\right) -f\left( x_{0}\right) =f\left( x_{0}+\Delta x\right) -f\left( x_{0}\right)\)。
则当\(\Delta x\neq 0\)时，商\(\dfrac {\Delta y}{\Delta x}=\dfrac {f\left( x_{0}+\Delta x\right) -f\left( x_{0}\right) }{\Delta x}\)称作函数\(y=f\left( x\right)\)在区间\(\left[ x_{0},x_{0}+\Delta x\right]\)或\(\left[ x_{0}+\Delta x,x_{0}\right]\)的平均变化率。
例1
求函数\(y=x^{2}\)在区间\(\left[ x_{0},x_{0}+\Delta x\right]\)的平均变化率。
解：\(\dfrac {\Delta y}{\Delta x}=\dfrac {\left( x_{0}+\Delta x\right) ^{2}-x^{2}_{0}}{\Delta x}=\Delta x+2x_{0}\)
例2
求函数\(y=\dfrac {1}{x}\)在区间\(\left[ x_{0},x_{0}+\Delta x\right]\)的平均变化率。
解：\(\dfrac {\Delta y}{\Delta x}=\dfrac {\dfrac {1}{x_{0}+\Delta x}-\dfrac {1}{x_{0}}}{\Delta x}=-\dfrac {1}{x_{0}\left( x_{0}+\Delta x\right) }\)
知识点2 瞬时变化率
如果\(\Delta x\)趋近于0，平均变化率\(\dfrac {\Delta y}{\Delta x}=\dfrac {f\left( x_{0}+\Delta x\right) -f\left( x_{0}\right) }{\Delta x}\)趋近于常数l，那么称常数l为函数\(y=f\left( x\right)\)在点\(x_{0}\)处的瞬时变化率，记作：
当\(\Delta x\rightarrow 0\)时，\(\dfrac {f\left( x_{0}+\Delta x\right) -f\left( x_{0}\right) }{\Delta x}\rightarrow l\)
也记作\[\lim _{\Delta x\rightarrow 0}\dfrac {f\left( x_{0}+\Delta x\right) -f\left( x_{0}\right) }{\Delta x}= l\]
知识点3
函数\(y=f\left( x\right)\)在点\(x_{0}\)处的瞬时变化率，通常称为\(f\left( x\right)\)在点\(x_{0}\)处的导数。
记作：\(f'\left( x_{0}\right)\)
即\[\lim _{\Delta x\rightarrow 0}\dfrac {f\left( x_{0}+\Delta x\right) -f\left( x_{0}\right) }{\Delta x}= f'\left( x_{0}\right)\]
知识点4 导数
如果\(y=f\left( x\right)\)在开区间\(\left( a,b\right)\)内每一点都是可导的，区间\(\left( a,b\right)\)内的每一个值都对应一个确定的导数\(f'\left( x\right)\)，称区间\(\left( a,b\right)\)内\(f'\left( x\right)\)可构成一个新的函数，称为\(y=f\left( x\right)\)的导函数。
记作：\(f'\left( x\right)(\)或$ y'\(或\)y'x)$,通称为导数。
例3
火箭竖直向上发射，熄火时速度达到100m/，试问熄火多长时间，火箭的速度为0。
解：
\(h\left( t\right) =100t-\dfrac {1}{2}gt^{2}\)
平均变化率(平均速度)：\(\dfrac {h\left( t+\Delta t\right) -h\left( t\right) }{\Delta t}=100-\dfrac {1}{2}g\Delta t-gt\)
瞬时速度：\(\Delta t\rightarrow 0\)时
\(\dfrac {h\left( t+\Delta t\right) -h\left( t\right) }{\Delta t}=100-gt\)
令\(100-gt=0,g=9.8\)
\(t=\dfrac {100}{g}\approx10.2s\)
例4
圆，面积\(S=\pi r^{2}\)，周长\(l=2\pi r\)
解：\[S'\left( x\right) =\lim _{\Delta r\rightarrow 0}\dfrac {S\left(r+\Delta r\right) -S\left( r\right) }{\Delta r}\]
\[ =\lim _{\Delta r\rightarrow 0}\dfrac {\pi \left( r-\Delta r\right) ^{2}-r^{2}}{\Delta r}\]
\[ =\lim _{\Delta r\rightarrow 0}\dfrac {2\pi r\Delta r+\pi \Delta r^{2}}{\Delta r}\]
\[ =\lim _{\Delta r\rightarrow 0} {(2\pi r+\pi \Delta r)}=2\pi r\]
知识点5 导数的几何意义
\(A\left( x_{0},f\left( x_{0}\right) \right) , B\left( x_{0}+\Delta x,f\left( x_{0}+\Delta x\right) \right)\)
AB:割线（平均变化率）：\(\dfrac {\Delta y}{\Delta x}=\dfrac {f\left( x_{0}+\Delta x\right) -f\left( x_{0}\right) }{\Delta x}\)
切线（瞬时变化率）：\(\Delta x\rightarrow 0\)，\(\dfrac {\Delta y}{\Delta x}\rightarrow k\)
\(B\rightarrow A\)（转动）
割线\(AB\rightarrow\)切线$ AB’$
割线斜率\(\rightarrow\)切线斜率
\[f\left( x_{0}\right)=k=\lim _{\Delta x\rightarrow 0}\dfrac {f\left( x_{0}+\Delta x\right) -f\left( x_{0}\right) }{\Delta x} \]
函数在一点的切线的斜率就是函数在这点的导数：导数=斜率
例5
求抛物线\(y=x^{2}\)在点\(\left( x_{0},f\left( x_{0}\right) \right)\)切线的斜率。
解：求导数的几何意义
知\[k=f\left( x_{0}\right)\]
\[=\lim _{\Delta x\rightarrow 0}\dfrac {f\left( x_{0}+\Delta x\right) -f\left( x_{0}\right) }{\Delta x} \]
\[=\lim _{\Delta x\rightarrow 0}\dfrac {\left( x_{0}+\Delta x\right) ^{2}-x^{2}_{0}}{\Delta x}\]
\[=\lim _{\Delta x\rightarrow 0}(\Delta x+2x_{0})\]
\[=2x_{0}\]
求在（0,0）处切线的斜率
\[f\left( x\right)=k=0\]
例6
求双曲线\(y=\dfrac {1}{x}\)在点
\(\left( 2,\dfrac {1}{2}\right)\)的切线方程
方程\(y-y_{0}=k\left( x-x_{0}\right)\)
\(k=-\dfrac {1}{x^{2}}=-\dfrac {1}{4}\)
\(y=-\dfrac {1}{4}x+1\)
知识点6 导数的运算
(1)常值函数的导数
\(y=f\left( x\right) =c\)(c为常数)
\[y'=f'\left( x\right) =c'=\lim _{\Delta x\rightarrow 0}\dfrac {f\left( x_{0}+\Delta x\right) -f\left( x_{0}\right) }{\Delta x} \]
\[=\lim _{\Delta x\rightarrow 0} \dfrac {c-c}{\Delta x}=0\]
(2)
\(y=x\\ y'=x'=1\)
(3)
\(y=x^{2}\\ y'=\left( x^2\right) '=2x\)
(4)
\(y=x^{3}\\ y'=3x^{2}\)
(5)
\(y=\dfrac {1}{x}\\ y'=-\dfrac {1}{x^{2}}\)
(6)
\(y=\dfrac {1}{\sqrt {x}}\\ y'=\dfrac {1}{2\sqrt {x_{0}}}\)
\[y'=\lim _{\Delta x\rightarrow 0}\dfrac {f\left( x_{0}+\Delta x\right) -f\left( x_{0}\right) }{\Delta x} \]
\[=\lim _{\Delta x\rightarrow 0}\dfrac {\sqrt {x_{0}+\Delta x}-\sqrt {x_{0}}}{\Delta x} \](分母有理化)
\[=\lim _{\Delta x\rightarrow 0}\dfrac {1}{\sqrt {x_{0}+\Delta x}+\sqrt {x_{0}}}=\dfrac {1}{2\sqrt {x_{0}}} \]
知识点7 导数公式表
\(y=f\left( x\right) ,y'=f'\left( x\right)\)
$ y=c,y'=0$
$ y=x^{n}\left( n\in N+\right) ,y'=nx^{n-1}$
\(y=x^{\alpha }( \alpha \in Q,x>0), y'=\alpha x^{\alpha -1}\)
\(y= a^{x}(a > 0,a\neq 1),y'=a^{x}\ln a\)
\(y=\log a^{x}( a >0,a\neq 1),y'=\dfrac {1}{x\ln a}\)
特别地
\(y=e^{x},y'=e^{x}\\ y=\ln x,y'=\dfrac {1}{x}\\ y=\sin x,y'=\cos x\\ y=\cos x,y'=-\sin x\)
补充：
\(y=\ln \left| x\right| ,y'=\dfrac {1}{\left| x\right| }\\ y=\tan x,y'=\sec ^{2}x=\dfrac {1}{\cos ^{2}x}\)
\(y=\arcsin x,y'=\dfrac {1}{\sqrt {1-x^{2}}}\\ y=\arccos x,y'=-\dfrac {1}{\sqrt {1-x^{2}}}\)
\(y=\arctan x,y'=\dfrac {1}{1+x^{2}}\\ y=arccotx,y'=-\dfrac {1}{1+x^{2}}\)
证明\(y=\arcsin x,y'=\dfrac {1}{\sqrt {1-x^{2}}}\):
\[\arcsin x=\theta \\ \Downarrow \\ \sin \theta =x\]
\[\cos\Delta\theta =\Delta x\\ \theta '=\dfrac {\Delta \theta }{\Delta x}=\dfrac {1}{\cos \theta }=\dfrac {1}{\sqrt {1-\sin ^{2}\theta }}\\ \Downarrow \\ \left( \arcsin x\right) '=\dfrac {1}{\sqrt {1-x^{2}}}\]
极限：
“\(\dfrac {0}{0}\)”\[\lim _{x\rightarrow 0}\dfrac {\sin x}{x}=1\]
"\(\dfrac {\infty }{\infty }\)"\[\lim _{x\rightarrow 0}\left( 1+\dfrac {1}{x}\right) ^{x}=e\]
或\[\lim _{x\rightarrow 0}\left( 1+x\right) ^{\dfrac {1}{x}}=e\]
例7
求导
\(f\left( x\right) =x^{5}+2x^{4}+x^{3},g\left( x\right) =3^{x}+\ln x,h\left( x\right) =\cos x+\sin x\)
\(\Downarrow\)
\(f'\left( x\right) =5x^{4}+8x^{3}+3x^{2},g'\left( x\right) =3^{x}\ln 3+\dfrac {1}{x}, h'\left( x\right) =-\sin x+\cos x\)
知识点8 导数的四则运算
(1)函数和(或差)的求导法则
\(\left[ f\left( x\right) \pm g\left( x\right) \right]'=f'\left( x\right) \pm g'\left( x\right)\)
(2)函数的积的求导法则
\(\left[ f\left( x\right) \cdot g\left( x\right) \right] '=f'\left( x\right) \cdot g\left( x\right) +f\left( x\right) \cdot g'\left( x\right)\)
(3)导数商的求导法则
\(\left[ \dfrac {f\left( x\right) }{g\left( x\right) }\right] ^{'}=\dfrac {f'\left( x\right) g\left( x\right) -f\left( x\right) g'\left( x\right) }{g^{2}\left( x\right) }\)
证明：
\[\left[ \dfrac {f\left( x\right) }{g\left( x\right) }\right] ^{'}=\lim _{\Delta x\rightarrow 0}\dfrac {\dfrac {f\left( x+\Delta x\right) }{g\left( x+\Delta x\right) }-\dfrac {f\left( x\right) }{g\left( x\right) }}{\Delta x}\]
\[=\lim _{\Delta x\rightarrow 0}\dfrac {f\left( x+\Delta x\right) g\left( x\right) -f\left( x\right) g\left( x+\Delta x\right) }{g\left( x+\Delta x\right) g\left( x\right) \Delta x}\]
\[=\dfrac {f'\left( x\right) g\left( x\right) -f\left( x\right) g'\left( x\right) }{g^{2}\left( x\right) }\]
例8
求\(f\left( x\right) =a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\ldots +a_{n-1}x+a_{n}\)的导数
解：
\(f\left( x\right) =na_{0}x^{n-1}+(n-1)a_{1}x^{n-2}+\ldots +a_{n}-1\)
例9
\(y=x\cdot \sin x\)
解：
\(y'=x'\sin x+x\cdot \left( \sin x\right) '\\ =\sin x+x\cdot \cos x\)
例10
\(y=\sin 2x\)
解：
\(y=\sin 2x=2\sin x\cos x\\ y'=2(\cos ^{2}x-\sin ^{2}x )\\ =2\cos 2x\)
例11
\(y=\tan x=\dfrac {\sin x}{\cos x}\)
解：
\(y'=\dfrac {\left( \sin x\right) '\cos x-\sin x\left( \cos x\right)' }{\cos ^{2}x}\)
\(=\dfrac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}\\ =\dfrac {1}{\cos ^{2}x}\)
知识点9 利用导数判断函数的单调性
(1)如果(a,b)内\(f'\left( x\right) >0\)，函数单调递增
(2)如果(a,b)内\(f'\left( x\right) <0\)，函数单调递减
例13
y=x在\(\left( -\infty ,+\infty \right)\)上的单调性
解：\(y'=1 >0\)
例14
\(y=2x^{2}-2x+4\)在\(\left( -\infty ,+\infty \right)\)上的单调性
解：\(y'=2x-2\)
令\(y' >0\)，即\(x >1\)，在\(\left( 1,+\infty \right)\)上函数单调递增
令\(y' <0\)，即\(x <1\)，在\(\left( -\infty ,1\right)\)上函数单调递减
例15
\(y=x^{3}-4x^{2}+x-1\)
解：
\(y'=3x^{2}-8x+1\)
令\(y' >0\)，即在\(\left( -\infty ,\dfrac {4-\sqrt {13}}{3}\right) ,\left( \dfrac {4+\sqrt {13}}{3},+\infty \right)\)上函数单调递增
令\(y' <0\)，即在\(\left( \dfrac {4-\sqrt {13}}{3} ,\dfrac {4+\sqrt {13}}{3} \right)\)上函数单调递减
知识点10 利用导数研究函数的性质
\(y=\dfrac {1}{3}x^{3}-4x+4\\ y'=x^{2}-4\)
令\(y=0\)即\(x=\pm2\)
| x | (\(-\infty\),-2) | -2 | (-2,2) | 2 | (2,\(+\infty\)) |
|---|---|---|---|---|---|
| \(f'\left( x\right)\) | + | - | + | ||
| \(f\left( x\right)\) | \(\nearrow\) | 极大值 | \(\searrow\) | 极小值 | \(\nearrow\) |
知识点11 曲边梯形与定积分
例16
求曲线\(y=x^{2}\)与直线\(x=1,y=0\)所围成区域的面积。
解：
[0,1]等分。
\(S_{n}=0+\left( \dfrac {1}{n}\right) ^{2}\cdot \dfrac {1}{n}+\left( \dfrac {2}{n}\right) ^{2}\cdot \dfrac {1}{n}+\ldots +\left( \dfrac {n-1}{n}\right) ^{2}\cdot \dfrac {1}{n}=\dfrac {1}{n^{3}}\)
\(\left[ 1^{2}+2^{2}+\ldots +\left( n-1\right) ^{2}\right] =\dfrac {1}{n^{3}}\)
\(\dfrac {n\left( n-1)(2n-1\right) }{6}=\dfrac {1}{6}\left( 1-\dfrac {1}{n}\right) \left( 2-\dfrac {1}{n}\right)\)
\(n\rightarrow \infty\)时
\[\lim _{n\rightarrow \infty }\dfrac {1}{6}\left( 1-\dfrac {1}{n}\right) \left( 2-\dfrac {1}{n}\right)= \dfrac {1}{3}\]
例12
\(y=f\left( u\right) ,u=ax+b\)，求\(\dfrac {\Delta y}{\Delta x}\)
解：\(u'=a\)
\(\dfrac {\Delta y}{\Delta x}=\dfrac {\Delta y}{\Delta u}\cdot \dfrac {\Delta u}{\Delta x}\)
\(=f'\left( u\right) \cdot u'\left( x\right)\)
\(=a\cdot f'\left( u\right)\)
eg1
\(\mu =x^{-1},g\left( x\right) =4x^{2}+3x\)
解：
\(g\left( u\right) =4u^{2}+3u\\ g'\left( u\right) =8u+3\\ u'=-1\cdot x^{-2}=-\dfrac {1}{x^{2}}\)
eg2
\(\left[ \sin \left( 2x+\dfrac {\pi }{3}\right) \right]'\)
解：
\(u=2x+\dfrac {\pi }{3} \\y=\sin u\\ =\cos u\cdot 2\\ =2\cos \left( 2x+\dfrac {\pi }{3}\right)\)
eg3
\(\left[ \left( 5x+3\right) ^{5}\right] '\)
解：
\(y=u,u=5x+3,u'=5\)
\(y'=f'\left( u\right) \cdot u'\left( x\right) \\ =5u^{4}\cdot 5\\ =25\left( 5x+3\right) ^{4}\)
知识点12 [a,b]n分
\(a=x_{0} <x_{1} <x_{2} <\ldots <x_{n}=b\)
\[I\left( n\right) =\sum ^{n-1}_{i=0}f\left( \varepsilon i\right) \Delta xi\]
\(x_{i}=\max\){\(\Delta xi\)}
\[\lambda _{i}\rightarrow 0\\ \lim _{\lambda i\rightarrow 0}f\left( \varepsilon i\right) \Delta xi\rightarrow l\]定积分
记作\[\int ^{b}_{a}f\left( x\right) \Delta x= \lim _{\lambda i\rightarrow 0}\sum ^{n-1}_{i=0}f\left( \varepsilon i\right) \Delta xi\]
\(f\left( x\right)\)：倍积函数
例16（续）
\(S=\int ^{1}_{0}x^{2}\Delta x=\dfrac {1}{3}\)
知识点13 微积分基本定理
如果\(F\left( x\right) '=f\left( x\right)\)且\(f\left( x\right)\)在\([a,b]\)内可积
则
\(\int ^{b}_{a}f\left( x\right) \Delta x\)
$ =F\left( x\right) \int ^{b}_{a}$
$ =F\left( b\right) -F\left( a\right) $
例17
\(\int ^{1}_{0}x^{2}\Delta x=\)
解：
\(\because \left( \dfrac {1}{3}x^{3}\right) '=x^{2}\\ \therefore \int ^{1}_{0}x^{2}\Delta x\\ =\dfrac {x^{3}}{3}\int ^{1}_{0}=\dfrac {1}{3}\)
例18
\(\int ^{2}_{0}\left( x^{2}+1\right) \Delta x\\ =\left( \dfrac {x^{3}}{3}+x\right) \int ^{2}_{0}\)
例19
\(\int ^{\pi }_{0}\sin \Delta x\\ =-\cos x\int ^{\pi }_{0}\)