前置例题
1.
S1=1+2+4+8+16+···
2S1= 2+4+8+16+···
2S1-S1=-1
S1=-1
2.
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=fleft( x
ight)),(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}=fleft( x_{1}
ight) -fleft( x_{0}
ight) =fleft( x_{0}+Delta x
ight) -fleft( x_{0}
ight))。
则当(Delta x
eq 0)时,商(dfrac {Delta y}{Delta x}=dfrac {fleft( x_{0}+Delta x
ight) -fleft( x_{0}
ight) }{Delta x})称作函数(y=fleft( x
ight))在区间(left[ x_{0},x_{0}+Delta x
ight])或(left[ x_{0}+Delta x,x_{0}
ight])的平均变化率。
例1
求函数(y=x^{2})在区间(left[ x_{0},x_{0}+Delta x
ight])的平均变化率。
解:(dfrac {Delta y}{Delta x}=dfrac {left( x_{0}+Delta x
ight) ^{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
ight])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
ight) })
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 {fleft( x_{0}+Delta x
ight) -fleft( x_{0}
ight) }{Delta x})趋近于常数l,那么称常数l为函数(y=fleft( x
ight))在点(x_{0})处的瞬时变化率,记作:
当(Delta x
ightarrow 0)时,(dfrac {fleft( x_{0}+Delta x
ight) -fleft( x_{0}
ight) }{Delta x}
ightarrow l)
也记作$$lim {Delta x
ightarrow 0}dfrac {fleft( x{0}+Delta x
ight) -fleft( x_{0}
ight) }{Delta x}= l$$
知识点3
函数(y=fleft( x
ight))在点(x_{0})处的瞬时变化率,通常称为(fleft( x
ight))在点(x_{0})处的导数。
记作:(f'left( x_{0}
ight))
即$$lim {Delta x
ightarrow 0}dfrac {fleft( x{0}+Delta x
ight) -fleft( x_{0}
ight) }{Delta x}= f'left( x_{0}
ight)$$
知识点4 导数
如果(y=fleft( x
ight))在开区间(left( a,b
ight))内每一点都是可导的,区间(left( a,b
ight))内的每一个值都对应一个确定的导数(f'left( x
ight)),称区间(left( a,b
ight))内(f'left( x
ight))可构成一个新的函数,称为(y=fleft( x
ight))的导函数。
记作:(f'left( x
ight)()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:
(hleft( t
ight) =100t-dfrac {1}{2}gt^{2})
平均变化率(平均速度):(dfrac {hleft( t+Delta t
ight) -hleft( t
ight) }{Delta t}=100-dfrac {1}{2}gDelta t-gt)
瞬时速度:(Delta t
ightarrow 0)时
(dfrac {hleft( t+Delta t
ight) -hleft( t
ight) }{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=2pi r)
解:$$S'left( x
ight) =lim _{Delta r
ightarrow 0}dfrac {Sleft(r+Delta r
ight) -Sleft( r
ight) }{Delta r}$$
知识点5 导数的几何意义
(Aleft( x_{0},fleft( x_{0}
ight)
ight) , Bleft( x_{0}+Delta x,fleft( x_{0}+Delta x
ight)
ight))
AB:割线(平均变化率):(dfrac {Delta y}{Delta x}=dfrac {fleft( x_{0}+Delta x
ight) -fleft( x_{0}
ight) }{Delta x})
切线(瞬时变化率):(Delta x
ightarrow 0),(dfrac {Delta y}{Delta x}
ightarrow k)
(B
ightarrow A)(转动)
割线(AB
ightarrow)切线$ AB’(
割线斜率)
ightarrow$切线斜率
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},fleft( x_{0}
ight)
ight))切线的斜率。
解:求导数的几何意义
知$$k=fleft( x_{0}
ight)$$
求在(0,0)处切线的斜率
例6
求双曲线(y=dfrac {1}{x})在点
(left( 2,dfrac {1}{2}
ight))的切线方程
方程(y-y_{0}=kleft( x-x_{0}
ight))
(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=fleft( x
ight) =c)(c为常数)
(2)
(y=x\ y'=x'=1)
(3)
(y=x^{2}\ y'=left( x^2
ight) '=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}{2sqrt {x_{0}}})
知识点7 导数公式表
(y=fleft( x
ight) ,y'=f'left( x
ight))
$ y=c,y'=0(
) y=x^{n}left( nin N+
ight) ,y'=nx^{n-1}(
)y=x^{alpha }( alpha in Q,x>0), y'=alpha x^{alpha -1}(
)y= a^{x}(a > 0,a
eq 1),y'=a^{x}ln a(
)y=log a^{x}( a >0,a
eq 1),y'=dfrac {1}{xln 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
ight| ,y'=dfrac {1}{left| x
ight| } y= an 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}}}$:
极限:
“(dfrac {0}{0})”$$lim _{x
ightarrow 0}dfrac {sin x}{x}=1$$
"(dfrac {infty }{infty })"$$lim _{x
ightarrow 0}left( 1+dfrac {1}{x}
ight) ^{x}=e$$
或$$lim _{x
ightarrow 0}left( 1+x
ight) ^{dfrac {1}{x}}=e$$
例7
求导
(fleft( x
ight) =x^{5}+2x^{4}+x^{3},gleft( x
ight) =3^{x}+ln x,hleft( x
ight) =cos x+sin x)
(Downarrow)
(f'left( x
ight) =5x^{4}+8x^{3}+3x^{2},g'left( x
ight) =3^{x}ln 3+dfrac {1}{x}, h'left( x
ight) =-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[ fleft( x
ight) pm gleft( x
ight)
ight]'=f'left( x
ight) pm g'left( x
ight))
(2)函数的积的求导法则
(left[ fleft( x
ight) cdot gleft( x
ight)
ight] '=f'left( x
ight) cdot gleft( x
ight) +fleft( x
ight) cdot g'left( x
ight))
(3)导数商的求导法则
(left[ dfrac {fleft( x
ight) }{gleft( x
ight) }
ight] ^{'}=dfrac {f'left( x
ight) gleft( x
ight) -fleft( x
ight) g'left( x
ight) }{g^{2}left( x
ight) })
证明:
例8
求(fleft( x
ight) =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:
(fleft( x
ight) =na_{0}x^{n-1}+(n-1)a_{1}x^{n-2}+ldots +a_{n}-1)
例9
(y=xcdot sin x)
解:
(y'=x'sin x+xcdot left( sin x
ight) '\ =sin x+xcdot cos x)
例10
(y=sin 2x)
解:
(y=sin 2x=2sin xcos x\ y'=2(cos ^{2}x-sin ^{2}x )\ =2cos 2x)
例11
(y= an x=dfrac {sin x}{cos x})
解:
(y'=dfrac {left( sin x
ight) 'cos x-sin xleft( cos x
ight)' }{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
ight) >0),函数单调递增
(2)如果(a,b)内(f'left( x
ight) <0),函数单调递减
例13
y=x在(left( -infty ,+infty
ight))上的单调性
解:(y'=1 >0)
例14
(y=2x^{2}-2x+4)在(left( -infty ,+infty
ight))上的单调性
解:(y'=2x-2)
令(y' >0),即(x >1),在(left( 1,+infty
ight))上函数单调递增
令(y' <0),即(x <1),在(left( -infty ,1
ight))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}
ight) ,left( dfrac {4+sqrt {13}}{3},+infty
ight))上函数单调递增
令(y' <0),即在(left( dfrac {4-sqrt {13}}{3} ,dfrac {4+sqrt {13}}{3}
ight))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 ight)) | + | - | + | ||
(fleft( x ight)) | ( earrow) | 极大值 | (searrow) | 极小值 | ( earrow) |
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}
ight) ^{2}cdot dfrac {1}{n}+left( dfrac {2}{n}
ight) ^{2}cdot dfrac {1}{n}+ldots +left( dfrac {n-1}{n}
ight) ^{2}cdot dfrac {1}{n}=dfrac {1}{n^{3}})
(left[ 1^{2}+2^{2}+ldots +left( n-1
ight) ^{2}
ight] =dfrac {1}{n^{3}})
(dfrac {nleft( n-1)(2n-1
ight) }{6}=dfrac {1}{6}left( 1-dfrac {1}{n}
ight) left( 2-dfrac {1}{n}
ight))
(n
ightarrow infty)时
例12
(y=fleft( u
ight) ,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
ight) cdot u'left( x
ight))
(=acdot f'left( u
ight))
eg1
(mu =x^{-1},gleft( x
ight) =4x^{2}+3x)
解:
(gleft( u
ight) =4u^{2}+3u\ g'left( u
ight) =8u+3\ u'=-1cdot x^{-2}=-dfrac {1}{x^{2}})
eg2
(left[ sin left( 2x+dfrac {pi }{3}
ight)
ight]')
解:
(u=2x+dfrac {pi }{3} \y=sin u\ =cos ucdot 2\ =2cos left( 2x+dfrac {pi }{3}
ight))
eg3
(left[ left( 5x+3
ight) ^{5}
ight] ')
解:
(y=u,u=5x+3,u'=5)
(y'=f'left( u
ight) cdot u'left( x
ight) \ =5u^{4}cdot 5\ =25left( 5x+3
ight) ^{4})
知识点12 [a,b]n分
(a=x_{0} <x_{1} <x_{2} <ldots <x_{n}=b)
(x_{i}=max){(Delta xi)}
(fleft( x ight)):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(Fleft( x
ight) '=fleft( x
ight))且(fleft( x
ight))在([a,b])内可积
则
(int ^{b}_{a}fleft( x
ight) Delta x)
$ =Fleft( x
ight) int ^{b}_{a}(
) =Fleft( b
ight) -Fleft( a
ight) $
例17
(int ^{1}_{0}x^{2}Delta x=)
解:
(ecause left( dfrac {1}{3}x^{3}
ight) '=x^{2}\ herefore 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 ight) Delta x\ =left( dfrac {x^{3}}{3}+x ight) int ^{2}_{0})
例19
(int ^{pi }_{0}sin Delta x\ =-cos xint ^{pi }_{0})