經(jīng)濟數(shù)學(導數(shù)與微分習題及答案)

1、1第三章第三章函數(shù)的導數(shù)與微分函數(shù)的導數(shù)與微分習題習題3－11.根據(jù)定義求下列函數(shù)的導數(shù):(1)xy1?(2)xycos?(3)baxy??(ab為常數(shù))(4)xy?解(1)因為00()()limlimxxyfxxfxyxx????????????=xxxxx??????11lim0=01lim()xxxx?????=21x?所以21yx???.(2)因為00cos()coslimlimxxyxxxyxx????????????02si

2、n()sin22limsinxxxxxx??????????所以sinyx???(3)因為00[()][]limlimxxyaxxbaxbyxx??????????????=xxax????0lim=a所以ya??(4)因為00limlimxxyxxxyxx????????????=)(lim0xxxxxx???????011lim2xxxxx???????所以12yx??.2.下列各題中假定)(0xf存在按照導數(shù)的定義觀察下列極限指出

3、A表示什么(1)Axxfxxfx???????)()(lim000(2)Axxfx??)(lim0(其中0)0(?f且)0(f)存在)(3)Axftxfx???)0()(lim0(其中)0(f存在)35.討論下列函數(shù)在0?x處的連續(xù)性和可導性：(1)21sin0xyx??????00??xx(2)cosyx?(3)2xyx?????00??xx解(1)因為0()(0)(0)lim0xfxffx????2001sin1limlimsin0

4、xxxxxxx?????所以函數(shù)21sin0xyx??????00??xx在0?x處可導，從而也連續(xù).(2)因為0()(0)(0)lim0xfxffx????0coscos0limxxx???2002sincos12limlim0xxxxxx???????所以函數(shù)cosyx?在x=0處可導，從而也連續(xù).(3)因為200lim()lim0(0)xxfxxf???????00lim()lim()0(0)xxfxxf????????所以函數(shù))

5、(xf在0?x處連續(xù).又因為200()(0)0(0)limlim000xxfxfxfxx????????????00()(0)0(0)limlim100xxfxfxfxx??????????????(0)(0)ff???故(0)f不存在即函數(shù))(xf在0?x不可導.6.設(shè)函數(shù)21()1xxfxaxbx???????，為使函數(shù)f(x)在x=1處連續(xù)且可導，ab應取什么值？解由題意，有11lim()lim()(1)(1)(1)xxfxfxf