利用復(fù)合函數(shù)公式求導(dǎo)數(shù)

1、11.1.利用復(fù)合函數(shù)公式求導(dǎo)數(shù)利用復(fù)合函數(shù)公式求導(dǎo)數(shù)(1)(1)??xe?(2)(2)??7xe?(3)(3)??cos4x(4)(4)??2xe?(5)(5)??sinln3x(6)(6)??2sin5x2.2.求導(dǎo)數(shù)求導(dǎo)數(shù)(1)(1)設(shè)函數(shù)設(shè)函數(shù)則=______________.=______________.2()(1)sinfxxx??)0(f(2)(2)函數(shù)函數(shù)的導(dǎo)數(shù)的導(dǎo)數(shù)=__________________.=_____

2、_____________.xey?y(3)(3)設(shè)函數(shù)設(shè)函數(shù)則=_____________________=_____________________.3sin33xyx???dxdy(4)(4)設(shè)函數(shù)設(shè)函數(shù)求.()ln1cosxfxxx???)(?f(1)(1).1cos2??xxy(3)(3)xy1tan2?(4)(4).22cosln1exxy????(5)(5)設(shè)則=______________.=______________

3、.xxxfarctan)1()(2??)0(f4.4.下列函數(shù)在下列函數(shù)在處是否可導(dǎo)？若可導(dǎo)，求出在處是否可導(dǎo)？若可導(dǎo)，求出在處的導(dǎo)數(shù)值處的導(dǎo)數(shù)值.0x?0x?3(2)(2)(3)(3)12xex233ln3xx?(4)(4)221cossin11cossin111()()(1cos)(1cos)2xxxfxfxx???????????????????3(1)3(1).(2)(2)222sin(1)2cos(1)xxxxx????21x

4、xee?(3)(3)11tantan222211ln22ln2()211coscosxxyxxxx?????(4)(4).222sintancos211xxxyxxxx???????(5)(5)設(shè)221()2arctan(1)2arctan1(1)fxxxxxxx??????(0)1f?4(1)4(1)22000(0)lim(2)0(0)lim1(0)01(0)xxxfxxfeeff?????????????????所以在所以在處不連續(xù)

5、，從而在處不連續(xù)，從而在處不可導(dǎo)處不可導(dǎo).0x?0x?(2)(2)00(0)lim0(0)limln(1)ln(10)0(0)ln(10)0xxfxfxf????????????????三者相等，故在三者相等，故在處連續(xù)處連續(xù).0x?0000(0)lim()lim111(0)lim(ln(1))lim11xxxxfxfxx??????????????????(0)(0)1.(0)1fff?????所以在x=0處可導(dǎo)，且.5.5.（用對(duì)數(shù)