4道三元函數(shù)的二階偏導(dǎo)數(shù)計(jì)算練習(xí)題及參考答案習(xí)題練習(xí)1.設(shè)z=f(52x2+24y2)+48x,求z對(duì)x,y的所有二階偏導(dǎo)數(shù)。2.設(shè)z=xln(35x+37y),求z的二階偏導(dǎo)數(shù)eq\f(?2z,?x2)、eq\f(?2z,?y2)、eq\f(?2z,?x?y)。3.z=ln(62-22x+19y)+58x2y的二階偏導(dǎo)數(shù)計(jì)算。4.求z(x,y)=exyln(26x2+4y2)的二階偏導(dǎo)數(shù)。參考答案1.設(shè)z=f(52x2+24y2)+48x,求z對(duì)x,y的所有二階偏導(dǎo)數(shù)。eq\f(?2z,?x2)=104f'(52x2+24y2)+1042x2f''(52x2+24y2);eq\f(?2z,?y2)=48f'(52x2+24y2)+482y2f''(52x2+24y2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=4*52*24xyf''(52x2+24y2)。2.設(shè)z=xln(35x+37y),求z的二階偏導(dǎo)數(shù)eq\f(?2z,?x2)、eq\f(?2z,?y2)、eq\f(?2z,?x?y).eq\f(?2z,?x2)=eq\f(35*(35x+2*37y),(35x+37y)2);eq\f(?2z,?y2)=-eq\f(372x,(35x+37y)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=eq\f(372y,(35x+37y)2)。3.z=ln(62-22x+19y)+58x2y的二階偏導(dǎo)數(shù)計(jì)算.eq\f(?2z,?x2)=2*58y-eq\f(222,(62-22x+19y)2);eq\f(?2z,?y2)=-eq\f(192,(62-22x+19y)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=2*58x+eq\f(22*19,(62-22x+19y)2)。4.求z(x,y)=exyln(26x2+4y2)的二階偏導(dǎo)數(shù)eq\f(?2z,?x2)=z(eq\f(?z,?x))2+eq\f(2*26xyz(26x2+3*4y2),(26x2+4y2)2);eq\f(?2z,?y2)=z(eq\f(?z,?y))2+eq\f(2*4yz*(3*26x2+4y2),(26x2+4y2)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=eq\f(1,z)*eq\f(?z,?y)*eq\f(?z,?x)+z[ln(26x2+4y2)+2*eq\f(262x?+42y?,(26x2+4y2)2)]。詳細(xì)步驟：1.設(shè)z=f(52x2+24y2)+48x,求z對(duì)x,y的所有二階偏導(dǎo)數(shù)。解：首先求一階偏導(dǎo)數(shù)，使用全微分法有：∵dz=f'(52x2+24y2)*(2*52xdx+2*24ydy)+48dx,∴eq\f(?z,?x)=104xf'(52x2+24y2)+48;eq\f(?z,?y)=48yf'(52x2+24y2)。進(jìn)一步求二階偏導(dǎo)數(shù)，有：eq\f(?2z,?x2)=104f'(52x2+24y2)+104xf''(52x2+24y2)*2*52x=104f'(52x2+24y2)+1042x2f''(52x2+24y2);eq\f(?2z,?y2)=48f'(52x2+24y2)+48yf''(52x2+24y2)*2*24y=48f'(52x2+24y2)+482y2f''(52x2+24y2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=104xf''(52x2+24y2)*2*24y=4*52*24xyf''(52x2+24y2)。2.設(shè)z=xln(35x+37y),求z的二階偏導(dǎo)數(shù)eq\f(?2z,?x2)、eq\f(?2z,?y2)、eq\f(?2z,?x?y)。解：先使用全微分求出一階偏導(dǎo)數(shù)：dz=ln(35x+37y)dx+eq\f(x(35dx+37dy),(35x+37y))=[ln(35x+37y)+eq\f(35x,35x+37y)]dx+eq\f(37x,35x+37y)dyeq\f(?z,?x)=ln(35x+37y)+eq\f(35x,35x+37y),eq\f(?z,?y)=eq\f(37x,35x+37y)，進(jìn)一步求偏導(dǎo)數(shù)有：eq\f(?2z,?x2)=eq\f(35,35x+37y)+eq\f(35[(35x+37y)-35x],(35x+37y)2)=eq\f(35,35x+37y)+eq\f(35*37y,(35x+37y)2)=eq\f(35*(35x+2*37y),(35x+37y)2);eq\f(?2z,?y2)=-eq\f(37x*37,(35x+37y)2)=-eq\f(372x,(35x+37y)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=eq\f(37(35x+37y–35x),(35x+37y)2)=eq\f(372y,(35x+37y)2).3.z=ln(62-22x+19y)+58x2y的二階偏導(dǎo)數(shù)計(jì)算解：先使用全微分求出一階偏導(dǎo)數(shù)：dz=eq\f((62-22x+19y)',62-22x+19y)+2*58xydx+58x2dy=eq\f(-22dx+19dy,62-22x+19y)+2*58xydx+58x2dy=(2*58xy-eq\f(22,62-22x+19y))dx+(58x2+eq\f(19,62-22x+19y))dy;根據(jù)全微分，則一階偏導(dǎo)數(shù)有：eq\f(?z,?x)=2*58xy-eq\f(22,62-22x+19y);eq\f(?z,?y)=58x2+eq\f(19,62-22x+19y).再求二階偏導(dǎo)數(shù)如下：eq\f(?2z,?x2)=2*58y+eq\f(22*(-22),(62-22x+19y)2)=2*58y-eq\f(222,(62-22x+19y)2);eq\f(?2z,?y2)=0-eq\f(192,(62-22x+19y)2)=-eq\f(192,(62-22x+19y)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=2*58x-eq\f(22*(-19),(62-22x+19y)2)=2*58x+eq\f(22*19,(62-22x+19y)2).4.求z(x,y)=exyln(26x2+4y2)的二階偏導(dǎo)數(shù)(1)計(jì)算一階偏導(dǎo)數(shù)1)先求對(duì)x的偏導(dǎo)數(shù)，此時(shí)將y看成常數(shù)，有：z=exyln(26x2+4y2)，eq\f(?z,?x)=z[yln(26x2+4y2)+eq\f(xy*2*26x,26x2+4y2)=z[yln(26x2+4y2)+eq\f(2*26x2y,26x2+4y2)],2)再求對(duì)y的偏導(dǎo)數(shù)，此時(shí)將x看成常數(shù)，有：eq\f(?z,?y)=z[xln(26x2+4y2)+eq\f(xy*2*4y,26x2+4y2)]=z[xln(26x2+4y2)+eq\f(2*4xy2,26x2+4y2)]。(2)對(duì)eq\f(?z,?x)再次對(duì)x求導(dǎo)，有：eq\f(?2z,?x2)=z*(eq\f(?z,?x))2+z{eq\f(y*2*26x,26x2+4y2)+eq\f(2*26[2xy*(26x2+4y2)-x2y*2*26x],(26x2+4y2)2)},=z(eq\f(?z,?x))2+2*26xyz[eq\f(1,26x2+4y2)+eq\f(2*4y2,(26x2+4y2)2)]=z(eq\f(?z,?x))2+eq\f(2*26xyz(26x2+3*4y2),(26x2+4y2)2)。(3)對(duì)eq\f(?z,?y)再次對(duì)y求導(dǎo)，有：eq\f(?2z,?y2)=z(eq\f(?z,?y))2+z{eq\f(2*4xy,(26x2+4y2))+eq\f(2*4[2xy(26x2+4y2)-xy2*2*4y],(26x2+4y2)2)}=z(eq\f(?z,?y))2+z[eq\f(2*4xy,(26x2+4y2))+eq\f(2*4*2xy*26x2,(26x2+4y2)2)]=z(eq\f(?z,?y))2+2*4xyz[eq\f(1,26x2+4y2)+eq\f(2*26x2,(26x2+4y2)2)]=z(eq\f(?z,?y))2+eq\f(2*4yz*(3*26x2+4y2),(26x2+4y2)2)。(4)對(duì)eq\f(?z,?x)再次對(duì)y求導(dǎo)，或eq\f(?z,?y)再次對(duì)x求導(dǎo)有：eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=z[xln(26x2+4y2)+eq\f(2*4xy2,26x2+4y2)][yln(26x2+4y2)+eq\f(2*26x2y,26x2+4y2)]+z{ln(26x2+4y2)+eq\f(y*2*4y,(26x2+4y2))+eq\f(2*26*x2[(26x2+4y2)-y*2*4y],(26x2+4y2)2)}=eq\f(?z,?y)*