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(10x2+46y2)+4x,求z對(duì)x,y的所有二階偏導(dǎo)數(shù)。2.設(shè)z=xln(4x+38y),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(8-16x+3y)+33x2y的二階偏導(dǎo)數(shù)計(jì)算。4.求z(x,y)=exyln(30x2+14y2)的二階偏導(dǎo)數(shù)。參考答案1.設(shè)z=f(10x2+46y2)+4x,求z對(duì)x,y的所有二階偏導(dǎo)數(shù)。eq\f(?2z,?x2)=20f'(10x2+46y2)+202x2f''(10x2+46y2);eq\f(?2z,?y2)=92f'(10x2+46y2)+922y2f''(10x2+46y2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=4*10*46xyf''(10x2+46y2)。2.設(shè)z=xln(4x+38y),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(4*(4x+2*38y),(4x+38y)2);eq\f(?2z,?y2)=-eq\f(382x,(4x+38y)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=eq\f(382y,(4x+38y)2)。3.z=ln(8-16x+3y)+33x2y的二階偏導(dǎo)數(shù)計(jì)算.eq\f(?2z,?x2)=2*33y-eq\f(162,(8-16x+3y)2);eq\f(?2z,?y2)=-eq\f(32,(8-16x+3y)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=2*33x+eq\f(16*3,(8-16x+3y)2)。4.求z(x,y)=exyln(30x2+14y2)的二階偏導(dǎo)數(shù)eq\f(?2z,?x2)=z(eq\f(?z,?x))2+eq\f(2*30xyz(30x2+3*14y2),(30x2+14y2)2);eq\f(?2z,?y2)=z(eq\f(?z,?y))2+eq\f(2*14yz*(3*30x2+14y2),(30x2+14y2)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(30x2+14y2)+2*eq\f(302x?+142y?,(30x2+14y2)2)]。詳細(xì)步驟：1.設(shè)z=f(10x2+46y2)+4x,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'(10x2+46y2)*(2*10xdx+2*46ydy)+4dx,∴eq\f(?z,?x)=20xf'(10x2+46y2)+4;eq\f(?z,?y)=92yf'(10x2+46y2)。進(jìn)一步求二階偏導(dǎo)數(shù)，有：eq\f(?2z,?x2)=20f'(10x2+46y2)+20xf''(10x2+46y2)*2*10x=20f'(10x2+46y2)+202x2f''(10x2+46y2);eq\f(?2z,?y2)=92f'(10x2+46y2)+92yf''(10x2+46y2)*2*46y=92f'(10x2+46y2)+922y2f''(10x2+46y2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=20xf''(10x2+46y2)*2*46y=4*10*46xyf''(10x2+46y2)。2.設(shè)z=xln(4x+38y),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(4x+38y)dx+eq\f(x(4dx+38dy),(4x+38y))=[ln(4x+38y)+eq\f(4x,4x+38y)]dx+eq\f(38x,4x+38y)dyeq\f(?z,?x)=ln(4x+38y)+eq\f(4x,4x+38y),eq\f(?z,?y)=eq\f(38x,4x+38y)，進(jìn)一步求偏導(dǎo)數(shù)有：eq\f(?2z,?x2)=eq\f(4,4x+38y)+eq\f(4[(4x+38y)-4x],(4x+38y)2)=eq\f(4,4x+38y)+eq\f(4*38y,(4x+38y)2)=eq\f(4*(4x+2*38y),(4x+38y)2);eq\f(?2z,?y2)=-eq\f(38x*38,(4x+38y)2)=-eq\f(382x,(4x+38y)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=eq\f(38(4x+38y–4x),(4x+38y)2)=eq\f(382y,(4x+38y)2).3.z=ln(8-16x+3y)+33x2y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((8-16x+3y)',8-16x+3y)+2*33xydx+33x2dy=eq\f(-16dx+3dy,8-16x+3y)+2*33xydx+33x2dy=(2*33xy-eq\f(16,8-16x+3y))dx+(33x2+eq\f(3,8-16x+3y))dy;根據(jù)全微分，則一階偏導(dǎo)數(shù)有：eq\f(?z,?x)=2*33xy-eq\f(16,8-16x+3y);eq\f(?z,?y)=33x2+eq\f(3,8-16x+3y).再求二階偏導(dǎo)數(shù)如下：eq\f(?2z,?x2)=2*33y+eq\f(16*(-16),(8-16x+3y)2)=2*33y-eq\f(162,(8-16x+3y)2);eq\f(?2z,?y2)=0-eq\f(32,(8-16x+3y)2)=-eq\f(32,(8-16x+3y)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=2*33x-eq\f(16*(-3),(8-16x+3y)2)=2*33x+eq\f(16*3,(8-16x+3y)2).4.求z(x,y)=exyln(30x2+14y2)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(30x2+14y2)，eq\f(?z,?x)=z[yln(30x2+14y2)+eq\f(xy*2*30x,30x2+14y2)=z[yln(30x2+14y2)+eq\f(2*30x2y,30x2+14y2)],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(30x2+14y2)+eq\f(xy*2*14y,30x2+14y2)]=z[xln(30x2+14y2)+eq\f(2*14xy2,30x2+14y2)]。(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*30x,30x2+14y2)+eq\f(2*30[2xy*(30x2+14y2)-x2y*2*30x],(30x2+14y2)2)},=z(eq\f(?z,?x))2+2*30xyz[eq\f(1,30x2+14y2)+eq\f(2*14y2,(30x2+14y2)2)]=z(eq\f(?z,?x))2+eq\f(2*30xyz(30x2+3*14y2),(30x2+14y2)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*14xy,(30x2+14y2))+eq\f(2*14[2xy(30x2+14y2)-xy2*2*14y],(30x2+14y2)2)}=z(eq\f(?z,?y))2+z[eq\f(2*14xy,(30x2+14y2))+eq\f(2*14*2xy*30x2,(30x2+14y2)2)]=z(eq\f(?z,?y))2+2*14xyz[eq\f(1,30x2+14y2)+eq\f(2*30x2,(30x2+14y2)2)]=z(eq\f(?z,?y))2+eq\f(2*14yz*(3*30x2+14y2),(30x2+14y2)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(30x2+14y2)+eq\f(2*14xy2,30x2+14y2)][yln(30x2+14y2)+eq\f(2*30x2y,30x2+14y2)]+z{ln(30x2+14y2)+eq\f(y*2*14y,(30x2+14y2))+eq\f(2*30*x2[(30x2+14y2)-y*2*14y],(30x2+14y2)2)}=eq\f(?z,?y)*[yln(30x2+1