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(47x2+45y2)+15x,求z對(duì)x,y的所有二階偏導(dǎo)數(shù)。2.設(shè)z=xln(50x+53y),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(25-27x+42y)+23x2y的二階偏導(dǎo)數(shù)計(jì)算。4.求z(x,y)=exyln(34x2+42y2)的二階偏導(dǎo)數(shù)。參考答案1.設(shè)z=f(47x2+45y2)+15x,求z對(duì)x,y的所有二階偏導(dǎo)數(shù)。eq\f(?2z,?x2)=94f'(47x2+45y2)+942x2f''(47x2+45y2);eq\f(?2z,?y2)=90f'(47x2+45y2)+902y2f''(47x2+45y2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=4*47*45xyf''(47x2+45y2)。2.設(shè)z=xln(50x+53y),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(50*(50x+2*53y),(50x+53y)2);eq\f(?2z,?y2)=-eq\f(532x,(50x+53y)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=eq\f(532y,(50x+53y)2)。3.z=ln(25-27x+42y)+23x2y的二階偏導(dǎo)數(shù)計(jì)算.eq\f(?2z,?x2)=2*23y-eq\f(272,(25-27x+42y)2);eq\f(?2z,?y2)=-eq\f(422,(25-27x+42y)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=2*23x+eq\f(27*42,(25-27x+42y)2)。4.求z(x,y)=exyln(34x2+42y2)的二階偏導(dǎo)數(shù)eq\f(?2z,?x2)=z(eq\f(?z,?x))2+eq\f(2*34xyz(34x2+3*42y2),(34x2+42y2)2);eq\f(?2z,?y2)=z(eq\f(?z,?y))2+eq\f(2*42yz*(3*34x2+42y2),(34x2+42y2)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(34x2+42y2)+2*eq\f(342x?+422y?,(34x2+42y2)2)]。詳細(xì)步驟：1.設(shè)z=f(47x2+45y2)+15x,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'(47x2+45y2)*(2*47xdx+2*45ydy)+15dx,∴eq\f(?z,?x)=94xf'(47x2+45y2)+15;eq\f(?z,?y)=90yf'(47x2+45y2)。進(jìn)一步求二階偏導(dǎo)數(shù)，有：eq\f(?2z,?x2)=94f'(47x2+45y2)+94xf''(47x2+45y2)*2*47x=94f'(47x2+45y2)+942x2f''(47x2+45y2);eq\f(?2z,?y2)=90f'(47x2+45y2)+90yf''(47x2+45y2)*2*45y=90f'(47x2+45y2)+902y2f''(47x2+45y2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=94xf''(47x2+45y2)*2*45y=4*47*45xyf''(47x2+45y2)。2.設(shè)z=xln(50x+53y),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(50x+53y)dx+eq\f(x(50dx+53dy),(50x+53y))=[ln(50x+53y)+eq\f(50x,50x+53y)]dx+eq\f(53x,50x+53y)dyeq\f(?z,?x)=ln(50x+53y)+eq\f(50x,50x+53y),eq\f(?z,?y)=eq\f(53x,50x+53y)，進(jìn)一步求偏導(dǎo)數(shù)有：eq\f(?2z,?x2)=eq\f(50,50x+53y)+eq\f(50[(50x+53y)-50x],(50x+53y)2)=eq\f(50,50x+53y)+eq\f(50*53y,(50x+53y)2)=eq\f(50*(50x+2*53y),(50x+53y)2);eq\f(?2z,?y2)=-eq\f(53x*53,(50x+53y)2)=-eq\f(532x,(50x+53y)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=eq\f(53(50x+53y–50x),(50x+53y)2)=eq\f(532y,(50x+53y)2).3.z=ln(25-27x+42y)+23x2y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((25-27x+42y)',25-27x+42y)+2*23xydx+23x2dy=eq\f(-27dx+42dy,25-27x+42y)+2*23xydx+23x2dy=(2*23xy-eq\f(27,25-27x+42y))dx+(23x2+eq\f(42,25-27x+42y))dy;根據(jù)全微分，則一階偏導(dǎo)數(shù)有：eq\f(?z,?x)=2*23xy-eq\f(27,25-27x+42y);eq\f(?z,?y)=23x2+eq\f(42,25-27x+42y).再求二階偏導(dǎo)數(shù)如下：eq\f(?2z,?x2)=2*23y+eq\f(27*(-27),(25-27x+42y)2)=2*23y-eq\f(272,(25-27x+42y)2);eq\f(?2z,?y2)=0-eq\f(422,(25-27x+42y)2)=-eq\f(422,(25-27x+42y)2);eq\f(?2z,?x?y)=eq\f(?2z,?y?x)=2*23x-eq\f(27*(-42),(25-27x+42y)2)=2*23x+eq\f(27*42,(25-27x+42y)2).4.求z(x,y)=exyln(34x2+42y2)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(34x2+42y2)，eq\f(?z,?x)=z[yln(34x2+42y2)+eq\f(xy*2*34x,34x2+42y2)=z[yln(34x2+42y2)+eq\f(2*34x2y,34x2+42y2)],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(34x2+42y2)+eq\f(xy*2*42y,34x2+42y2)]=z[xln(34x2+42y2)+eq\f(2*42xy2,34x2+42y2)]。(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*34x,34x2+42y2)+eq\f(2*34[2xy*(34x2+42y2)-x2y*2*34x],(34x2+42y2)2)},=z(eq\f(?z,?x))2+2*34xyz[eq\f(1,34x2+42y2)+eq\f(2*42y2,(34x2+42y2)2)]=z(eq\f(?z,?x))2+eq\f(2*34xyz(34x2+3*42y2),(34x2+42y2)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*42xy,(34x2+42y2))+eq\f(2*42[2xy(34x2+42y2)-xy2*2*42y],(34x2+42y2)2)}=z(eq\f(?z,?y))2+z[eq\f(2*42xy,(34x2+42y2))+eq\f(2*42*2xy*34x2,(34x2+42y2)2)]=z(eq\f(?z,?y))2+2*42xyz[eq\f(1,34x2+42y2)+eq\f(2*34x2,(34x2+42y2)2)]=z(eq\f(?z,?y))2+eq\f(2*42yz*(3*34x2+42y2),(34x2+42y2)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(34x2+42y2)+eq\f(2*42xy2,34x2+42y2)][yln(34x2+42y2)+eq\f(2*34x2y,34x2+42y2)]+z{ln(34x2+42y2)+eq\f(y*2*42y,(34x2+42y2))+eq\f(2*34*x2[(34x2+42y2)-y*2*42y],(34x2+42y2)2)}=eq\f(?z,?y)*[yln(34