设z=ln根号x^2 y^2全微分-搜答案-大师作文网

∵y＝ln［x＋√（x^2＋a^2）］,∴e^y＝x＋√（x^2＋a^2）,∴（e^y－x）^2＝x^2＋a^2,∴2（e^y－x）（e^y－x）′＝2x,∴［x＋√（x^2＋a^2）－x］［（e^y

z=x/ln(y/2)z′（x）=1/ln(y/2)z′（y）=-x/ln(y/2)^2*(1/(y/2))*1/2=-2x/(y*ln(y/2）^2)

方程x^2－z^2+lny-lnz＝0两端对x求导得2x-2zz'x-z'x/z=0z'x=2x/(2z+1/z)两端对y求导得-2zz'y+1/y-z'y/z=0z'y=1/[y(2z+1/z)]因

∂z/∂x把y看成常数所以1+0+∂z/∂x-2/[2√(xyz)]*y*(1*z+x*∂z/∂x)=01+∂z/&

δz=2xδx/(x^2*y^2)+2yδy/(x^2*y^2)代入求证的式子左边就知道了,等于0

z=(1/2)ln(x^2+y^2)az/ax=x/(x^2+y^2)(az/ax)^2=x^2/(x^2+y^2)^2az/ay=y/(x^2+y^2)(az/ay)^2=y^2/(x^2+y^2)

设z=ln(x^2+y),求

∂z/∂x=(1/(x²+y))(2x)=2x/(x²+y)∂²f/∂x∂y=∂[∂z

∂z/∂x=2x/(1+x^2+y^2)∂z/∂y=2y/(1+x^2+y^2)dz=∂z/∂xdx+∂z/W

z=ln[x+a^(-y^2)],以下'表示对y求偏导,z'=[a^(-y^2)]'/[x+a^(-y^2)]=(-y^2)'a^(-y^2)lna/[x+a^(-y^2)],z'=-2ya^(-y^

ux=2x/(x^2+y^2+z^2)uy=2y/(x^2+y^2+z^2)uz=2z/(x^2+y^2+z^2)故du=uxdx+uydy+uzdz=2x/(x^2+y^2+z^2)dx+2y/(x

z=(x^2)*ln(2xy),Zx=(2x)ln(2xy)+(x^2)/2xy*(2xy)'=(2x)ln(2xy)+xZxx=2ln(2xy)+(2x)/2xy*(2xy)'+1=2ln(2xy)

设z=ln(x^z×y^x),求dz

z=lnx^z+lny^x=zlnx+xlnyz=xlny/(1-lnx)先关于x求偏导,把y看做常数,再对y求偏导,把x看做常数dz=0dx+x/y(1-lnx)dy（此处省略了一些计算过程,）dz

两边取e的指数：e^(x+y²+z)=(x+y²+z)/2对x求导：[e^(x+y²+z)]*(1+ðz/ðx)=(1+ðz/ðx

应该是∂z/∂x吧!令u=x+y^2+z=>du/dx=1+dz/dxu=lnu^(1/2)=1/2*lnudu/dx=1/2*1/u*du/dx=>du/dx=u/(1/2+

y-x^2>01-y-x>=0所以x^2

u'x=2x/(x^2+y^2+z^2)u'y=2y/(x^2+y^2+z^2)u'z=2z/(x^2+y^2+z^2)du=2xdx/(x^2+y^2+z^2)+2ydy/(x^2+y^2+z^2)

z=1/2*ln(x^2+y^2+4)Z'x=1/2*1/(x^2+y^2+4)*(2x)=x/(x^2+y^2+4)Z'y=1/2*1/(x^2+y^2+4)*(2y)=y/(x^2+y^2+4)所

对等式两边求全微分du=【1/(2x+3y+4z^2)】【2dx+3dy+8zdz】

设z=ln(x+y),则dz=

dz=dx/(x+y)+dy/(x+y)