二阶偏导数z=x(e^y)
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/>d(sinxy-2z+e^z)=0dsinxy-d2z+de^z=0ycosxydx+xcosxydy-2dz+e^zdz=0ycosxydx+xcosxydy=2dz-e^zdz=(2-e^z)
Z=U*V则∂Z/∂U=V∂Z/∂V=UX=e^UsinV则∂X/∂U=e^UsinV=X∂X/∂V=e
数学之美团为你解答y'=-e^(-x)*sin(2x)+e^(-x)*2cos(2x)=e^(-x)*[2cos(2x)-sin(2x)]y''=-e^(-x)[2cos(2x)-sin(2x)]+e
两边对z微分e^zdz-d(xyz)=0=e^zdz-xydz-zd(xy)=e^zdz-xydz-zxdy-zydx所以,整理两边:(e^z-xy)dz=zxdy+zydx所以:dz=zx/(e^z
根据一阶全微分形式不变得dz=d(xf(x^y,e^xy)=f(x^y,e^xy)dx+xd(f(x^y,e^xy))=f(x^y,e^xy)dx+x[f1'd(x^y)+f2'(de^xy)]=f(
对x求导,e^z*z'(x)=yz+xyz'(x),z'(x)=yz/(e^z-xy)对y求导,e^z*z'(y)=xz+xyz'(y),z'(y)=xz/(e^z-xy)
δz/δx=f1·cosx+f2δ^2z/δxδy=cosx﹙f11+f12·e^y﹚+f21+f22e^y再问:大哥,能在详细点吗再答:δz/δx=f1·cosx+f2(把x当常数,把y当未知数求导
可以拆分成先对x的偏导数.再对y的偏导数,原函数是复合函数,可以令m=sinx,n=e^x-y&Z/&x=&Z/&m*&m/&x+&Z/&n*&n/&x符号太难找我就这么代替了,希望能让你看懂啊...
z=x^y,lnz=ylnx;(1/z)∂z/∂x=y/x,∂z/∂x=yz/x=yx^(y-1);(1/z)∂z/∂y=lnx
1.z'x=3x²y²z'y=2x³y2.z'x=4x³z'y=3y³3.z'x=ye^(xy)+2xyz'y=xe^(xy)+x²4.u'
先等会,十分钟再问:嗯嗯,谢谢再答:你确定括号里面是e-xy?再问:是e^(-xy)再答:哦再问:再答:图片发不过去再答:我告诉你怎么做吧再问:啊?QQ邮箱再问:可以吗再问:嗯嗯再问:62630868
偏Z比偏Y=xf(x+y,e^xsiny)+xy(f1'+f2'e^xcosy),偏Z比偏x=z=yf(x+y,e^xsiny)+xy(f1'+f2'e^xsiny).
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)
令u=xy,v=e^(x+y)Z'x=Z'u*U'x+Z'v*V'x=f'u*y+f'v*e^(x+y)Z'y=Z'u*U'y+Z'v*V'y=f'u*x+f'v*e^(x+y)
xyz=e^(x+y)两边求关于x的偏导数(把z当成常数)∂(xyz)/∂x=∂e^(x+y)/∂xz∂(xy)/∂x=e^(x
your answer here
F(x,y,z)=xy+e^xz-zlny-1.Fx=y+ze^xzFy=x-z/yFz=xe^xz-lnyz对x的偏导:-Fx/Fz=-(y+ze^xz)/(xe^xz-lny)z对y的偏导:-Fy
(2^sinx)cosxlg2e^arctan√z*(1/(1+z))*(1/(2√z))e^(-x^2)*(-2x)tanxsecx
二阶偏导数有四个Z''xx=(lin(x+y)+x/(x+y))'=1/(x+y)+y/(x+y)^2Z''yy=(x/(x+y))'=-x/(x+y)^2Z''yx=Z''xy=(x/(x+y))'