求函数z=xy的条件下x y=1下的极值
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我来试试吧...z=e^xy*cos(x+y)Z'x=ye^xycos(x+y)-e^xysin(x+y)Z'y=xe^xycos(x+y)-e^xysin(x+y)故dZ=[ye^xycos(x+y
e^y-e^x=xy两边求导,得e^y*y'-e^x=y+xy'(e^y-x)y'=(e^x+y)所以y'=(e^x+y)/(e^y-x)x=0时,e^y-e^0=0,则e^y=1,则y=0所以y'(
Z=f'x(x,y)=xy*[x^(xy-1)]*yZ=f'y(x,y)=xy*[x^(xy-1)]*x再问:答案是Z=f'x(x,y)=yx^xy(lnx+1),Z=f'y(x,y)=x^(xy+1
首先z'(x)=x*(a-x-2*y)=0z'(y)=y(a-y-2*x)=0计算得到四组解(0,0)(a,0)(0,a)(a/3,a/3)1.(0,0)时,f''xx=0,f''xy=a,f''yy
一阶dz/dx=ycosxydz/dy=xcosxy二阶d^2z/dx^2=y^2cosxyd^2z/dy^2=x^2cosxy还有混合导数相等就写一个了=cosxy-xcosy
z=xy+x/y对x的偏导数=y+1/y对y的偏导数=x-x/y^2
Z=e^xy在x处的导函数为ye^(xy)在y处的导函数为xe^(xy)dz=ye^(xy)dx+xe^(xy)dy=2e^2dx+e^2dy
x+y=1.===>y=1-x.===>z=xy=x(1-x)=-x^2+x.===>z=-x^2+x=-[x-(1/2)]^2+(1/4).===>当x=y=1/2时,zmax=1/4.
x+y=1=>y=1-xz=xy=x(1-x)=x-x^2对x求导z'=1-2x令z'=0=>1-2x=0=>x=0.5所以,x=y=0.5时z有是大值0.25再问:嗯。thankyou
z=xy=x(1-x)=-x^2+x=-(x-1/2)^2+1/4,z最大为1/4也可以用求导的方法:对z=-x^2+x求导并令其等于0得:-2x+1=0,x=1/2时,z去极大值并是最大值1/4
全微分啊dz=(1+xy)^x[ln(1+xy)+xy/(1+xy)]dx+(1+xy)^xx^2/(1+xy)dy
Zx=ycos(xy)-2ycos(xy)sin(xy)=ycos(xy)-ysin(2xy)Zy=xcos(xy)-xsin(2xy)
lnz=yln(1+xy)z=e^{yln(1+xy)}dz/dy=e^[yln(1+xy)]{ln(1+xy)+xy/(1+xy)}dz/dx=e^[yln(1+xy)]{y^2/(1+xy)}
令u=xy,则z对x的偏导就变为(dz/du)*(偏u/偏x),然后按这样的顺序算就行了,同理,对y也一样,不知道这样说你明不明白
z=y+cosx+x再问:偏导数,不是导数再答:这不就是偏导数吗再问:哦,有全过程吗,谢谢再答:ðz/ðx=y+cosxðz/ðy=x
x+y=1=(x+y)^2=x^2+y^2+2xy≥4xy(xy>0)z=xy=xy/1=xy/(x+y)^2≤1/4,这个是极大值,
x=-2:0.1:2;y=x;[x,y]=meshgrid(x,y);z=x.*y;surf(x,y,z);grid on;xlabel('x.axis');ylabel(&
dz=(y+1/y)dx+(x-x/y^2)dy
再问:就是这个吗?再答:是的。如还有不懂请追问,懂了请采纳。再问:还有这三题