x²y+xy²=6微分
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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
隐函数求导设z=x²y²-cos(xy)dy/dx=-(δz/δx)/(δz/δy)=-(2xy²+ysin(xy))/(2x²y+xsin(xy))=-y/x
-sin(xy)[ydx+xdy]=2xy^2*dx+x^2*2ydy-sin(xy)ydx-sin(xy)xdy=2xy^2*dx+2x^2*ydy-2x^2*ydy-sin(xy)xdy=2xy^
两边对x求导:-(y+xy')sin(xy)=2xy^2+2x^2yy'解得:y'=-[ysin(xy)+2xy^2]/[2x^2y+xsin(xy)]所以dy=-[ysin(xy)+2xy^2]/[
设u=xy,v=lnx+g(xy),则x(∂z/∂x)-y(∂z/∂y)=∂f/∂v.原因如下:dz=(∂f/
令u=2x^2-y^2,v=xy然后链导法则!再问:请您把详细过程给我好吗?再答:偏导数符号打不上去啊du=(4xfu+yfv)dx+(-2yfu+xfv)dy其中fu、fv是偏导数符号
dy/dx==-(2e^x)/x^3+(e^x)/x^2我用数学软件算的,绝对不会错.
如果对x求导,则ln|x|=yln|y|,1/x=y'/y+yy'/y=y'/y+y',.对数求导法.如果对y求导,则ln|x|=yln|y|,x'/x=ln|y|+y/y,x'=y^y(1+ln|y
y'+xy=0的通解y.=Ce^(-x).特解y=x^2-2x.通解y=Ce^(-x)+x^2-2x.再问:不好意思啊,之前一直在忙别的。没有及时回复,首先谢谢你的回答。但是我觉得你的回答有点问题。‘
∵x²dy+(y-2xy)dx=0==>x²dy/dx+(1-2x)y=0==>dy/y+(1/x²-2/x)dx=0==>ln|y|-1/x-2ln|x|=ln|C|(
xy'+y+sin(πy)πy'=0y'=-y/[x+πsin(πy)]
dx+dy-3^(xy)•ln3(dx•y+dy•x)=0dx+dy-ln3•3^(xy)•(dx•y+dy•x)=
dy/dx=3xy=xy^2dy/(3y+y^2)=xdx1/3*ln(y/3+y)=1/2*x^2+c1ln(y/3+y)=3/2*x^2+c2(c2=3c1)y/3+y=e^(3/2*x^2+c2
两边即对数得:lnz=xy*ln(lnu),不妨记u=x^2+y^2z'x/z=yln(lnu)+2x^2y/lnu,z'x=z[yln(lnu)+2x^2y/lnu]z'y/z=xln(lnu)+2
dz=(y+y/(X^2))dx+(x-1/x)dy,
隐函数求导的问题,由F(x,y)=0确定隐函数y=y(x),对方程两边求导时,其中含y的式子要始终注意y是一个x的函数,如(siny)'=cosy*y',(e^siny)'=e^siny*(siny)
az/ax=y^3-2xy^6az/ay=3xy^2-6x^2y^5所以dz=(y^3-2xy^6)dx+(3xy^2-6x^2y^5)dy在点(1,1)的全微分为dz=-dx-3dy
这个题要用复合函数求导公式算:记m=xy,n=x/y,那么有:∂m/∂x=y∂m/∂y=x∂n/∂x=1/y∂n/&
求二元函数全微分z=f[x²-y²,e^(xy)]设z=f(u,v),u=x²-y²,v=e^(xy)则dz=(∂f/∂u)du+(
dz=(y+1/y)dx+(x-x/y^2)dy