y=ln的平方(1-x)的微分-搜答案-神马作文网

Z=（1/2）ln(1+x²+y²)dz=(1/2)2x/(1+x²+y²)dx+(1/2)2y/(1+x²+y²)dy=x/(1+x&su

2X/(X^2+1)的零点在X=0所以X=0是极值,为0(一眼其实就能看出来)

第一题,这是个隐函数,两边对x求导得：2y'-1=(1-y')*ln(x-y)+(x-y)*(1-y')/(x-y)=(1-y')*ln(x-y)+(1-y')所以[3+ln(x-y)]y'=ln(x

等式两边同时求导得:2y*y'+y'/y=4*x^3-->y'=4y*x^3/(2y^2+1)y'=dy/dx-->dy=y'*dx=dx*4y*x^3/(2y^2+1)

F(x,y)=x+lny-y=0dF(x,y)=0=(∂F(x,y)dx/∂x)+(∂F(x,y)dy/∂y)dy/dx=-(∂F(x,y)

全微分,分别对x和y求微分,再综合起来

dz=[-3ysin3xy+1/(1+x+y)]dx+[-3xsin3xy+1/(1+x+y)]dy

z偏x=-sin3xy*3y+1/(x+y+1)z偏y=-sin3xy*3x+1/(x+y+1)dz=[-sin3xy*3y+1/(x+y+1)]dx+[sin3xy*3x+1/(x+y+1)]dy

再答：我的回答你还满意吗?如有疑问请继续追问我

这是复合函数的求导：y=√u,u=lnv,v=3x^2则y'=1/(2√u)*u'=1/(2√u)*1/v*v'=1/(2√u)*1/v*6x=1/(2√u)*1/(3x^2)*6x=1/(x√u)=

dy=1/(x²+1)*d(x²+1)=1/(x²+1)*2xdx=2xdx/(x²+1)

dz(x=0,y=-1)=-2dy 详解如图

解y=ln²(1-2x)y'=dy/dx=[ln²(1-2x)]'=2ln(1-2x)[ln(1-2x)]'(1-2x)'=2ln(1-2x)[1/(1-2x)(-2)=[-4ln

y=[ln(1-x)^2]^2y'=2[ln(1-x)^2]*[ln(1-x)^2]'=2[ln(1-x)^2]*[2ln(1-x)]'=2[ln(1-x)^2]*2*1/(1-x)=4*[ln(1-

答：y=sin²x求导：y'=dy/dx=2sinxcosx=sin2x所以：微分dy=(sin2x)dx

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】

点击放大,荧屏放大再放大：