u=ln(x^2 y^2 z^2),求du
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u'x=1/(x+y^2+z^3)u'y=2y/(x+y^2+z^3)u'z=3z^2/(x+y^2+z^3)du=u'xdx+u'ydy+u'zdz=1/(x+y^2+z^3)dx+2y/(x+y^
y'=f'(ln(x+√(a+x²)))·ln(x+√(a+x²))‘=f'(ln(x+√(a+x²)))·1/(x+√(a+x²))·(x+√(a+x
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)
题目不清楚,有两个变量,是求偏导还是全微分表达式?求偏导的话,将其中一个变量看做常数,按一元函数的方法求
δz/δx=1/(xy+x/y)*(y+1/y)=(y²+1)/(xy²+x)=1/xδ^2z/δxδy=δ(δz/δx)/δy=0
由柯西不等式(a^2+b^2+c^2)(x^2+y^2+z^2)>=(ax+by+cz)^2,得((1/√2)^2+(1/√3)^2+1)(2x^2+3y^2+z^2)>=(x+y+z)^22x^2+
这是求偏导数.偏u/偏x=fx'dx+fz'*偏z/偏x=fx'dx+fz'*x/[(x^2+y^2)^0.5],偏u/偏y=fy'dy+fz'*偏z/偏y=fy'dy+fz'*y/[(x^2+y^2
令x+y^2+z=t那么x+y^2+z=ln(x+y^2+z)^1/2可以转化为2t=lnt根据图象,s1=2t以及s2=lnt这两条曲线是不会相交的!所以2t=lnt没有实根所以x+y^2+z=t没
由柯西-黎曼条件:对u(x,y)=1/2ln(x^2+y^2)求x的偏导x/(x^2+y^2),对u(x,y)=1/2ln(x^2+y^2)求x的偏导y/(x^2+y^2),f'(z)=x/(x^2+
∂z/∂x=(1/(x²+y))(2x)=2x/(x²+y)∂²f/∂x∂y=∂[∂z
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^
σu/σx=(z+y)+x(σz/σx+0)=z+y+xcos(x+y)σ2u/σxσy=σz/σy+1-xsin(x+y)=cos(x+y)+1-xsin(x+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
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+
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)
对等式两边求全微分du=【1/(2x+3y+4z^2)】【2dx+3dy+8zdz】