u-ln(u)-ln(x)-ln(c)=0,求u怎么算
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![u-ln(u)-ln(x)-ln(c)=0,求u怎么算](/uploads/image/f/846607-31-7.jpg?t=u-ln%28u%29-ln%28x%29-ln%28c%29%3D0%2C%E6%B1%82u%E6%80%8E%E4%B9%88%E7%AE%97)
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^
(lnx)'=1/x,这是公式.
y'=f'(ln(x+√(a+x²)))·ln(x+√(a+x²))‘=f'(ln(x+√(a+x²)))·1/(x+√(a+x²))·(x+√(a+x
令u=secA,du=dsecA=sinA/(cosA)^2*dA∫du/(u^2-1)^(1/2)=∫sinAdA/(cosA)^2*tanA=∫dA/cosA=∫cosAdA/(1-sinA^2)
lim[ln(1+u)/u]=u→0lim[ln(1+u)^(1/u)]=u→0=lne=1
这是复合函数求导,把u^2-1看做整体,设u^2-1=y,则lny的导数为(1/y)*dy,在对u^2-1=y求导则dy=(2u)du,所以dx={2u/(u^2-1)}du
题目没有写完吧.再问:x^2+y^2+z^2=1,x,y,z>0
左边对u积分,右边对x积分∫du/(u^2-1)^(1/2)=ln[u+(u^2-1)^(1/2)]+C1∫dx/x=lnx+C2所以ln[u+(u^2-1)^(1/2)]=lnx+C题目是不是写错了
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
u=ln(xy+z)du=d[ln(xy+z)]/dx*dx+d[ln(xy+z)]/dy*dy+d[ln(xy+z)]/dz*dz=y/(xy+z)*dx+x/(xy+z)*dy+1/(xy+z)*
Fy(Y)=P(Ye^(-y))=1-P(x=0)
令a=lnxx=e^adx=e^ada原式=∫a²*e^ada=∫a²de^a=a²*e^a-∫e^ada²=a²*e^a-2∫ade^a=a
dy/dx=dy/du*du/dx+dy/dv*dv/dx=v*e^(x+y)+u*y/x=ln(xy)*e^(x+y)+e^(x+y)*y/x=e^(x+y)[ln(xy)+y/x]所以dy=e^(
结果是t的8次方
Y=-2ln(X)在X~(0,1)上是相互一对一的函数关系所以可以使用密度函数乘上导数的方法fy(y)=fx(x(y))*|dx/dy|=1|dx/dy|Y=-2ln(X)lnX=-0.5YX=e^(
美国设计,中国制造
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】