y等于e右上角是x减1ncosx,求dy
设y=Ex -1ncosx,求dy.
求微分方程的通解 {[e^(x+y)]-e^x}dx+{[e^(x+y)]+ey}dy=0 答案是(e^x+1)(e^y
设y等于Y括号X是由方程e的x方减e的y方等于sin括号xy所确定的隐函数,求微分dy
求dy/dx y=e^(2x+1)
y=sin(e∧x+1),求dy
求(x-e^-y)dy/dx=1通解
设y=ln(1+e^-x)求dy
设y=e∧1/x+sin2x,求dy
设y=[e^x+e^(-x)]^2,求dy
求由方程y=1-xe右上角y所确定的隐函数y=y(X)的导数dy/dx=( )怎么填
设e^(x+y)+cos(xy)=0确定y是x的函数求dy
[e^(x+y)-e^x]dx+[e^(x+y)-e^y]dy=0求通解