Question

设y=y(x)由方程y=1-ln(x+y)+e^y所确定求y的导数

Answer 1

y=1-ln(x+y)+e^y

y'=0-1/(x+y) *(x+y)' +e^y*y'

y'=-(1+y')/(x+y)+e^yy'

y'+y'/(x+y)-e^yy'=-1/(x+y)

y'[1+1/(x+y)-e^y]=-1/(x+y)

y'[(x+y)+1-(x+y)e^y]=-1

y'=1/[(x+y)e^y -(x+y)-1]