Question

求隐函数导数ln（xy）=e^(x+y)所确定的隐函数y=y(x)的导数dy/dx.

Answer 1

[ln(xy)]' = [e^(x+y)]'

(xy)'/(xy) = e^(x+y) * (x+y)'

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

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