Question

if log (x+y) = x+y prove that dy/dx= - 1

Answer 1

log (x+y) =x+y
d/dx log (x+y) = d/dx x+y
d/dx log (x+y) d/dx (x+y) d/dx x + d/dx y
1/(x+y) (1+dy/dx ) = 1+ dy/dx
1 (x+y) + 1/(x+y) dy/dx = 1 +dy/dx
[1 /(x+y) -1 ] = dy/dx - 1/ (x+y) dy/dx
[1/(x+y) - 1 ] = -dy/dx ( 1/ (x+y) -1)
1 = - dy/dx
dy/dx = - 1