implicit differentation means you assume y is a function of x: Dx(xe^y) -Dx(10x) +Dx(3y) = Dx(0)
Dx(xe^y) use the product rule: (dx/dx)e^y + (dy/dx)xe^y = e^y + y'xe^y Dx(10x) = (dx/dx)(10) = 10 Dx(3y) = 3(dy/dx) = 3y' Dx(0) = (d0/dx)0 = 0
Get all your y' on one side and everything else to the other: e^y + y'xe^y -10 +3y' = 0 y'xe^y + 3y' = 10 -e^y y'(xe^y +3) = 10 -e^y (dy/dx) =y' =(10-e^y) / (xe^y +3)
