How do you find #dy/dx# if #x+y=e^(xy)#?

Question

4589 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-06T20:38:17

#dy/dx = -(y^2+xy-1)/(x^2+xy-1)#

Differentiate both sides of the equation with respect to #x#:

#d/dx(x+y) = d/dx e^(xy)#

#1+ y' = (y+xy')e^(xy)#

substitute #e^(xy)# from the original equation:

#1+ y' = (y+xy')(x+y)#

#1+ y' = xy +x^2y'+y^2+xyy'#

Solve for #y'#

#y'(1-xy-x^2) =y^2+xy-1#

#y' = -(y^2+xy-1)/(x^2+xy-1)#