If x2ey+2xyex+23=0 then dydx=
2xey-x+2y(x+1)
2xex-y-3y(x+1)
2xey-x-2y(x+1)x(xey-x+2)
2xey-x-y(x+1)
Explanation for the correct option
Let y=x2ey+2xyex+23=0, then
2xey+x2eydydx+2dxyexdx+xyex=0⇒2xey+x2eydydx+2exy+xdydx+xy=0⇒2xey+x2eydydx+2yex+2xexdydx+2xy2ex=0⇒dydx=2xey-x-2y1+xxxey-x+2
Hence, option C is correct.