## anonymous 5 years ago Find implict differentiation √xy = 2 + x2y

1. anonymous

I'm assuming you want dy/dx and that the y is not under the square root and the x is squared. $1/2(x^{-1/2})y+x ^{1/2}y'=0+2xy+x^2y'$ $x ^{1/2}y'-x^2y'=2xy-y/(2x ^{1/2})$ $y'(x ^{1/2}-x^2)=2xy-y/(2x ^{1/2})$ $y'=(2xy-0.5x ^{-0.5}y)/(x ^{0.5}-x^2)$ $y'=[(0.5x ^{-0.5}(4x ^{1.5}y-y))/(x ^{-0.5}(x-x^2.5)]$ $y'=(4x \sqrt{x}y-y)/(2x-2x^2\sqrt{x})$ Sorry, that took a while. I put it in the neatest form I think I could... though you could factor out a 2x on the bottom.

2. anonymous

Thank You So much for your timely help.