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Find implict differentiation √xy = 2 + x2y
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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.
Thank You So much for your timely help.
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