First, write it as (xy)^{1/2}=x-2y(xy)12=x−2y or x^{1/2}y^{1/2}=x-2yx12y12=x−2y.
Next, differentiate both sides with respect to xx, assuming that yy is a function of xx. You'll need the Product Rule and the Chain Rule:
\frac{1}{2}x^{-1/2}y^{1/2}+\frac{1}{2}x^{1/2}y^{-1/2}\cdot dy/dx=1-2 dy/dx12x−12y12+12x12y−12⋅dydx=1−2dydx.
Finally, solve this equation for dy/dxdydx:
dy/dx(\frac{1}{2}x^{1/2}y^{-1/2}+2)=1-\frac{1}{2}x^{-1/2}y^{1/2}dydx(12x12y−12+2)=1−12x−12y12
dy/dx=\frac{1-\frac{1}{2}x^{-1/2}y^{1/2}}{\frac{1}{2}x^{1/2}y^{-1/2}+2}=\frac{1-\frac{\sqrt{y}}{2\sqrt{x}}}{\frac{\sqrt{x}}{2\sqrt{y}}+2}=\frac{2\sqrt{xy}-y}{x+4\sqrt{xy}}dydx=1−12x−12y1212x12y−12+2=1−√y2√x√x2√y+2=2√xy−yx+4√xy
