University Calculus: Early Transcendentals (3rd Edition)

$$y'=-\frac{y}{x}$$
$$\sqrt{xy}=1$$ $$(xy)^{1/2}=1$$ Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$\frac{1}{2}(xy)^{-1/2}\times(xy)'=0$$ $$\frac{1}{2}(xy)^{-1/2}\times(y+xy')=0$$ $$\frac{1}{2}(xy)^{-1/2}\times y+\frac{1}{2}(xy)^{-1/2}\times(xy')=0$$ $$\frac{y}{2\sqrt{xy}}+\frac{x}{2\sqrt{xy}}\times y'=0$$ Now we set all elements with $y'$ to one side and the rest to the other side: $$\frac{x}{2\sqrt{xy}}\times y'=-\frac{y}{2\sqrt{xy}}$$ Finally, we calculate $y'$: $$y'=-\frac{\frac{y}{2\sqrt{xy}}}{\frac{x}{2\sqrt{xy}}}=-\frac{y\times2\sqrt{xy}}{x\times2\sqrt{xy}}$$ $$y'=-\frac{y}{x}$$