![\bf \cfrac{x}{2\sqrt{xy}}\cdot \cfrac{dy}{dx}=\cfrac{dy}{dx}-\cfrac{y}{2\sqrt{xy}} \implies \cfrac{x}{2\sqrt{xy}}\cdot \cfrac{dy}{dx}-\cfrac{dy}{dx}=-\cfrac{y}{2\sqrt{xy}} \\\\\\ \stackrel{\textit{common factor}}{\cfrac{dy}{dx}\left( \cfrac{x}{2\sqrt{xy}}-1 \right)}=-\cfrac{y}{2\sqrt{xy}} \implies \cfrac{dy}{dx}=-\cfrac{y}{\left( \frac{x}{2\sqrt{xy}}-1 \right)2\sqrt{xy}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \cfrac{dy}{dx}=-\cfrac{y}{x-2\sqrt{xy}}~\hfill](/tpl/images/0016/7116/4a65f.png)
I’m not sure how to do this question, if you could explain your work that’d b
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Mathematics, 25.06.2019 19:10 paisley13
I’m not sure how to do this question, if you could explain your work that’d b
Answers: 3
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