Question

Hello!! i’m not sure how to do this question, if you could explain your work that’d b great!!

Answer 1

`\bf √(xy)=y\implies \left( xy \right)^{(1)/(2)}=y\implies \stackrel{\textit{chain rule~\hfill }}{\cfrac{1}{2}(xy)^{-(1)/(2)}\stackrel{\textit{product rule}}{\left(y+x\cfrac{dy}{dx} \right)}}=\cfrac{dy}{dx} \\\\\\ \cfrac{1}{2√(xy)}\left(y+x\cfrac{dy}{dx} \right)=\cfrac{dy}{dx}\implies \cfrac{y}{2√(xy)}+\cfrac{x}{2√(xy)}\cdot \cfrac{dy}{dx}=\cfrac{dy}{dx}`

`\bf \cfrac{x}{2√(xy)}\cdot \cfrac{dy}{dx}=\cfrac{dy}{dx}-\cfrac{y}{2√(xy)} \implies \cfrac{x}{2√(xy)}\cdot \cfrac{dy}{dx}-\cfrac{dy}{dx}=-\cfrac{y}{2√(xy)} \\\\\\ \stackrel{\textit{common factor}}{\cfrac{dy}{dx}\left( \cfrac{x}{2√(xy)}-1 \right)}=-\cfrac{y}{2√(xy)} \implies \cfrac{dy}{dx}=-\cfrac{y}{\left( (x)/(2√(xy))-1 \right)2√(xy)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \cfrac{dy}{dx}=-\cfrac{y}{x-2√(xy)}~\hfill`