Respuesta :

[tex]\bf \sqrt{x+y}=x-2y\implies (x+y)^{\frac{1}{2}}=x-2y \\\\\\ \cfrac{1}{2}(x+y)^{-\frac{1}{2}}\left(1+\frac{dy}{dx} \right)=1-2\frac{dy}{dx} \implies \cfrac{1+\frac{dy}{dx} }{2(x+y)^{\frac{1}{2}}}=1-2\frac{dy}{dx} \\\\\\ 1+\frac{dy}{dx} =2(x+y)^{\frac{1}{2}}-2(x+y)^{\frac{1}{2}}\cdot 2\frac{dy}{dx} \\\\\\ 1+\frac{dy}{dx} =2(x+y)^{\frac{1}{2}}-4(x+y)^{\frac{1}{2}}\frac{dy}{dx} \\\\\\ \frac{dy}{dx}+4(x+y)^{\frac{1}{2}}\frac{dy}{dx} =2(x+y)^{\frac{1}{2}}-1[/tex]

[tex]\bf \cfrac{dy}{dx}\left[ 1+4(x+y)^{\frac{1}{2}} \right]=2(x+y)^{\frac{1}{2}}-1\implies \cfrac{dy}{dx}=\cfrac{2(x+y)^{\frac{1}{2}}-1}{1+4(x+y)^{\frac{1}{2}} } \\\\\\ \cfrac{dy}{dx}=\cfrac{2\sqrt{x+y}-1}{1+4\sqrt{x+y}}[/tex]

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