Respuesta :

[tex]\bf y=x\sqrt{x+4}\implies y=x\left( x+4 \right)^{\frac{1}{2}} \\\\\\ \cfrac{dy}{dx}=1\cdot \left( x+4 \right)^{\frac{1}{2}}~~+~~x\cdot \cfrac{1}{2}\left( x+4 \right)^{-\frac{1}{2}}\implies \cfrac{dy}{dx}=\sqrt{x+4}+\cfrac{x}{2\sqrt{x+4}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2(x+4)~~+~~x}{2\sqrt{x+4}}\implies \cfrac{dy}{dx}=\cfrac{2x+8+x}{2\sqrt{x+4}}\implies \cfrac{dy}{dx}=\cfrac{3x+8}{2\sqrt{x+4}}[/tex]

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