Question Number 184720 by mustafazaheen last updated on 10/Jan/23
$$ \\ $$$${y}=\left(\sqrt{\mathrm{3}{xy}^{\mathrm{2}} }\right)\left(\sqrt[{\mathrm{5}}]{{x}^{\mathrm{5}} {y}^{\mathrm{2}} }\right) \\ $$$${y}'=?\:\:\:\:\:\:{y}^{''} =? \\ $$
Answered by Frix last updated on 10/Jan/23
$${y}=\sqrt{\mathrm{3}{xy}^{\mathrm{2}} }×\sqrt[{\mathrm{5}}]{{x}^{\mathrm{5}} {y}^{\mathrm{2}} }=\sqrt{\mathrm{3}{x}}\mid{y}\mid×{x}\sqrt[{\mathrm{5}}]{{y}^{\mathrm{2}} } \\ $$$$\mathrm{If}\:{x},\:{y}\in\mathbb{R}\:\Rightarrow\:{x}\geqslant\mathrm{0}\wedge{y}\geqslant\mathrm{0}\:\Rightarrow \\ $$$${y}=\sqrt{\mathrm{3}}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} {y}^{\frac{\mathrm{7}}{\mathrm{5}}} \\ $$$$\Rightarrow \\ $$$${y}=\frac{\mathrm{1}}{\:\mathrm{3}{x}^{\mathrm{3}} \sqrt[{\mathrm{4}}]{\mathrm{3}{x}^{\mathrm{3}} }} \\ $$$$\Rightarrow \\ $$$${y}'=−\frac{\mathrm{5}}{\:\mathrm{4}{x}^{\mathrm{4}} \sqrt[{\mathrm{4}}]{\mathrm{3}{x}^{\mathrm{3}} }} \\ $$$${y}''=\frac{\mathrm{95}}{\:\mathrm{16}{x}^{\mathrm{5}} \sqrt[{\mathrm{4}}]{\mathrm{3}{x}^{\mathrm{3}} }} \\ $$