[tex]y=\sqrt{3x^{4}-5x^{3}}, \hspace{2em} 3x^{4}-5x^{3}\ge 0 \Leftrightarrow x^{3}(3x-5)\ge 0 \Leftrightarrow x\in (-\infty;0] \cup \left[\frac{5}{3};+\infty\right)[/tex]
[tex]u(x)=3x^{4}-5x^{3}, v(x)=\sqrt{u(x)} \\ u'(x)=12x^{3}-15x^{2}, v'(x)=\frac{1}{2\sqrt{u(x)}}=\frac{1}{2\sqrt{3x^{4}-5x^{3}}} \Rightarrow[/tex]$$ y'=v'(x)\cdot u'(x)=\frac{12x^{3}-15x^{2}}{2\sqrt{3x^{4}-5x^{3}}} $$
[tex]y'=v'(x)\cdot u'(x)=\frac{12x^{3}-15x^{2}}{2\sqrt{3x^{4}-5x^{3}}}=\frac{3x^{2}(4x-5))}{2|x|\sqrt{3x^{2}-5x}}=\frac{3|x|(4x-5)}{2\sqrt{3x^{2}-5}}[/tex]
[tex]\color{lightseagreen}\text{''Който никога не е правил грешка, никога не е опитвал нещо ново.''} \\
\hspace{21em}\text{(Алберт Айнщайн)}[/tex]