Ответ:

[tex]\dfrac{4\sqrt[6]{x^{5}}}{\cos^{2}x}+\dfrac{10tgx}{3\sqrt[6]{x}}[/tex]

Пошаговое объяснение:

[tex]y=4tgx \cdot \sqrt[6]{x^{5}}=4tgx \cdot x^{\tfrac{5}{6}};[/tex]

[tex]y'=(4tgx \cdot x^{\tfrac{5}{6}})'=(4tgx)' \cdot x^{\tfrac{5}{6}}+4tgx \cdot (x^{\tfrac{5}{6}})'=4 \cdot (tgx)' \cdot \sqrt[6]{x^{5}}+4tgx \cdot \dfrac{5}{6} \cdot x^{\tfrac{5}{6}-1}=[/tex]

[tex]=\dfrac{4\sqrt[6]{x^{5}}}{\cos^{2}x}+\dfrac{20}{6}tgx \cdot x^{-\tfrac{1}{6}}=\dfrac{4\sqrt[6]{x^{5}}}{\cos^{2}x}+\dfrac{10tgx}{3} \cdot \bigg (\dfrac{1}{x} \bigg )^{\tfrac{1}{6}}=\dfrac{4\sqrt[6]{x^{5}}}{\cos^{2}x}+\dfrac{10tgx}{3} \cdot \sqrt[6]{\dfrac{1}{x}}=[/tex]

[tex]=\dfrac{4\sqrt[6]{x^{5}}}{\cos^{2}x}+\dfrac{10tgx}{3\sqrt[6]{x}};[/tex]