Question

Найдите:
$y= \frac{5 x^{2} }{ \sqrt[5]{ x^{2} } } +30 \sqrt[15]{x} + \frac{6}{ \sqrt[3]{x} } ; y'=?$

Answer 1

$y=\frac{5x^2}{\sqrt[5]{x^2}}+30\sqrt[15]{x}+\frac{6}{\sqrt[3]{x}}=5*\sqrt[10]{\frac{x^{20}}{x^4}}+30*x^{\frac{1}{15}}+6*x^{-\frac{1}3}=\\\\=5*x^{\frac{8}5}+30*x^{\frac{1}{15}}+6*x^{-\frac{1}3}\\\\y'=5*\frac{8}5*x^{\frac{8}5-1}+30*\frac{1}{15}*x^{\frac{1}{15}-1}-6*\frac{1}3*x^{-\frac{1}3-1}=\\\\=8x^{\frac{3}5}+2x^{-\frac{14}{15}}-2*x^{-\frac{4}3}=2(4\sqrt[5]{x^3}+\frac{1}{\sqrt[15]{x^{14}}}-\frac{1}{\sqrt[3]{x^4}})$