Question

Дифференциал функции

Answer 1

$y={\rm tg}\ x\cdot \sqrt[3]{x};\ dy=y'\cdot dx=(\frac{1}{\cos^2 x}\cdot\sqrt[3]{x}+{\rm tg}\ x\cdot \frac{1}{3\sqrt[3]{x^2}})\ dx$

Answer 2

Ответ:

$y=tgx\cdot \sqrt[3]{x}\\\\\\y'=\dfrac{1}{cos^2x}\cdot \sqrt[3]{x}+tgx\cdot \dfrac{1}{3}\, x^{-\frac{2}{3}}=\dfrac{\sqrt[3]{x}}{cos^2x}+\dfrac{tgx}{3\sqrt[3]{x^2}}\\\\\\dy=\Big(\dfrac{\sqrt[3]{x}}{cos^2x}+\dfrac{tgx}{3\sqrt[3]{x^2}}\Big)\, dx$