Question

Помогите найти производную, очень нужно
1)$y=(\frac{1}{5}x^{5}-3x\sqrt[3]{x}-4)^{4}$
2)$y-ln\sqrt[3]{\frac{x^{3}-3 }{x^{3}-2 } }$

Answer 1

Ответ:

$1)\ \ y=\Big(\dfrac{1}{5}\, x^5-3x\sqrt[3]{x}-4\Big)^4\\\\\\y'=4\cdot \Big(\dfrac{1}{5}\, x^5-3x\sqrt[3]{x}-4\Big)^3\cdot \Big(\dfrac{1}{5}\cdot 5x^4-3\cdot \dfrac{4}{3}\cdot x^{\frac{1}{3}}\Big)=\\\\\\=4\cdot \Big(\dfrac{1}{5}\, x^5-3x\sqrt[3]{x}-4\Big)^4\cdot \Big(x^4-4\sqrt[3]{x}\Big)$

$2)\ \ y=ln\sqrt[3]{\dfrac{x^3-3}{x^3-2}}\ \ ,\ \ \ y=ln\sqrt[3]{1-\dfrac{1}{x^3-2}}\\\\\\y'=\sqrt[3]{\dfrac{x^3-2}{x^3-3}}\cdot \dfrac{1}{3}\cdot \Big(\dfrac{x^3-3}{x^3-2}\Big)^{-\frac{2}{3}}\cdot \dfrac{3x^2}{(x^3-2)^2}=\dfrac{1}{3}\cdot \dfrac{x^3-2}{x^3-3}\cdot \dfrac{3x^2}{(x^3-2)^2}=\\\\\\=\dfrac{1}{3}\cdot \dfrac{1}{x^3-3}\cdot \dfrac{3x^2}{x^3-2}=\dfrac{x^2}{(x^3-3)(x^3-2)}$