Question

найти производную у=1/(х^3-1)^4, у = 3^корень 3х^-2х+1, у=1/^3корень sin^5 2x

Answer 1

$1)\; \; y=\frac{1}{(x^3-1)^4}\; \; ,\; \; \; \Big [\; (\frac{1}{u})'=\frac{-u'}{u^2}\; \Big ]\\\\y'=\frac{-4(x^3-1)^3\cdot 3x^2}{(x^3-1)^8}=-\frac{12x^2}{(x^3-1)^5}\\\\2)\; \; y=3^{\sqrt{3x^2-2x+1}}\\\\y'=3^{\sqrt{3x^2-2x+1}}\cdot ln3\cdot \frac{1}{2\sqrt{3x^2-2x+1}}\cdot (6x-2)\\\\3)\; \; y=\frac{1}{\sqrt[3]{sin^52x}}=(sin2x)^{-5/3}\\\\y'=-\frac{5}{3}\cdot (sin2x)^{-\frac{8}{3}}\cdot cos2x\cdot 2=-\frac{10\, cos2x}{3\sqrt[3]{sin^82x}}$