Question

Найдите производные : y=10x^2-4/x^4-3корень 3 степни из x^4-5 y=7/(x+2)^3 y=2x/ctgx

Answer 1

$1)\; \; y=10x^2-\frac{4}{x^4}-3\sqrt[3]{x^4-5}\\\\y'=20x-4\cdot (-4)\cdot x^{-5}-3\cdot \frac{1}{3}\cdot (x^4-5)^{-\frac{2}{3}}\cdot 4x^3=\\\\=20x+\frac{16}{x^5}-\frac{4x^3}{\sqrt[3]{(x^4-5)^2}}\\\\2)\; \; y= \frac{7}{(x+2)^3} \\\\y'=-\frac{7\cdot 3(x+2)^2}{(x+2)^6}=- \frac{21}{(x+2)^4} \\\\3)\; \; y= \frac{2x}{ctgx} \\\\y'= \frac{2\cdot ctgx-2x\cdot \frac{-1}{sin^2x}}{ctg^2x} = \frac{2ctgx\cdot sin^2x+2x}{sin^2x\cdot ctg^2x} = \frac{2cosx\cdot sinx+2x}{cos^2x} = \frac{sin2x+2x}{cos^2x}$