Question

Найти производную функции (рус)
Знайти похідну функції (укр)
За фотографией
СРОЧНО!!!! (даю 100баллов)

Answer 1

$a)\ \ y=\dfrac{3\cdot ctg2x}{sin4x}\\\\\\y'=\dfrac{3\cdot \dfrac{-2}{sin^22x}\cdot sin4x-3\cdot ctg2x\cdot 4cos4x}{sin^24x}=\dfrac{-12\, ctg2x-12\, ctg2x\cdot cos4x}{sin^24x}=\\\\\\=\dfrac{-12\, ctg2x\cdot (1+cos4x)}{sin^24x}=-\dfrac{12\, ctg2x\cdot 2cos^22x}{(2\, sin2x\cdot cos2x)^2}=-\dfrac{6cos2x}{sin^32x}\\\\\\b)\ \ y=arctg\dfrac{2x}{1-x^2}\\\\\\y'=\dfrac{1}{1+\dfrac{4x^2}{(1-x^2)^2}}\cdot \dfrac{2(1-x^2)+2x\cdot 2x}{(1-x^2)^2}=\dfrac{2+2x^2}{1+2x^2+x^4}=\dfrac{2(1+x^2)}{(1+x^2)^2}=\dfrac{2}{1+x^2}$

$c)\ \ y=\dfrac{cos(sin5)\cdot sin^22x}{2\cdot cos4x}\\\\\\y'=\dfrac{cos(sin5)\cdot 2\, sin2x\cdot cos2x\cdot 2\cdot 2\, cos4x-cos(sin5)\cdot sin^22x\cdot (-2\, sin4x\cdot 4)}{4\cdot cos^24x}=\\\\\\=\dfrac{2\, cos(sin5)\cdot (sin8x+4\, sin^22x\cdot sin4x)}{4\cdot cos^24x}$

$d)\ \ y=\dfrac{1}{4}\, x\sqrt[3]{x}+\dfrac{\sqrt[4]{x}}{x}+\sqrt[5]7=\dfrac{1}{4}\cdot x^{\frac{4}{3}}+x^{-\frac{3}{4}}+\sqrt[5]{7}\\\\\\y'=\dfrac{1}{4}\cdot \dfrac{4}{3}\cdot x^{\frac{1}{3}}-\dfrac{3}{4}\cdot x^{-\frac{7}{4}}=\dfrac{1}{3}\cdot \sqrt[3]{x}-\dfrac{3}{4\sqrt[4]{x^7}}$

$e)\ \ y=2\, arcsin\dfrac{2}{3x+1}+\sqrt{9x^2+6x-3}\\\\\\y'=2\cdot \dfrac{1}{\sqrt{1-\dfrac{4}{(3x+1)^2}}}\cdot \dfrac{-2\cdot 3}{(3x+1)^2}+\dfrac{18x+6}{2\sqrt{9x^2+6x-3}}=\\\\\\=-\dfrac{12}{(3x+1)\sqrt{9x^2+6x-3}}+\dfrac{6\, (3x+1)}{\sqrt{9x^2+6x-3}}=\dfrac{-12+6(3x+1)^2}{(3x+1)\sqrt{9x^2+6x-3}}=\\\\\\=\dfrac{-12+6(9x^2+6x+1)}{(3x+1)\sqrt{9x^2+6x-3}}=\dfrac{6\, (9x^2+6x-1)}{(3x+1)\sqrt{9x^2+6x-3}}$