Question

Первые 4 задание,помогите! Даю 67 баллов

Answer 1

$1)\; \; y=(2^{cos3x}+sin3x)^3\\\\y'=3\cdot (2^{cos3x}+sin3x)^2\cdot (2^{cos3x}\cdot ln2\cdot (-sin3x)\cdot 3+3\, cos3x)\\\\\\2)\; \; y=ln\sqrt[5]{\frac{1-x}{1+x}}\\\\y'=\sqrt[5]{\frac{1+x}{1-x}}\cdot \frac{1}{5}\cdot (\frac{1-x}{1+x})^{-\frac{4}{5}}\cdot \frac{1-x-(1+x)}{(1+x)^2}=\sqrt[5]{\frac{1+x}{1-x}}\cdot \frac{1}{5}\cdot (\frac{1+x}{1-x})^{\frac{4}{5}}\cdot \frac{-2x}{(1+x)^2}=\\\\=-\frac{1}{5}\cdot \frac{1+x}{1-x}\cdot \frac{2x}{(1+x)^2}=-\frac{1}{5}\cdot \frac{2x}{1-x^2}$

$3)\; \; y=sin^32x\cdot cos8x^5\\\\y'=3\cdot sin^22x\cdot cos2x\cdot 2\cdot cos8x^5+sin^32x\cdot (-sin8x^5)\cdot 40x^4=\\\\=3\cdot sin2x\cdot sin4x\cdot cos8x^5-40x^4\cdot sin^32x\cdot sin8x^5$