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Решите задачу:

$1)\quad y= \frac{1-cos2x}{1+2cos2x} \\\\y'= \frac{2sin2x(1+cos2x)-(1-cos2x)\cdot (-2sin2x)}{(1+cos2x)^2} = \frac{4sin2x}{(1+cos2x)^2} \\\\2)\quad y=\sqrt[3]{7x^3-7x^2+1}\\\\y'=\frac{1}{3}(7x^3-7x^3+1)^{-\frac{2}{3}}\cdot (21x^2-14x)\\\\3)\quad y=x^2\cdot cos\frac{1}{x}\\\\y'=2x\cdot cos\frac{1}{x} +x^2\cdot (-sin\frac{1}{x})\cdot (-\frac{1}{x^2})=2x\cdot cos\frac{1}{x}+\frac{1}{x^2}\cdot sin\frac{1}{x}\\\\4)\quad y=sin2x\cdot tgx\\\\y'=2cos2x\cdot tgx+sin2x\cdot \frac{1}{cos^2x}$