Помогите, срочно нужно!!

Решите задачу:

$1)\; \; y= \frac{6x}{\sqrt{x^2+1}} \\\\y'= \frac{6\sqrt{x^2+1}-6x\cdot \frac{2x}{2\sqrt{x^2+1}}}{x^2+1} = \frac{6(x^2+1)-6x^2}{\sqrt{(x^2+1)^3}} = \frac{6}{\sqrt{ (x^2+1)^3}}\\\\y'(\sqrt3)= \frac{6}{\sqrt{4^3}} = \frac{6}{4\sqrt4} = \frac{3}{4} \\\\2)\; \; y=ln\sqrt{ \frac{1-x}{1+x} }\\\\y'=\sqrt{ \frac{1+x}{1-x} }\cdot \frac{1}{2\sqrt{\frac{1-x}{1+x}}} \cdot \frac{-(1+x)-(1-x)}{(1+x)^2} = \frac{1+x}{2(1-x)}\cdot \frac{-2}{(1+x)^2} =-\frac{1}{1-x^2}\\\\y'(2)=- \frac{1}{-3} =\frac{1}{3}$

$3)\; \; y=sin^46x\\\\y'=4sin^36x\cdot cos6x\cdot 6\\\\y'( \frac{\pi }{3} )=24sin^32\pi \cdot cos2\pi =0\\\\4)\; \; y=arcctg3x\\\\y'=- \frac{1}{1+9x^2} \cdot 3\\\\y'(0)=-3\\\\5)\; \; y= \frac{e^{e^x}}{2} +x^3\\\\y'= \frac{1}{2}\cdot e^{e^{x}} \cdot e^{x}+3x^2\\\\y'(0)= \frac{1}{2} \cdot e^{e^0}\cdot e^0+0=\frac{e}{2}$