Помогите найти производную функцию: 1) y=arcsin ((2x^2)/(1+x^4)) 2) y=корень из 1+lnx 3)...

Решите задачу:

$1)\; \; y=arcsin \frac{2x^2}{1+x^4}\\\\y'= \frac{1}{\sqrt{1-( \frac{2x^2}{1+x^4} })^2} \cdot \frac{4x(1+x^4)-2x^2\cdot 4x^3}{(1+x^4)^2}= \frac{1+x^4}{\sqrt{1+2x^4+x^8-4x^4}}\cdot \frac{4x+4x^5-8x^5}{(1+x^4)^2}$

$= \frac{1+x^4}{\sqrt{x^8-2x^4+1}} \cdot \frac{4x-4x^5}{(1+x^4)^2} = \frac{1}{\sqrt{(1-x^4)^2}} \cdot \frac{4x(1-x^4)}{1+x^4} = \frac{4x(1-x^4)}{(1-x^4)(1+x^4)} =\\\\=\frac{4x}{1+x^4}$

$2)\; \; y=\sqrt{1+lnx}\\\\y'= \frac{1}{2\sqrt{1+lnx}} \cdot \frac{1}{x} \\\\3)\; \; y= \frac{1+\sqrt{x}}{1-\sqrt{x}} \\\\y'= \frac{\frac{1}{2\sqrt{x}}\cdot (1-\sqrt{x})-(1+\sqrt{x})\cdot (-\frac{1}{2\sqrt{x}})}{(1-\sqrt{x})^2} = \frac{1-\sqrt{x}+1+\sqrt{x}}{2\sqrt{x}(1-\sqrt{x})^2} = \frac{1}{\sqrt{x}(1-\sqrt{x})^2}$