Пожалуйста картинкой!!!!!!! найдите производную функции

Решите задачу:

$y=\arcsin\frac{2x^{3}}{1+x^{6}}\\(\arcsin Z)'=\frac1{\sqrt{1-Z^2}}\cdot Z'\\\left(\arcsin\frac{2x^{3}}{1+x^{6}}\right)'=\frac1{\sqrt{1-\frac{4x^6}{(1+x^6)^2}}}\cdot \left( \frac{2x^{3}}{1+x^{6}}\right)'\\\frac1{\sqrt{1-\frac{4x^6}{(1+x^6)^2}}}=\frac1{\sqrt{\frac{(1+x^6)^2-4x^6}{(1+x^6)^2}}}=\sqrt{\frac{(1+x^6)^2}{(1+x^6)^2-4x^6}}=\sqrt{\frac{(1+x^6)^2}{1+2x^6+x^{12}-4x^6}}=$

$=\sqrt{\frac{(1+x^6)^2}{x^{12}-2x^6+1}}=\sqrt{\frac{(1+x^6)^2}{(1-x^6)^2}}=\sqrt{\left(\frac{1+x^6}{1-x^6}\right)^2}=\pm\frac{1+x^6}{1-x^6}\right)\\\left( \frac{2x^{3}}{1+x^{6}}\right)'=\frac{6x^2(1+x^6)-2x^3\cdot6x^5}{(1+x^6)^2}=\frac{6x^2+6x^8-12x^8}{(1+x^6)^2}=\frac{6x^2-6x^8}{(1+x^6)^2}=\frac{6x^2(1-x^6)}{(1+x^6)^2}\\\pm\frac{1+x^6}{1-x^6}\right)\cdot\frac{6x^2(1-x^6)}{(1+x^6)^2}=\pm\frac{6x^2}{1+x^6}$