Question

Пожалуйста картинкой!!!!!!!

 

найдите производную функции $y=arcsinfrac{2x^{3}}{1+x^{6}}$

Answer 1

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

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