Найти производную функции. Листочек с решением скиньте пжл.

Решите задачу:

$\displaystyle y=\frac{arccosx}{\sqrt{1-x^2}}*(1+x^2)\\\\\\lny=ln\bigg(\frac{arccosx}{\sqrt{1-x^2}}*(1+x^2)\bigg)\\\\\\lny=ln(arccosx)+ln(1+x^2)-ln\sqrt{1-x^2}\\\\\\lny=ln(arccosx)+ln(1+x^2)-\frac{1}2ln(1-x^2)\\\\\\(lny)'=\bigg(ln(arccosx)+ln(1+x^2)-\frac{1}2ln(1-x^2)\bigg)'\\\\\\\frac{1}{y}*y'=\frac{1}{arccosx}*(arccosx)'+\frac{(1+x^2)'}{1+x^2}-\frac{(1-x^2)'}{2(1-x^2)}\\\\\\y'=y\bigg(-\frac{1}{arccosx*\sqrt{1-x^2}}+\frac{2x}{1+x^2}-\frac{-2x}{2(1-x^2)}\bigg)$

$\displaystyle y'=\frac{arccosx}{\sqrt{1-x^2}}*(1+x^2)\bigg(-\frac{1}{arccosx*\sqrt{1-x^2}}+\frac{2x}{1+x^2}+\frac{2x}{2(1-x^2)}\bigg)\\\\\\y'=-\frac{1+x^2}{1-x^2}+\frac{2x*arccosx}{\sqrt{1-x^2}}+\frac{x(1+x^2)arccosx}{\sqrt{1-x^2}(1-x^2)}\\\\\\y'=\boxed{-\frac{1+x^2}{1-x^2}+\frac{2x*arccosx}{\sqrt{1-x^2}}+\frac{x(1+x^2)arccosx}{\sqrt{(1-x^2)^3}}}$