Срочна помогите дам 99 баллов!+ Отмечу лучшим! Найти производную: 1)y=arctg(tg²x) 2)y=ln²...

1)
$\displaystyle y=arctg(tg^2x)\\\\y`= \frac{1}{1+tg^4x}* \frac{2tgx}{cos^2x}$

2)
$\displaystyle y=ln^2(cos^3(4x-1))\\\\y`= \frac{2ln(cos^3(4x-1))}{cos^3(4x-1)}*3cos^2(4x-1)*(-sin(4x-1))*4=\\\\=- \frac{24*ln(cos^3(4x-1))*cos^2(4x-1)*sin(4x-1)}{cos^3(4x-1)}=\\\\=- \frac{24(ln(cos^3(4x-1))*sin(4x-1)}{cos(4x-1)}=\\\\=-24ln(cos^3(4x-1))*tg(4x-1)$

3)
$\displaystyle y=x*e^x(cosx+sinx)\\\\y`=e^x(cosx+sinx)+x*e^x(cosx+sinx)+x*e^x(-sinx+cosx)=\\\\=e^x(x*(cosx+sinx-sinx+cosx)+cosx+sinx)=\\\\=e^x(2x*cosx+cosx+sinx)$