Помогите пожалуйста)

Решите задачу:

$\boxed{(uv)'=u'v+uv'}$
$1) y= \sqrt{x} \cdot 7x^3$
$\displaystyle y'= \frac{1}{2 \sqrt x} \cdot 7x^3+ \sqrt{x} \cdot 21x^2= \frac{7x^3}{2 \sqrt{x}} +21x^2 \sqrt{x}= \frac{7x^3+42x^3}{2 \sqrt{x}}=$
$\displaystyle = \frac{49x^3}{2 \sqrt{x}}= \frac{49x^{7/2}}{2x}= \frac{x(49x^{5/2})}{2x}= \frac{49x^{5/2}}{2}$

$2) y=cos3x$
$y'=-3sin3x$

$\boxed{( \frac{u}{v})'= \frac{u'v-uv'}{v^2} }$
$\displaystyle 3) y= \frac{x^6+3x^4}{1-x^2}$
$\displaystyle y'= \frac{(6x^5+12x^3)(1-x^2)-(x^6+3x^4)(-2x)}{(1-x^2)^2}=$
$\displaystyle = \frac{6x^5+12x^3-6x^7-12x^5-(-2x^7-6x^5)}{(1-x^2)^2}=$
$\displaystyle = \frac{6x^5+12x^3-6x^7-12x^5+2x^7+6x^5}{(1-x^2)^2} = \frac{12x^3-4x^7}{(1-x^2)^2}$

$\boxed{f(g(x))'=f(x)' \cdot g(x)'}$
$4) y=arccos \sqrt{x-1}$
$\displaystyle y'=- \frac{1}{\sqrt{1-x^2}}\cdot \sqrt{x-1}+arccos \frac{1}{2 \sqrt{x-1}}= -\frac{\sqrt{x-1}}{\sqrt{1-x^2}}+\frac{arccos}{2\sqrt{x-1}}$