Помогите, пожалуйста. узтест, производные

$y= \frac{4x^2-4x^5}{x^2+5x}= \frac{x(4x-4x^4)}{x(x+5)}= \frac{4x-4x^4}{x+5} \\ \\ y'= \frac{(4-16x^3)(x+5)-(4x-4x^4)}{(x+5)^2}= \frac{4x-16x^4+20-80x^3-4x+4x^4}{(x+5)^2}= \\ \\ = \frac{20-12x^4-80x^3}{x^2+10x+25}$

y=-5cosx-3
y ' = 5sinx

y=tg(2-4x)
$y'= - \frac{4}{cos^2(2-4x)}$

$f(x)= \frac{cosx}{1-x} \\ \\ f'(x)= \frac{-(1-x)sinx+cosx}{(1-x)^2}= \frac{(x-1)sinx+cosx}{(1-x)^2} \\ \\ f'(0)= \frac{(0-1)sin0+cos0}{(1-0)^2}= \frac{1}{1^2}=1$