Question

найти производную dy/dx
y=arctg2x/4+x

Answer 1

$y=\frac {arctg2x}{4+x} \\ \frac{dy}{dx} = \frac{ \frac{2(4+x)}{1+4x^2} -arctg2x }{(4+x)^2} = \frac{ 2(4+x) - (1+4x^2)arctg2x }{(4+x)^2(1+4x^2)}=\frac {2x+8-arctg2x-4x^2*arctg2x}{(4+x)^2 ( 1+4x^2)}$