#### Answer

$g'(0)=0$

#### Work Step by Step

Take the derivative as is on either side of the equation:
$g'(x)+xcos(g(x))g'(x)+sin(g(x))=2x$
Move all terms with g'(x) onto one side of the equal sign and distribute the g'(x) out of each term:
$g'(x)(1+xcos(g(x))=2x-sin(g(x))$
Isolate g'(x) by dividing both sides by the terms:
$g'(x)=\frac{2x-sin(g(x))}{1+xcos(g(x)}$
Plug in 0 to find g'(0):
$g'(0)=\frac{2(0)-sin(g(0))}{1+(0)cos(g(0)}=\frac{-sin(g(0))}{1}=-sin(g(0))=-sin(0)=0$
(We can see that $g(0)=0$ from the original equation.) Thus $g'(0)=0$.