Answer
$y'=2-x(2\tan{x}+x\sec^2{x}).$
Work Step by Step
$y=f(x)+g(x)\rightarrow f(x)=-x^2\tan{x}$ ; $g(x)=2x$
Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=-x^2 ;u’(x)=-2x $
$v(x)=\tan{x} ;v’(x)=\sec^2{x} $
$f'(x)=(-2x)(\tan{x})+(\sec^2{x})(-x^2)$
$=-x(2\tan{x}+x\sec^2{x}).$
$g'(x)=2$
$y'=f'(x)+g'(x)=2-x(2\tan{x}+x\sec^2{x}).$
