Answer
$f'(x)=-\dfrac{2(x^2+7x-4)}{(x^2+4)^2}.$
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=2x+7; u'(x)=2$
$v(x)=x^2+4; v'(x)=2x$
$f'(x)=\dfrac{2(x^2+4)-2x(2x+7)}{(x^2+4)^2}$
$=-\dfrac{2(x^2+7x-4)}{(x^2+4)^2}.$