Answer
$g'(x)=\dfrac{(x+5)(4-2x^2-20x)}{(x^2+2)^3}$
Work Step by Step
Using the quotient rule: $g’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=(x+5)^2; u'(x)=u'(t)\times t'=2(x+5)$
$v(x)=(x^2+2)^2; v'(x)=v'(z)\times z'=4x(x^2+2)$
$g'(x)=\dfrac{(2(x+5))(x^2+2)^2-(4x(x^2+2))(x+5)^2}{(x^2+2)^4}$
$=\dfrac{(x^2+2)(x+5)(2(x^2+2)-4x(x+5))}{(x^2+2)^4}$
$=\dfrac{(x+5)(4-2x^2-20x)}{(x^2+2)^3}$
