Answer
$y'=\dfrac{4-x^4}{\sqrt{(x^4+4)^3}}.$
Work Step by Step
Using the quotient rule: $y'=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=x; u'(x)=1$
$v(x)=\sqrt{x^4+4}.$
$v'(x)$ is found using the Chain Rule:
$u=x^4+4$; $\dfrac{du}{dx}=4x^3$
$\dfrac{d}{du}v(u)=\dfrac{1}{2\sqrt{u}}$
$\dfrac{d}{dx}v(x)=\dfrac{d}{du}v(u)\times\dfrac{du}{dx}=(4x^3)(\dfrac{1}{2\sqrt{x^4+4}})=\dfrac{2x^3}{\sqrt{x^4+4}}.$
$y'=\dfrac{(1)(\sqrt{x^4+4})-(x)(\dfrac{2x^3}{\sqrt{x^4+4}})}{(\sqrt{x^4+4})^2}$
$=\dfrac{(\dfrac{1}{\sqrt{x^4+4}})(x^4+4-2x^4)}{x^4+4}$
$=\dfrac{4-x^4}{\sqrt{(x^4+4)^3}}.$
