Answer
$g'(x)=24x(2+(x^2+1)^4)^2(x^2+1)^3.$
Work Step by Step
Using the Chain Rule:
$u=2+(x^2+1)^4=f(x)+h(x)\rightarrow f(x)=2$; $h(x)=(x^2+1)^4$
$f'(x)=0$.
$h'(x)=(4)(2x)(x^2+1)^{4-1}=8x(x^2+1)^3.$
$\dfrac{du}{dx}=8x(x^2+1)^3+0=8x(x^2+1)^3.$
$g(u)=u^3;\dfrac{d}{du}g(u)=3u^2$.
$\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}$.
$=3(2+(x^2+1)^4)^2(8x(x^2+1)^3)$ $=24x(2+(x^2+1)^4)^2(x^2+1)^3$.
