Answer
$f'(x)=2(10x(x^2+3)^4+1)((x^2+3)^5+x)$
Work Step by Step
Using the Chain Rule:
$u=g(x)+h(x)=(x^2+3)^5+x\rightarrow g(x)=(x^2+3)^5$; $h(x)=x$
$g'(x)=(5)(2x)(x^2+3)^{5-1}=10x(x^2+3)^4$
$h'(x)=1$
$\dfrac{du}{dx}=g'(x)+h'(x)=10x(x^2+3)^4+1$
$\dfrac{d}{du}f(u)=2u$
$\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}$
$=(10x(x^2+3)^4+1)(2((x^2+3)^5+x))$
$=2(10x(x^2+3)^4+1)((x^2+3)^5+x)$
