## Calculus 10th Edition

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