If $f'(x)=\sqrt{x+1}$ and $h(x)=f(x^2+x)$, what is $h'(x)$? I was thinking it could be plugging in $x^2+x$ to the $f'(x)$, then using the Chain Rule to solve it, but I'm not sure if it is right. Please help!

Put $u = x^2+x$ then the Chain Rule says that: $h’(x) = f’(u)\cdot u’(x) = \sqrt{u+1}\cdot (2x+1) = (2x+1)\cdot \sqrt{x^2+x+1}$.
