<s>[INST] 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!
 [/INST] If $f'(x) = \sqrt{x+1}$, then $f(x) = \int \sqrt{x+1} dx = \frac{2}{3}(x+1)^{\frac{3}{2}} + c$. You can solve that integral using u-substitution. Then we also know that $h(x) = f(x^2 + x)$. So then using function notation, we know that $h(x) = f(x^2 + x) = \frac{2}{3}(x^2 + x+1)^{\frac{3}{2}} + c$. We can now differentiate this using chain rule as follows:
$h(x) = \frac{2}{3}(x^2 + x+1)^{\frac{3}{2}} + c$
$h'(x) = \frac{2}{3} \cdot \frac{3}{2}(x^2 + x+1)^{\frac{1}{2}} \cdot (2x+1) = (x^2 + x+1)^{\frac{1}{2}} \cdot (2x+1) $
Notice that when we do so, the derivative of $c$ will just be $0$, so it doesn't matter that we don't know the specific value of $c$.
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