<s>[INST] I need to find the second derivative! I need to find the second derivative of
$$f(x)=\frac{\sqrt{x}}{x+4}.$$
I think the first derivative is 
$$f'(x)=\frac{4-x} {2 \sqrt{x} (x+4)^2}.$$
EDIT: I got the second derivative to be: 
$$f''(x)=-\frac{1}{4 x^{\frac{3}{2}} (x+4) }-\frac{1}{\sqrt{x}(x+4)^2 }+\frac{2 \sqrt{x}}{(x+4)^3}.$$
I'm asking because I need to find the concavity and am very confused.
((just a remark: I have no idea how to comment on other people's posts other than edit my main questions or create more questions, I sincerely apologize for any illicit posts on here or if it goes against guidelines, I am just a college freshman struggling in Calculus, and genuinely appreciate all answers and tips! Thanks again to ANYONE who's helped!!))
Does this mean the only real root is 8.62 because -0.62 would be eliminated due to restrictions of domain? 
 [/INST] Hint.
$$\begin{align}
 \frac{d}{dx} \frac{4 - x}{2\sqrt{x}(x + 4)^2} ~ = ~ & \frac{2\sqrt{x}(x + 4)^2\frac{d}{dx}(4 - x) - (4 - x)\frac{d}{dx}(2\sqrt{x}(x + 4)^2)}{(2\sqrt{x}(x + 4)^2)^2} \\
 ~ = ~ & \frac{2\sqrt{x}(x + 4)^2\frac{d}{dx}(4 - x) - 2(4 - x)((x + 4)^2\frac{d}{dx}\sqrt{x} + \sqrt{x}\frac{d}{dx}(x + 4)^2)}{(2\sqrt{x}(x + 4)^2)^2}
\end{align}$$
</s>