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? 
 A: I see that your first derivative is correct, and your second derivative is correct as well. To apply the concavity test it will help to simplify your second derivative. Apply your algebra skills to find that
$$
\begin{align*}
f''(x)&=-\frac{1}{4 x^{\frac{3}{2}} (x+4) }-\frac{1}{\sqrt{x}(x+4)^2 }+\frac{2 \sqrt{x}}{(x+4)^3}\\
&=\frac{3 x^2-24 x-16}{4 x^{\frac{3}{2}} (x+4)^3}.
\end{align*}$$
Simplifying to this form will make it much easier to apply the concavity test.
To find concavity of the function $f(x)$ at points on $f(x)$ where the function is twice differentiable using the second derivative, you first find the values where $f''(x)=0$. You can do this by solving $3 x^2-24 x-16=0$. There are two real roots. You then test around and between these roots mindful of the places where $f(x)$, $f'(x)$, and $f''(x)$ are defined. The graph is concave up wherever $f''(x)>0$, and concave down wherever $f''(x)<0$.
The points where $f''(x)=0$ are called inflection points on $f(x)$, this is where the concavity changes. 
Be sensitive to the domains of these functions. For example, on $f(x)$, $x \geq 0$ due to the $\sqrt{x}$ term in the numerator, and so the function is not even defined for $x<0$. The functions $f'(x)$ and $f''(x)$ are not defined for $x \leq 0$ due to factors in their denominators.
A: Hint.
$$\begin{align}
 \frac{d}{dx} \frac{4 - x}{2\sqrt{x}(x + 4)^2} ~ = ~ & \frac{2\sqrt{x}(x + 4)^2\frac{d}{dx}(4 - x) - (4 - x)\frac{d}{dx}(2\sqrt{x}(x + 4)^2)}{(2\sqrt{x}(x + 4)^2)^2} \\
 ~ = ~ & \frac{2\sqrt{x}(x + 4)^2\frac{d}{dx}(4 - x) - 2(4 - x)((x + 4)^2\frac{d}{dx}\sqrt{x} + \sqrt{x}\frac{d}{dx}(x + 4)^2)}{(2\sqrt{x}(x + 4)^2)^2}
\end{align}$$
A: You can also see my way of taking the second derivative of $f'$ as follows:
$$f'=\frac{4-x}{2\sqrt{x}(x+4)^2}=\underbrace{\frac{4-x}2}_{ a(x)}\times \underbrace{x^{-1/2}}_{ b(x)}\times\underbrace{(x+4)^{-2}}_{ c(x)}$$ Now, we know that:


*

*$a'(x)=-1/2,~~b'(x)=\frac{-1}{2}x^{\frac{-1}{2}-1},~~c'(x)=-2(x+4)^{-2-1}$ and,

*$f''=\frac{df'}{dx}=a'(x)b(x)c(x)+a(x)b'(x)c(x)+a(x)b(x)c'(x)$.

A: Try with Wolfram: second derivative of sqrt(x)/(x+4)
It will give you the second derivative of your function.
A: Your first derivative is correct.
To obtain the second derivative, just take the derivative of the first derivative.
One possible way to do this is as follows: (from WA obviously)

