<s>[INST] Differentiate $f(x)=\frac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}$, $-2
Differentiate $f(x)=\frac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}$, $-2<x<2$
My Attempt
$$
\begin{align}
f'(x)&=\frac{\sqrt{2x+7}\frac{d}{dx}\cos^{-1}\big(\frac{x}{2}\big)-\cos^{-1}\big(\frac{x}{2}\big)\frac{d}{dx}\sqrt{2x+7}}{(\sqrt{2x+7})^2}\\
&=\frac{\sqrt{2x+7}\frac{\frac{-1}{2}}{\sqrt{1-\frac{x^2}{4}}}-\cos^{-1}\big(\frac{x}{2}\big)\frac{2}{2\sqrt{2x+7}}}{(\sqrt{2x+7})^2}\\
&=\frac{\frac{-\sqrt{2x+7}.\sqrt{4}}{2\sqrt{4-x^2}}-\frac{\cos^{-1}\big(\frac{x}{2}\big)}{\sqrt{2x+7}}}{2x+7}\\
&=\frac{\pm2}{2.\sqrt{2x+7}\sqrt{4-x^2}}-\frac{\cos^{-1}\big(\frac{x}{2}\big)}{(2x+7)^{3/2}}\\
&\color{blue}{=\frac{\pm1}{\sqrt{2x+7}\sqrt{4-x^2}}-\frac{\cos^{-1}\big(\frac{x}{2}\big)}{(2x+7)^{3/2}}\\}
\end{align}
$$
Result by Mathematica
D[(ArcCos[x/2]/Sqrt[2 x + 7]),x]

$$
f'(x)={\frac{-1}{\sqrt{2-x}\sqrt{2+x}\sqrt{7+2x}}-\frac{\arccos\big(\frac{x}{2}\big)}{(7+2x)^{3/2}}}\\
\color{blue}{=\frac{-1}{\sqrt{2x+7}\sqrt{4-x^2}}-\frac{\cos^{-1}\big(\frac{x}{2}\big)}{(2x+7)^{3/2}}\\}
$$
Mathematica seems to use only the principal root, ie. $\sqrt{4}=+2$ but in that case can I say the given solution is incomplete ?
Or is there any other domain or range considerations which eliminate the other case ?
 [/INST] $\sqrt{4}=|2|=+2$ in line $3-5$ of your solution
Edit (After OP edited the question):
The derivativeof $\arccos(x)$ is found to be $\frac{-1}{\sqrt{1-x^2}}$, where we must take the positive sign so that the derivative always has negative sign (Otherwise why even bother with the minus sign too?). So, $\frac{d}{dx}\arccos(x/2)= \frac{-1/2}{\sqrt{1-x^2/4}}$.As you can see, the $\sqrt{4}$ causing all the trouble for you originates from here, and as I said we must only take the positive square root, because the derivative requires it.
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