<s>[INST] Differentiation of $2\arccos \left(\sqrt{\frac{a-x}{a-b}}\right)$ Okay so the question is:

Show that the function 
  $$2\arccos \left(\sqrt{\dfrac{a-x}{a-b}}\right)$$
  is equal to 
  $$\frac{1}{\sqrt{(a-x)(x-b)}} .$$

I started by changing the arccosine into inverse cosine, then attempted to apply chain rule but I didn't get very far.
Then I tried substituting the derivative for arccosine in and then applying chain rule. Is there another method besides chain rule I should use? Any help is appreciated.
 [/INST] $$\dfrac{d}{du} 2\arccos u = - 2\dfrac{1}{\sqrt{1 - u^2}} ~du$$
See the Proof Wiki for a proof of this.
In this problem, we have, $u = \sqrt{\dfrac{a-x}{a-b}}$, and we need to find $dx$, so we have:
$$ \dfrac{d}{dx} \left(\sqrt{\dfrac{a-x}{a-b}} \right) = -\dfrac{\sqrt{\dfrac{a-x}{a-b}}}{2 (a-x)} = -\dfrac{1}{2 \sqrt{(a - b)(a - x)}}$$
So, lets put these two together.
$\dfrac{d}{du}\left(2 \arccos u \right) =-2 \dfrac{1}{\sqrt{1 - u^2}} ~du = -\dfrac{2}{\sqrt{1 - \left(\sqrt{\dfrac{a-x}{a-b}}\right)^2}} \left(-\dfrac{1}{2 \sqrt{(a - b)(a - x)}} \right)$
We can reduce this to:
$$\dfrac{d}{dx} \left(2 \arccos \left(\sqrt{\dfrac{a-x}{a-b}}\right)\right)=\dfrac{1}{\sqrt{(a-x)(x-b)}}$$
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