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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.

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  • $\begingroup$ I'm sorry, I'm still completely clueless $\endgroup$ – user100893 Nov 10 '13 at 23:54
  • $\begingroup$ In the title you show product and in the problem you show division, which is it? $\endgroup$ – Amzoti Nov 11 '13 at 0:26
  • $\begingroup$ Sorry, my mistake. It's division $\endgroup$ – user100893 Nov 11 '13 at 1:02
  • $\begingroup$ Sorry, it's the squareroot of [(a-x)/(a-b)]. I've been working on it and denoted squareroot of [(a-x)/(a-b)] as t then using the derivative of of arccos as given in your hints and replacing x^2 with t^2. I've also taken the 2 out as a constant. $\endgroup$ – user100893 Nov 11 '13 at 1:35
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$$\dfrac{d}{du} 2\arccos u = - 2\dfrac{1}{\sqrt{1 - u^2}} ~du$$

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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  • $\begingroup$ Needs a TU to beautify the green! +1 $\endgroup$ – Namaste Nov 11 '13 at 14:50

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