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.

  • $\begingroup$ 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

$$\dfrac{d}{du} 2\arccos u = - 2\dfrac{1}{\sqrt{1 - u^2}} ~du$$

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:

$$\dfrac{d}{dx} \left(2 \arccos \left(\sqrt{\dfrac{a-x}{a-b}}\right)\right)=\dfrac{1}{\sqrt{(a-x)(x-b)}}$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.