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In a problem, we have to differentiate : $$ y = \arcsin (2 x \sqrt{1 - x^2}),$$ given: $$-{1/(\sqrt2)} < x <{1/(\sqrt2)}$$.

My approach :

Put $x = \cos (p)$, so y reduces to:- $y = 2p = 2 \arccos (x) \implies$ ${dy}/{dx}$ $ = -{2}/{\sqrt{1-x^2}}$

But the solution given is without the minus sign, so I am doing something wrong here but don't know what.

So please tell me what I am doing wrong and how to correct it.

Bernard
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HeWhoMustBeNamed
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1 Answers1

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$$-\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}$$

If $x=\cos P, \dfrac\pi4\le P\le\dfrac{3\pi}4$ as $0\le\arccos x\le\pi$

$\implies\dfrac\pi2\le2P\le\dfrac{3\pi}2$

But as $-\dfrac\pi2\le\arcsin(\sin2P)\le\dfrac\pi2$

$\arcsin(\sin2P)=\pi-2P=\pi-2\arccos x$

lab bhattacharjee
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  • Thanks for answering. Now I understood all but the last line, as I have learned that $\arcsin (\sin x) = x $ ; is that wrong? Why is it $\pi - 2P $ here? – HeWhoMustBeNamed Mar 15 '18 at 09:10
  • @MrReality, See https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values – lab bhattacharjee Mar 15 '18 at 09:12
  • I could find only $\sin ( \arcsin x) = x $ there, related to this in the Basic Properties section. – HeWhoMustBeNamed Mar 15 '18 at 09:16
  • @MrReality, True. But inverse trigonometric functions are multi valued. See http://mathworld.wolfram.com/InverseTrigonometricFunctions.html – lab bhattacharjee Mar 15 '18 at 09:18
  • Yes I know they are. But I don't understand how you got to the last step. ( In my course textbook all that's given is: For suitable values of domain, $\arcsin (\sin x) = x $. ) Can you please point me to somewhere where I can read up on what to do in such cases, when the domains are 'unsuitable' like here? – HeWhoMustBeNamed Mar 15 '18 at 09:26
  • @MrReality, Make sure that inverse trig function values lie in the principal ranges. See also :https://math.stackexchange.com/questions/502189/why-is-arctan-fracxy1-xy-arctan-x-arctan-y – lab bhattacharjee Mar 15 '18 at 09:28

  • Make sure that inverse trig function values lie in the principal ranges − How to do that? Isn't there a resource somewhere on internet that tells this? (Because I don't follow from where you got π in the last step).

     – HeWhoMustBeNamed Mar 15 '18 at 12:16
  • Let us continue this discussion in chat. – lab bhattacharjee Mar 15 '18 at 12:34